Answer:
2000 m
Explanation:
since the boat is 1000 m from the river bank, the beam must be focused at this point. This indicates that the focal length is 1000 m
for a spherical mirror, the focal length is given by
f = R/2
where R is the radius of curvature
1000 = R/2
R = 2000 m
R = 2000 m
this means that the radius of curvature must be 2000 m
A parallel-plate capacitor has square plates that are 7.20 cm on each side and 3.40 mm apart. The space between the plates is completely filled with two square slabs of dielectric, each 7.20 cm on a side and 1.70 mm thick. One slab is Pyrex glass and the other slab is polystyrene. If the potential difference between the plates is 96.0 V, find how much electrical energy (in nJ) can be stored in this capacitor.
Answer:
U = 218 nJ
Explanation:
We are given;
Spacing between the plates; d = 3.4 mm = 3.4 × 10^(-3) m
Voltage across the capacitor; V = 96 V
Dimension of the square plates is 7.2cm x 7.2cm.
So, Area = 7.2 × 7.2 = 51.84 cm² = 51.84 × 10^(-4) m²
Permittivity of free space; ε_o = 8.85 × 10^(-12) C²/N.m²
From relative permeability table;
Dielectric constant of Pyrex; k1 = 5.6
Dielectric constant of polystyrene; k2 = 2.56
Now, formula for capacitance of a capacitor with Dielectric is;
C = kC_o
Where, C_o = ε_o(A/d)
Since there are 2 capacitors, d will now be d/2 = (3.4 × 10^(-3))/2 m = 1.7 × 10^(-3)
Since we have 2 capacitor, thus ;
C1 = k1*ε_o*(A/d)
C1 = (5.6 × 8.85 × 10^(-12) × (51.84 × 10^(-4))/(1.7 × 10^(-3))
C1 = 1.51 × 10^(-10) F
Similarly;
C2 = (2.56 × 8.85 × 10^(-12) × (51.84 × 10^(-4))/(1.7 × 10^(-3))
C2 = 0.691 × 10^(-10) F
For capacitors in series, formula for total capacitance(Cs) is;
1/Cs = (1/C1) + (1/C2)
Simplifying this, we have;
Cs = (C1*C2)/(C1 + C2)
Plugging in the relevant values ;
Cs = (1.51 × 10^(-10)*0.691 × 10^(-10))/((1.51 × 10^(-10)) + (0.691 × 10^(-10)))
Cs = 0.474 × 10^(-10) F
The formula for energy stored in a capacitor with 2 Dielectrics is given as;
U = ½Cs*V²
So,
U = ½ × 0.474 × 10^(-10) × 96²
U = 2.18 × 10^(-7) J = 218 × 10^(-9) = 218 nJ
Which of these boxes will not accelerate!
30 Newtons
40 Newtons
50 kg
15 Newton
B.
10 kg
30 Newtons
C.
30 Newtons
80 kg
20 Newtons
20 Newtons
20 Newtons
D.
75 kg
Answer:
(possibly) Box D
Explanation:
The one that has balanced forces will not accelerate. The forces are unbalanced in figures A, B, C. We cannot tell about figure D, because the downward force is not shown. If that force is 20 N, box D will not accelerate.
A certain type of laser emits light that has a frequency of 4.6 x 1014 Hz. The light, however, occurs as a series of short pulses, each lasting for a time of 3.1 x 10-11s. The light enters a pool of water. The frequency of the light remains the same, but the speed of light slows down to 2.3 x 108 m/s. In the water, how many wavelengths are in one pulse
Answer:
14,260
Explanation:
Relevant data provided for computing the wavelengths are in one pulse is here below:-
The number of wavelengths in Ls = [tex]4.6\times 10_1_4[/tex]
Therefore the Number of in time = Δt = [tex]3.1\times 10_-_1_1[/tex]
The number of wavelengths are in one pulse is shown below:-
[tex]Number\ of\ wavelengths = \triangle t\times f[/tex]
[tex]= 3.1\times 10_-_1_1\times 4.6\times 10_1_4[/tex]
= 14,260
Therefore for computing the number of wavelengths are in one pulse we simply applied the above formula.
Parallel light rays with a wavelength of 610nm fall on a single slit. On a screen 3.10m away, the distance between the first dark fringes on either side of the central maximum is 4.00mm.
What is the width of the slit?
Answer:
The width of the slit will be ".946 mm".
Explanation:
The given values are:
Wavelength = 610 × 10⁻⁹
Length, L = 3 m
As we know,
⇒ [tex]\frac{y}{L} = \frac{m(wavelength)}{a}[/tex]
On putting the estimated values, we get
⇒ [tex]\frac{2\times 10^{-3}}{3.1} = \frac{(1)(610 X 10^{-9})}{a}[/tex]
On applying cross-multiplication, we get
⇒ [tex]a=9.46\times 10^{-4}[/tex]
⇒ [tex]a = .946 mm[/tex]
A large box containing your new computer sits on the bed of your pickup truck. You are stopped at a red light. When the light turns green, you stomp on the gas and the truck accelerates. To your horror, the box starts to slide toward the back of the truck. Draw clearly labeled free-body diagrams for the truck and for the box. Indicate pairs of forces, if any, that are third-law action–reaction pairs. (The horizontal truck bed is not frictionless.)
