Answer:
Delivery truck
Explanation:
What happens when the elevator is in free-fall, that is, what is the value for FN and what is the sensation experienced by the person in the elevator? Also, discuss the meaning of FN<0.
Answer:
Weightlessness
Explanation:
When the elevator is in free fall, this can only occur when the cord of the elevator breaks.
The acceleration of the elevator will be equal to the acceleration due to gravity. That is:
a = g
Then, the body will experience what we called WEIGHTLESSNESS
Where the normal reaction N of the person will tend to zero. That is
N = 0
The value of FN < 0 because the person inside the lift will experience weightlessness.
When does a magnet induce an electric current in a wire coil?
O A. When the wire is connected to the coil
O B. When the magnet is near the coil
O C. When the magnet is moving back and forth in the coil
D. When the magnet is very strong
Answer:
B I believe
Explanation:
Which statement best describes energy and matter in a closed system? (2 po
O Energy and matter flow into and out of the system.
Energy can flow into and out of the system but matter cannot.
Energy and matter are contained within a closed system.
O There is no energy in a closed system; there is matter.
Answer:
Energy can flow into and out of the system but matter cannot.
Explanation:
In a closed system, energy can flow in and out of the system but matter cannot.
A closed system prevents double way flow of matter. A closed system conserves matter.For an isolated system, energy and matter cannot flow out of the system.
For open systems, energy and matter can flow out of the system.
Such systems are used for certain thermodynamics experiment.
I need emergency help we only have 3 minutes left
Answer:
Option A
Explanation:
Ast the force is equal and the diayance is equal the beam is also balanced
A car is moving at an average speed of 20 meters per second. This is equivalent to
Answer:
44.73 MP/H or 71.98 KM/H
Explanation:
Imagine you are running PE class. You run the first 1000 meters in 3 minutes and then get tired and run the last 600 meters in 5 minutes. What was your *average* speed?
Answer:
200 meters per minute
Explanation:
You run the 1600 meters in a total of 8 minutes, so the average speed if 200 meters per minute.
A 1200 N force acts on an object, resulting in an acceleration of 8.0 m/s2. What is the mass of the object?
Answer:
The answer is 150 kgExplanation:
The mass of the object can be found by using the formula
[tex]m = \frac{f}{a} \\ [/tex]
f is the force
a is the acceleration
From the question we have
[tex]m = \frac{1200}{8} \\ [/tex]
We have the final answer as
150 kgHope this helps you
Calculate the distance covered by a bus moving at a rate of 11.5km/h with a time of 2060seconds
Answer:
6.56km
Explanation:
Given parameters:
Speed = 11.5km/hr
Time = 2060s
Unknown:
Distance covered = ?
Solution:
Speed is distance divided by the time taken.
Speed = [tex]\frac{distance}{time}[/tex]
Distance = Speed x time
Let us convert the seconds to hours;
3600s = 1hr
2060s = [tex]\frac{2060}{3600}[/tex] = 0.57hr
Now
Distance = 11.5km/hr x 0.57hr = 6.56km
A 4.00-kg particle moves along the x axis. Its position varies with time according to x= 5t +1 2.0t^3, where x is in meters and t is in seconds. Find:
a. the kinetic energy of the particle at any time t.
b. the acceleration of the particle and the force acting on it at time t.
c. the power being delivered to the particle at time t.
d. the work done on the particle in the interval t = 0 to t =5
Answer:
a) The kinetic energy of the particle at any time t is [tex]K = 50+120\cdot t^{2}+72\cdot t^{4}[/tex].
b) The acceleration of the particle at time t is [tex]a= 12\cdot t[/tex]. The force acting on the particle at time t is [tex]F = 48\cdot t[/tex].
c) The power being delivered to the particle at time t is [tex]\dot W = 240\cdot t +288\cdot t^{3}[/tex].
d) The work done on the particle in the interval t = 0 to t = 5 is 48000 joules.
Explanation:
a) The kinetic energy of the particle is entirely translational, whose formula is:
[tex]K = \frac{1}{2}\cdot m \cdot v^{2}[/tex] (1)
Where:
[tex]K[/tex] - Translational kinetic energy, measured in joules.
[tex]m[/tex] - Mass of the particle, measured in kilograms.
[tex]v[/tex] - Velocity of the particle, measured in meters per second.
The velocity of the particle is the rate of change of the position of the particle in time, that is:
[tex]v = 5+6\cdot t^{2}[/tex] (2)
Where [tex]t[/tex] is the time, measured in seconds.