Answer:
The description of that same situation has been listed throughout the explanation segment below.
Explanation:
When another huge box or container containing your new machine or device sits on someone's pick-up truck's bed, the third low portion of the operation response force. This same friction force of the box mostly on the truck bed as well as the friction force including its truck bed on either the box from either the immune response pair.So that the above seems to be the right answer.
If a metal rod is moved through magnetic field, the charged particles will feel a force, and if there is a complete circuit, a current will flow. We talk about the induced emf of the rod. The rod essentially acts like a battery, and the induced emf is the voltage of the battery. A magnetic field with a strength of 0.732 T is pointing into the page and a metal rod L=0.362 m in length is moved to the right at a speed v of 15.1m/s.
Required:
a. What is the induced emf in the rod?
b. Suppose the rod is sliding on conducting rails, and a complete circuit is formed. If the load resistance is 5.74Ω , what is the magnitude and direction (clockwise or counterclockwise) of the current flowing in the circuit?
Answer:
a. 4 V
b. 0.697 A
Explanation:
Magnetic field strength B = 0.732 T
length of rod l = 0.362 m
velocity of rod v = 15.1 m/s
a. EMF can be calculated as
E = Blv = 0.732 x 0.362 x 15.1 = 4 V
b. If the rod is connected to a conducting rail, with resistance R = 5.74Ω
current I = V/R = 4/5.74 = 0.697 A
the current flows in a clockwise direction
Representar con una escala de 1cm = 10N dos fuerzas que tengan igual dirección, distinto sentido y sus intensidades son de 40n y 60n, respectivamente.
Alguien que me lo hagaaaaaaa
Answer:
To solve this problem we just need to graph two forces with same direction, pointing to different sides with intensities of 40 N and 60 N.
The image attached shows these forces.
Notice that the vectors are parallel, that's because they have the same direction, but they point to different sides, and their magnitudes have a difference of 20 N.
A plastic rod of length d = 1.5 m lies along the x-axis with its midpoint at the origin. The rod carries a uniform linear charge density λ = 2.5 nC/m. The point P is located on the positive y-axis at a distance y0 = 15 cm from the origin. The z-axis points out of the screen. Integrate your correct choice in part (b) over the length of the rod and choose the correct expression for the y-component of the electric field at point P.
Answer:
Explanation:
Let the plastic rod extends from - L to + L .
consider a small length of dx on the rod on the positive x axis at distance x . charge on it = λ dx where λ is linear charge density .
It will create a field at point P on y -axis . Distance of point P
= √ x² + .15²
electric field at P due to small charged length
dE = k λ dx x / (x² + .15² )
Its component along Y - axis
= dE cosθ where θ is angle between direction of field dE and y axis
= dE x .15 / √ x² + .15²
= k λ dx .15 / (x² + .15² )³/²
If we consider the same strip along the x axis at the same position on negative x axis , same result will be found . It is to be noted that the component of field in perpendicular to y axis will cancel out each other . Now for electric field due to whole rod at point p , we shall have to integrate the above expression from - L to + L
E = ∫ k λ .15 / (x² + .15² )³/² dx
= k λ x L / .15 √( L² / 4 + .15² )
b) The length of the rod:
[tex]E = \int\limits dx . k \lambda .15 / (x^2 + .15^2 )^{1/2} dx\\\\E= \frac{k \lambda * L}{0.15} \sqrt{( L^2 / 4 + .15^2 )[/tex]
Given:
d = 1.5 mλ = 2.5 nC/m
Let the plastic rod extends from - L to + L .Consider a small length of dx on the rod on the positive x axis at distance x . charge on it = λ dx where λ is linear charge density .It will create a field at point P on y -axis.
Distance of point P =[tex]\sqrt{x^2 + 0.15^2}[/tex]
How to calculate Electric Field?E.F at P due to small charged length[tex]dE = \frac{ k \lambda x.dx}{(x^2 + .15^2 )}[/tex]
Its component along Y - axis = dE cosθ where θ is angle between direction of field dE and y axis
[tex]= \frac{dE x .15 }{\sqrt{x^2 + .15^2} }\\\\= \frac{k \lambda dx .15}{(x^2 + .15^2 )^{1/2}}[/tex]
If we consider the same strip along the x axis at the same position on negative x axis , same result will be found . We can say that the component of field in perpendicular to y axis will cancel out each other.
Now for electric field due to whole rod at point p , we shall have to integrate the above expression from - L to + L
[tex]E = \int\limits dx . k \lambda .15 / (x^2 + .15^2 )^{1/2} dx\\\\E= \frac{k \lambda * L}{0.15} \sqrt{( L^2 / 4 + .15^2 )}[/tex]
Find more information about Electric field here:
brainly.com/question/14372859
HELP, END OF SCHOOL YEAR, 30 POINTS Unit 9 lesson 15 astronomy unit test answers I ONLY HAVE ONE MORE DAY
1 As evidence supporting the Big Bang theory, what does the redshift of light from galaxies indicate?
The universe is mainly hydrogen.
The universe is 13.8 billion years old.
The universe is cooling off.
The universe is expanding.
2 Which evidence supports the idea that Cosmic Microwave Background radiation is a remnant of the Big Bang?(1 point)
Its temperature is uniform.
Its mass fluctuates greatly.
Its temperature fluctuates greatly.
Its mass is uniform.
3 Which of these items provide evidence supporting the Big Bang theory? Select the two correct items.(1 point)
rate of star formation
composition of matter in the universe
sizes and shapes of distant galaxies
cosmic background radiation
4 How does the change in the temperature of the universe provide evidence for universe expansion that supports the Big Bang Theory?(1 point)
The universe is warming which, according to the Big Bang Theory, is expected to happen as the cosmos disperses.
The universe is warming which, according to the Big Bang Theory, is expected to happen as the cosmos accumulates.
The universe is cooling which, according to the Big Bang Theory, is expected to happen as the cosmos accumulates.
The universe is cooling which, according to the Big Bang Theory, is expected to happen as the cosmos disperses.
5 How does weak background radiation coming from every direction in the sky support the Big Bang Theory?(1 point)
It provides evidence of the universe's increasing mass.
It provides evidence of universe expansion.
It provides evidence of universe contraction.
It provides evidence of the universe's decreasing mass.
6 Which statements describe ways that nuclear fission is different than nuclear fusion? Select the two correct answers.(1 point)
Nuclear fission is used to produce electricity at nuclear power plants.
Nuclear fission involves one large atom splitting into two smaller atoms.
Nuclear fission takes place in the nucleus of an atom.
Nuclear fission releases a huge amount of energy.
7 Blueshift is observed when(1 point)
a distant luminous object travels rapidly away from an observer.
a distant luminous object travels rapidly towards an observer.
a luminous object travels alongside an observer.
a luminous object is stationary compared to an observer.
8 Which statements about nuclear fusion are false? Select the two correct answers.(1 point)
The fuel for nuclear fusion is often uranium.
Nuclear fusion is used to generate electricity at nuclear power plants.
Nuclear fusion releases large amounts of energy.
Nuclear fusion takes place in the cores of stars.
9 Which of the following statements provide evidence to support the big bang theory? Select the two correct answers.
The ratios of hydrogen and helium in the universe match those of the early universe.
The universe began as a very high density singularity.
Dark matter makes up the majority of matter in the galaxy.
Small spiral galaxies become larger elliptical galaxies when they collide.
10 Which represents a correct match between ideas related to the formation of the universe? Select the two correct answers.(1 point)
accelerating expansion — dark energy
structures forming in the early universe — dark matter
greatest percent of mass of universe — dark matter
glowing nebulae — dark energy
11 How is dark energy related to the theory of the Big Bang?(1 point)
It causes the expansion of the universe to accelerate.
It causes the universe to expand.
It seeded the formation of galaxies and star clusters.
It causes the spinning of galaxies.
Answer:
1. The Universe is Expanding
2. It’s temperature is it’s uniform
3. Cosmic background radiation
4. I will give a hint for this one, since I don’t know, the hint is the universe is cooling.
5. It provides evidence of universe expansion.
6. Sorry I don’t know the rest
Explanation:
The universe is the collection of every item in space and time as well as the contents of those items
The correct options are as follows;
1. The universe is expanding
2. Its temperature is uniform
3. Composition of matter in the universe, cosmic background radiation
4. The universe is cooling which, according to the Big Bang Theory, is expected to happen as the cosmos disperses
5. It provides evidence of universe expansion
6. Nuclear fission involves one large atom splitting into two smaller atoms
Nuclear fission takes place in the nucleus of an atom
7. A distant object travels rapidly towards an observer
8. The fuel in nuclear fusion is often uranium
Nuclear fusion is used to generate electricity at nuclear power plants
9. The universe began as a very high density singularity
Dark matter makes up majority of the universe
10. Acceleration expansion — Dark energy
Structure forming in the early universe — Dark matter
11. It causes the expansion of the universe to accelerate
The reasons for selecting the above options are as follows;
1. The universe is expanding
The redshift of light from galaxies indicates that that are moving away
2. Its temperature is uniform
The uniform temperature of the microwave background suggest a common source
3. Composition of matter in the universe, cosmic background radiation
The matter present in the universe are characteristically similar in their origins
The cosmic background provides evidence of the existence of a singularity
4. The universe is cooling which, according to the Big Bang Theory, is expected to happen as the cosmos disperses
Based on the Big Bang Theory, the temperature of the universe is reducing as the universe expands, compared to the initial temperature
5. It provides evidence of universe expansion
The background radiation coming from a single source as the rest of the universe is expected to spread throughout the universe
6. Nuclear fission involves one large atom splitting into two smaller atoms
Nuclear fission takes place in the nucleus of an atom
Nuclear fusion involves the joining of small atoms to form a larger atom
7. A distant object travels rapidly towards an observer
The redshift is the opposite, indicating that the object is moving further away
8. The fuel in nuclear fusion is often uranium
Nuclear fusion is used to generate electricity at nuclear power plants
Nuclear fusion usually consists of joining small atoms together. It has not been used for commercial energy production
9. The universe began as a very high density singularity
According to the Big Bang Theory, the universe started from the dense, high temperature singularity
Dark matter makes up majority of the universe
10. Acceleration expansion — Dark energy
Dark energy causes expansion
Structure forming in the early universe — Dark matter
Dark matter is instrumental to the formation of structures in the universe
11. It causes the expansion of the universe to accelerate
Dark energy is seen as the cause of the accelerating expansion of the universe
Learn more about the universe here:
https://brainly.com/question/17525451
A pendulum on a planet, where gravitational acceleration is unknown, oscillates with a time period 5 sec. If the mass is increased six times, what is the time period of the pendulum?
Explanation:
We have, a pendulum on a planet, oscillates with a time period 5 sec. The formula used to find the time period is given by :
[tex]T=2\pi \sqrt{\dfrac{l}{g}}[/tex]
l is length of the pendulum
g is acceleration due to gravity on which it is placed
It is clear that, the time period of pendulum is independent of the mass. Hence, if the mass is increased six times, its time period remains the same.
The mass of a particular eagle is twice that of a hunted pigeon. Suppose the pigeon is flying north at ,2=17.1 m/s when the eagle swoops down, grabs the pigeon, and flies off. At the instant right before the attack, the eagle is flying toward the pigeon at an angle =52.7 ° below the horizontal and a speed of ,1=41.5 m/s.
Answer:
31.4 m/s
44.4°
Explanation:
Momentum is conserved in the horizontal direction:
pₓᵢ = pₓ
m vᵢ₂ + 2m vᵢ₁ cos θ = (m + 2m) vₓ
vᵢ₂ + 2 vᵢ₁ cos θ = 3 vₓ
17.1 m/s + 2 (41.5 m/s) (cos -52.7°) = 3 vₓ
vₓ = 22.5 m/s
Momentum is conserved in the vertical direction:
pᵧᵢ = pᵧ
2m vᵢ₁ sin θ = (m + 2m) vᵧ
2 vᵢ₁ sin θ = 3 vᵧ
2 (41.5 m/s) (sin -52.7°) = 3 vᵧ
vᵧ = -22.0 m/s
The speed is:
v = √(vₓ² + vᵧ²)
v = √((22.5 m/s)² + (-22.0 m/s)²)
v = 31.4 m/s
The direction is:
θ = atan(vᵧ / vₓ)
θ = atan(-22.0 m/s / 22.5 m/s)
θ = -44.4°
The speed of the eagle at that instant is 31.4 m/s while it moves off in the direction of 44.4°.
Since momentum is conserved horizontally;
17.1 m/s + 2 (41.5 m/s) (cos -52.7°) = 3 vx
vx = 17.1 m/s + 2 (41.5 m/s) (cos -52.7°)/3
vx = 22.5 m/s
Also, momentum is conserved vertically hence;
2 (41.5 m/s) (sin -52.7°) = 3 vy
vy = 2 (41.5 m/s) (sin -52.7°) /3
vy = -22.0 m/s
The effective speed therefore, is;
v = √((22.5 m/s)² + (-22.0 m/s)²)
v = 31.4 m/s
The direction of this effective speed is;
θ = tan-1(22.0 m/s / 22.5 m/s)
θ = 44.4°
Learn more: https://brainly.com/question/13322477
Our Sun shines bright with a luminosity of 3.828 x 1025 Watt. Her energies
responsible for many processes and the habitable temperatures on the earth that
make our life possible.
a) Calculate the amount of energy arriving on the Earth in a single day
b) To how many litres of heating oil (energy density 37.3 x 10^6 J/litre is the equivalent?
C) The Earth reflects 30% of this energy : Determine the temperature on Earth's sufact
d) what other factors should be considered to get an even more precisa temperature postiache
Note: The Earth's radius is 6370km; the Sun's sadius is 696 ×10^3km, I AU is 1.495 × 10^8km)
Answer:
a) E = 1.58 10²¹ J , b) Oil = 4,236 107 liter , e) T = 54.3 C
Explanation:
a) To calculate the energy that reaches Earth, let us combine that the power emitted by the Sun is distributed uniformly on a spherical surface
I = P / A
A = 4π r²
in this case the radius of the sphere is the distance from the Sun to Earth r = 1.5 10¹¹ m
I = P / A
I = P / 4π r²
let's calculate
I = 3,828 10²⁵/4 pi (1.5 10¹¹)²
I = 1.3539 10²W / m² = 135.4 W / m2
the energy that reaches the disk of the Earth is
E = I A
the area of a disc
A = π r²
E = I π r²
where r is the radius of the Earth 6.37 10⁶ m
E = 135.4 π(6.37 10⁶)
E = 1,726 10¹⁶ W
This is the energy per unit of time that reaches Earth
t = 1 dai (24h / 1day) (3600s / 1h) = 86400 s
E = 1,826 10¹⁶ 86400
E = 1.58 10²¹ J
b) for this part we can use a direct proportions rule
Oil = 1.58 10²¹ (1 / 37.3 10⁶)
Oil = 4,236 10⁷ liter
c) to silence the surface temperature of the Earth we use the Stefan-Bolztman Law
P = σ A e T⁴
T = [tex]\sqrt[4]{P/Ae}[/tex]
nos indicate the refect, therefore the amount of absorbencies
P_absorbed = 0.7 P
let's calculate
T = REA (0.7 1.58 1021 / [pi (6.37 106) 2 1)
T = RER (8,676 106)
T = 54.3 C
b) Among the other factors that must be taken into account is the greenhouse effect, due to the absorption of gases from the atmosphere
Friction is a force that acts in an ___________ direction of movement.
a) similar
b) opposite
c) parallel
d) west
Answer:
the answer is opposite.
plz mark brainliest
Explanation:
help yall 13 points!!
Answer:
Explanation:
12.)
A. Opposite poles attract
B. Same poles repel
13.)
IDK
Positive charge Q is placed on a conducting spherical shell with inner radius R1 and outer radius R2. A particle with charge q is placed at the center of the cavity. The net charge on the inner surface of the conducting shell is
Answer: in this question, the only charge in the cavity is Q. Inside the conducting spherical shell, the electric field is zero.
While outside the shell, the electric field is given by: k(q + Q)/r²
Where;
K= is a constant which is given as, 8.99 x 10^9 N m² / C².
Q= source charge which creates the electric field
q= is the test charge which is used to measure the strength of the electric field at a given location.
r= is the radius
Explanation: Inside the conducting spherical shell, the electric field is zero since the Electric field vanishes everywhere inside the volume of a good conductor.
A 110-kg football player running at 8.00 m/s catches a 0.410-kg football that is traveling at 25.0 m/s. Assuming the football player catches the ball with his feet off the ground with both of them moving horizontally, calculate: the final velocity if the ball and player are going in the same directio
Answer:[tex]8.062\ m/s[/tex]
Explanation:
Given
masss of football player [tex]M=110\ kg[/tex]
Velocity of football player [tex]u_1=8\ m/s[/tex]
mass of football [tex]m=0.41\ kg[/tex]
velocity of football [tex]u_2=25\ m/s[/tex]
Final velocity will be given by applying conservation of linear momentum
After catching the ball Player and ball moves with same velocity
[tex]\Rightarrow Mu_1+mu_2=(M+m)v[/tex]
[tex]\Rightarrow 110\times 8+0.41\times 25=(110+0.41)v[/tex]
[tex]\Rightarrow 880+10.25=110.41\times v[/tex]
[tex]\Rightarrow v=\frac{890.25}{110.41}=8.063\ m/s[/tex]
So, final velocity will be [tex]8.062\ m/s[/tex]
A subatomic particle X spontaneously decays into two particles, A and B, each of rest energy 1.40 × 10^2 MeV. The particles fly off in opposite directions, each with speed 0.827c relative to an inertial reference frame S. What is the total energy of particle A?
Answer:
E = 389 MeV
Explanation:
The total energy of particle A, will be equal to the sum of rest mass energy and relative energy of particle A. Therefore,
Total Energy of A = E = Rest Mass Energy + Relative Energy
Using Einstein's Equation: E = mc²
E = m₀c² + mc²
From Einstein's Special Theory of Relativity, we know that:
m = m₀/[√(1-v²/c²)]
Therefore,
E = m₀c² + m₀c²/[√(1-v²/c²)]
E = m₀c²[1 + 1/√(1-v²/c²)]
where,
m₀c² = rest mass energy = 140 MeV
v = relative speed = 0.827 c
Therefore,
E = (140 MeV)[1 + 1/√(1 - (0.827c)²/c²)]
E = (140 MeV)(2.78)
E = 389 MeV
Your new toaster has two separate toasting units, each of which consumes 600 watts of power when it is in use. When you operate one unit, a current of 5 amperes flowsthrough the wiring in your home and the wires waste about 1 watt of power handling that current. If you operate both toasting units at once, your toaster consumes 1200 watts and the current flowing through the wiring in your home doubles to 10 amperes. How much power will the wires in your home waste now
Answer:
1.92 Watt lost
Explanation:
Power rating of each toaster = 600 Watts
Current that flows = 5 Amperes
Wasted power = 1 Watt
Voltage of toaster can be gotten from P = [tex]I^{2}[/tex]R
where I = current
and R = Resistance
600 = [tex]5^{2}[/tex] x R
R = 600/25 = 24 Ohms.
According to joules loss due to heating of wire
Power loss P ∝ [tex]I^{2}[/tex]R
imputing values,
1 ∝ [tex]5^{2}[/tex] x 24
1 ∝ 600
to remove the proportionality sign, we introduce a constant k
1 = 600k
k = 1/600 = 0.00167
For the case where the current is doubled to 10 ampere, as the power doubles to 1200 W.
The resistance across the wire becomes
1200 = [tex]10^{2}[/tex]R
R = 1200/100 = 12 Ohms
power loss P = k x [tex]I^{2}[/tex]R
P = 0.0016 x [tex]10^{2}[/tex] x 12
P = 1.92 Watt lost
This question involves the concepts of power, current, and resistance.
The power wasted by the wires in the home for two units will be "4 watt".
POWER WASTAGEThe power wasted by the wires can be given in terms of current and resistance by the following formula:
[tex]P=I^2R\\\\\frac{P}{I^2}=R=Constant\\\\\frac{P_1}{I_1^2}=\frac{P_2}{I_2^2}[/tex]
where,
P₁ = Power wasted for one unit = 1 wattI₁ = current through wires for one unit = 5 AR = Resistance of wires = constantP₂ = Power wasted for two units = ?I₂ = Current through wires for two units = 10 ATherefore,
[tex]\frac{1\ watt}{(5\ A)^2}=\frac{P_2}{(10\ A)^2}\\\\P_2=\frac{(1\ watt)(100\ A^2)}{25\ A^2}[/tex]
P₂ = 4 watt
Learn more about power here:
https://brainly.com/question/7963770
How can socialism
impact populations?
Answer:
it represents a fundamental difference. (more info below)
Explanation:
Production is incessantly developing and expanding in socialist countries, and employment is guaranteed for the entire productive population. Consequently, the relative overpopulation problem has been eliminated. This represents the fundamental difference between socialism's demographic law and capitalism's law.
hope this helped!
When a fuel is burned in a cylinder fitted with a piston, the volume expands from an initial value of 0.250 L against an external pressure of 2.00 atm. The expansion does 288 J of work on the surroundings. What is the final volume of the cylinder
Answer:
Vf = 0.0017 m³ = 1.7 L
Explanation:
The work done by the system on the surrounding at constant pressure is given by the following formula:
W = PΔV
W = P(Vf - Vi)
where,
W = Work done = 288 J
P = Constant Pressure = (2 atm)(101325 Pa/atm) = 202650 Pa
Vf = Final Volume f Cylinder = ?
Vi = Initial Volume of Cylinder = (0.25 L)(0.001 m³/ 1 L) = 0.00025 m³
Therefore,
288 J = (202650 Pa)(Vf - 0.00025 m³)
Vf = 288 J/202650 Pa + 0.00025 m³
Vf = 0.0017 m³ = 1.7 L
Olaf is standing on a sheet of ice that covers the football stadium parking lot in Buffalo, New York; there is negligible friction between his feet and the ice. A friend throws Olaf a ball of mass 0.400 kg that is traveling horizontally at 11.3 m/s. Olaf's mass is 75.0 kg. (a) If Olaf catches the ball, with what speed v_f do Olaf and the ball move afterward
Answer:
v = 0.059 m/s
Explanation:
To find the final speed of Olaf and the ball you use the conservation momentum law. The momentum of Olaf and the ball before catches the ball is the same of the momentum of Olaf and the ball after. Then, you have:
[tex]mv_{1i}+Mv_{2i}=(m+M)v[/tex] (1)
m: mass of the ball = 0.400kg
M: mass of Olaf = 75.0 kg
v1i: initial velocity of the ball = 11.3m/s
v2i: initial velocity of Olaf = 0m/s
v: final velocity of Olaf and the ball
You solve the equation (1) for v and replace the values of all variables:
[tex]v=\frac{mv_{1i}}{m+M}=\frac{(0.400kg)(11.3m/s)}{0.400kg+75.0kg}=0.059\frac{m}{s}[/tex]
Hence, after Olaf catches the ball, the velocity of Olaf and the ball is 0.059m/s
0.92 kg of R-134a fills a 0.14-m^3 weighted piston–cylinder device at a temperature of –26.4°C. The container is now heated until the temperature is 100°C. Determine the final volume of R-134a.
Answer:
The final volume of R-134a is 0.212m³Explanation:
Using one of the general gas equation to find the final volume of the R-134a.
According to pressure law; The volume of a given mas of gas is directly proportional to its temperature provided that the pressure remains constant.
VαT
V = kT
k = V/T
V1/T1 = V2/T2 = k
Given V1 = 0.14-m³ at T1 = –26.4°C = –26.4° + 273 = 246.6K
V2 = ? at T = 100°C = 100+273 = 373K
On substituting this values for T2;
0.14/246.6 = V2/373
373*0.14 = 246.6V2
V2 = 373*0.14 /246.6
V2 = 0.212m³
The final volume of R-134a is 0.212m³
A ball thrown horizontally from the top of a building hits the ground in 0.600 s. If it had been thrown with twice the speed in the same direction, it would have hit the ground in:________.
a. 4.0 s.
b. 1.0 s.
c. 0.50 s.
d. 0.25 s.
e. 0.125 s.
Answer:
none of the answers is correct, the time is the same t₁ = t₂ = 0.600 s
Explanation:
This is a kinematics exercise, analyze the situation a bit. The vertical speed in both cases is the same is zero, the horizontal speed in the second case is double (vₓ₂ = 2 vₓ₁)
let's find the time to hit the ground
y = y₀ + I go t - ½ g t²
0 = y₀ - ½ g t²
t = √ 2y₀ / g
with the data from the first launch
y₀i = ½ g t²
y₀ = ½ 9.8 0.6²
y₀ = 1,764 m
with this is the same height the time to descend in the second case is the same
t₂ = 0.600 s
this is because the horizontal velocity change changes the offset on the x axis, but does not affect the offset on the y axis
Therefore, none of the answers is correct, the time is the same
t₁ = t₂ = 0.600 s
Q) A particle in simple harmonic motion starts its motion from its mean position. If T be the time period, calculate the ratio of kinetic energy and potential energy of the particle at the instant when t = T/12.
t\12 and the parties are spreading ever
Explanation:
my point is that you can get sick if
you sont wash your ha
nds or be
save
A machinist turns the power on to a grinding wheel, which is at rest at time t = 0.00 s. The wheel accelerates uniformly for 10 s and reaches the operating angular velocity of 25 rad/s. The wheel is run at that angular velocity for 37 s and then power is shut off. The wheel decelerates uniformly at 1.5 rad/s2 until the wheel stops. In this situation, the time interval of angular deceleration (slowing down) is closest to: A machinist turns the power on to a grinding wheel, which is at rest at time t = 0.00 s. The wheel accelerates uniformly for 10 s and reaches the operating angular velocity of 25 rad/s. The wheel is run at that angular velocity for 37 s and then power is shut off. The wheel decelerates uniformly at 1.5 rad/s2 until the wheel stops. In this situation, the time interval of angular deceleration (slowing down) is closest to:__________.
a) 19 s
b) 17 s
c) 21 s
d) 23 s
e) 15 s
Starting from rest, the wheel attains an angular velocity of 25 rad/s in a matter of 10 s, which means the angular acceleration [tex]\alpha[/tex] is
[tex]25\dfrac{\rm rad}{\rm s}=\alpha(10\,\mathrm s)\implies\alpha=2.5\dfrac{\rm rad}{\mathrm s^2}[/tex]
For the next 37 s, the wheel maintains a constant angular velocity of 25 rad/s, meaning the angular acceleration is zero for the duration. After this time, the wheel undergoes an angular acceleration of -1.5 rad/s/s until it stops, which would take time [tex]t[/tex],
[tex]0\dfrac{\rm rad}{\rm s}=25\dfrac{\rm rad}{\rm s}+\left(-1.5\dfrac{\rm rad}{\mathrm s^2}\right)t\implies t=16.666\ldots\,\mathrm s[/tex]
which makes B, approximately 17 s, the correct answer.
The time interval of angular deceleration is 16.667 seconds, whose closest integer is 17 seconds. (B. 17 s.)
Let suppose that the grinding wheel has uniform Acceleration and Deceleration. In this question we need to need to calculate the time taken by the grinding wheel to stop, which is found by means of the following Kinematic formula:
[tex]t = \frac{\omega - \omega_{o}}{\alpha}[/tex] (1)
Where:
[tex]\omega_{o}[/tex] - Initial angular velocity, in radians per second.
[tex]\omega[/tex] - Final angular velocity, in radians per second.
[tex]\alpha[/tex] - Angular acceleration, in radians per square second.
[tex]t[/tex] - Time, in seconds.
If we know that [tex]\omega = 0\,\frac{rad}{s}[/tex], [tex]\omega_{o} = 25\,\frac{rad}{s}[/tex] and [tex]\alpha = -1.5\,\frac{rad}{s^{2}}[/tex], then the time taken by the grinding wheel to stop:
[tex]t = \frac{0\,\frac{rad}{s}-25\,\frac{rad}{s}}{-1.5\,\frac{rad}{s^{2}} }[/tex]
[tex]t = 16.667\,s[/tex]
The time interval of angular deceleration is 16.667 seconds. (Answer: B)
Please this related question: https://brainly.com/question/10708862
Which formation is one feature of karst topography?
Sinkholes formation is one feature of karst topography. The top of a cave falls if it develops large enough and its top extends near enough to the surface.
What is karst topography?Karst topography is a type of natural environment formed mostly by chemical weathering by water, resulting in caves, sinkholes, cliffs, and steep-sided hills known as towers.
The top of a cave falls if it develops large enough and its top extends near enough to the surface. Sinkholes are formed as a result of this, and they are one of the most distinguishing aspects of karst terrain.
When water absorbs carbon dioxide from the atmosphere and ground, it becomes carbonic acid.
Hence, sinkholes formation is one feature of karst topography
To learn more about the karst topography, refer to the link;
https://brainly.com/question/1167881
#SPJ2
Answer: A) Caves
Explanation:
A jet plane is flying at a constant altitude. At time t1=0t 1=0, it has components of velocity vx=90m/s,vy=110m/sv x = 90m/s,v y=110m/s. At time t2=30.0st 2=30.0s, the components are vx=−170m/s,vy=40m/sv x =−170m/s,v y=40m/s.
(a) Sketch the velocity vectors at t1and t2.
How do these two vectors differ? For this time interval calculate
(b) the components of the average acceleration, and
(c) the magnitude and direction of the average acceleration.
The average acceleration [tex]\vec a_{\rm ave}[/tex] over some time interval [tex][t_1,t_2][/tex] is equal to the ratio of the change in velocity [tex]\vec v_2-\vec v_1[/tex] over the duration of the interval [tex]t_2-t_1[/tex], or
[tex]\vec a_{\rm ave}=\dfrac{\Delta\vec v}{\Delta t}=\dfrac{\vec v_2-\vec v_1}{t_2-t_1}[/tex]
which can be split into the [tex]x[/tex] and [tex]y[/tex] components as
[tex]a_{\rm{ave},x}=\dfrac{v_{2,x}-v_{1,x}}{t_2-t_1}=\dfrac{-170\frac{\rm m}{\rm s}-90\frac{\rm m}{\rm s}}{30.0\,\mathrm s-0}\approx-8.67\dfrac{\rm m}{\mathrm s^2}[/tex]
[tex]a_{\rm{ave},y}=\dfrac{v_{2,y}-v_{1,y}}{t_2-t_1}=\dfrac{40\frac{\rm m}{\rm s}-110\frac{\rm m}{\rm s}}{30.0\,\mathrm s-0}\approx-2.33\dfrac{\rm m}{\mathrm s^2}[/tex]
The magnitude of this average acceleration is
[tex]\left\|\vec a_{\rm ave}\right\|=\sqrt{{a_{\rm{ave},x}}^2+{a_{\rm{ave},y}}^2}\approx8.98\dfrac{\rm m}{\mathrm s^2}[/tex]
and its direction is [tex]\theta[/tex] such that
[tex]\tan\theta=\dfrac{a_{\rm{ave},y}}{a_{\rm{ave},x}}\implies\theta\approx-164.9^\circ[/tex]
which corresponds to a direction of about 15.1º South of West.
If you were to drop a rock from a tall building, assuming that it had not yet hit the ground, and neglecting air resistance. What is its vertical displacement (in m) after 4 s? (g = 10 m/s2)
Answer:
d = 80 m
its vertical displacement (in m) after 4 s is 80 m
Explanation:
From the equation of motion;
d = vt + 0.5at^2 ......1
Where;
d = displacement
v = initial velocity = 0 (dropped with no initial speed)
t = time of flight = 4s
a = g = acceleration due to gravity = 10 m/s^2
Substituting the given values into equation 1;
d = 0(4) + 0.5(10 × 4^2)
d = 0.5(10×16)
d = 80 m
its vertical displacement (in m) after 4 s is 80 m
b) A non-inductive load takes a current of 15 A at 125 V. An inductor is then connected
in series in order that the same current shall be supplied from 240 V, 50 Hz mains.
Ignore the resistance of the inductor and calculate:
i. the inductance of the inductor;
ii. the impedance of the circuit;
iii. the phase difference between the current and the applied voltage.
Assume the waveform to be sinusoidal.
Answer:
i. 43.5 mH ii. 16 Ω. In phasor form Z = (8.33 + j13.66) Ω iii 58.64°
Explanation:
i. The resistance , R of the non-inductive load R = 125 V/15 A = 8.33 Ω
The reactance X of the inductor is X = 2πfL where f = frequency = 50 Hz.
So, x = 2π(50)L = 100πL Ω = 314.16L Ω
Since the current is the same when the 240 V supply is applied, then
the impedance Z = √(R² + X²) = 240 V/15 A
√(R² + X²) = 16 Ω
8.33² + X² = 16²
69.3889 + X² = 256
X² = 256 - 69.3889
X² = 186.6111
X = √186.6111
X = 13.66 Ω
Since X = 314.16L = 13.66 Ω
L = 13.66/314.16
= 0.0435 H
= 43.5 mH
ii. Since the same current is supplied in both circuits, the impedance Z of the circuit is Z = 240 V/15 A = 16 Ω.
So in phasor form Z = (8.33 + j13.66) Ω
iii. The phase difference θ between the current and voltage is
θ = tan⁻¹X/R
= tan⁻¹(314.16L/R)
= tan⁻¹(314.16 × 0.0435 H/8.33 Ω)
= tan⁻¹(13.66/8.33)
= tan⁻¹(1.6406)
= 58.64°
A ball rolls over the edge of a platform with only a horizontal velocity. The height of the platform is 1.60m and the horizontal range of the ball from the base of the platform is 20.0m. What is the velocity of the ball just before it touches the ground
Answer:
v = 46.99 m/s
Explanation:
The velocity of the ball just before it touches the ground, is given by the following formula:
[tex]v=\sqrt{v_x^2+v_y^2}[/tex] (1)
vx: horizontal component of the velocity
vy: vertical component of the velocity
The vertical component vy is calculated by using the following formula:
[tex]v_y^2=v_{oy}^2+2gh[/tex] (2)
vy: final velocity
voy: initial vertilal velocity = 0m/s (because it is a semi parabolic motion)
g: gravitational acceleration = 9.8 m/s^2
h: height = 1.60m
You replace the values of the parameters in the equation (2):
[tex]v_y=2(9.8m/s^2)(1.60m)=31.36\frac{m}{s}[/tex]
vx is calculated by using the information about the horizontal range of the ball:
[tex]R=v_o\sqrt{\frac{2h}{g}}[/tex] (3)
R: horizontal range of the ball = 20.0 m
You solve the previous equation for vo, the initial horizontal velocity:
[tex]v_o=R\sqrt{\frac{g}{2h}}=(20.0m)\sqrt{\frac{9.8m/s^2}{2(1.60m)}}\\\\v_o=35\frac{m}{s}[/tex]
The horizontal component of the velocity is constant in the complete trajectory, hence, you have that
vx = vo = 35 m/s
Finally, you replace the values of vx and vy in the equation (1):
[tex]v=\sqrt{(35m/s)^2+(31.36m/s)^2}=46.99\frac{m}{s}[/tex]
The velocity of the ball just before it touches the ground is 46.99 m/s