By substituting on (1), we have the following expression: ([tex]m = 4\,kg[/tex])
[tex]K = 2\cdot (5+6\cdot t^{2})^{2}[/tex]
[tex]K = 2\cdot (25+60\cdot t^{2} +36\cdot t^{4})[/tex]
[tex]K = 50+120\cdot t^{2}+72\cdot t^{4}[/tex]
The kinetic energy of the particle at any time t is [tex]K = 50+120\cdot t^{2}+72\cdot t^{4}[/tex].
b) The acceleration of the particle is the rate of change of the velocity of the particle in time, that is:
[tex]a= 12\cdot t[/tex] (3)
Where [tex]a[/tex] is the acceleration of the particle, measured in meters per square second.
The acceleration of the particle at time t is [tex]a= 12\cdot t[/tex].
The force is obtained by multiplying (3) by the mass of the particle. That is to say: ([tex]m = 4\,kg[/tex])
[tex]F = m\cdot a[/tex] (4)
[tex]F = 48\cdot t[/tex]
The force acting on the particle at time t is [tex]F = 48\cdot t[/tex].
c) According to the Work-Energy Theorem, the change in kinetic energy of the particle equals the change in the net work done on the particle. In this case, the power is equal to the rate of change in kinetic energy.
[tex]\dot W = \dot K[/tex] (5)
[tex]\dot W = \frac{d}{dt}(50+120\cdot t^{2}+72\cdot t^{4})[/tex]
[tex]\dot W = 240\cdot t +288\cdot t^{3}[/tex]
The power being delivered to the particle at time t is [tex]\dot W = 240\cdot t +288\cdot t^{3}[/tex].
d) The work done on the particle ([tex]W[/tex]), measured in joules, is equal to the change of the kinetic energy of the particle:
[tex]W = K(5)-K(0)[/tex] (6)
[tex]W = [50+120\cdot (5)^{2}+72\cdot (5)^{4}]-[50+120\cdot (0)^{2}+72\cdot (0)^{4}][/tex]
[tex]W = 48000\,J[/tex]
The work done on the particle in the interval t = 0 to t = 5 is 48000 joules.
In a warehouse, the workers sometimes slide boxes along the floor to move them. Two boxes were sliding toward each other and crashed. The crash caused both boxes to change speed. Based on the information in the diagram, which statement is correct? In your answer, explain what the forces were like and why the boxes changed speed.
Box 1 has more mass than Box 2.
Box 1 and Box 2 are the same mass.
Box 1 has less mass than Box 2.
**YOU MUST BE DESCRIPTIVE! Any short answers not explaining it wont get brainliest!**
Answer:
box 1 has larger mass than box 2 (which agrees with the first answer option given.
Explanation:
We need to consider the linear momentum of the boxes immediately before and after they crash.
Recall that momentum is defined as mass times velocity.
So for before the collision, the linear momentum of the system of two boxes is:
m1 * 4km/h - m2 * 8km/h
with m1 representing mass "1" on the left, and m2 representing mass 2 on the right.
Notice the sign of the linear momentum (one positive (moving towards the right) and the other one negative (moving towards the left)
For after the collision, we have or the linear momentum of the system:
- m1 * 2km/h - m2 * 1km/h
Then, since the linear momentum is conserved in the collision, we make the initial momentum equal the final and study the mass relationship between m1 and m2:
4 m1 - 8 m2 = - 2 m1 - m2
combining like terms for each mas on one side and another of the equal sign, we get;
4 m1 + 2 m1 = 8 m2 - m2
6 m1 = 7 m2
therefore m1 = (7/6) m2
which (since 7/6 is a number larger than one) tells us that m1 is larger than m2 by a factor of 7/6
Therefore, the first answer option is the correct answer.
if you increase the frequency of a wave by 5x whats it’s period?
We know that Period of a wave is the inverse of its Frequency
So, Period = 1 / Frequency
From the above, we can say that Period is inversely proportional to Frequency and hence, any change in Frequency will be the inverse change in the period
Therefore, we can say that if the frequency is increased by 5 times, the period will increase by 1/5 times
A wire carries a current of 11.4 A in a direction that makes an angle of 11.4° with a magnetic field of magnitude 11.4 à 10-3 T. The magnitude of the force on 11.4 cm of this wire is:____.a) 11.4 * 10^-3 N
b) 0.130 N
c) 1.48 * 10^-2 N
d) 2.93 * 10^-3 N
Answer:
(d) 2.93 x 10⁻³ N
Explanation:
Given;
current in the wire, I = 11.4 A
angle of inclination, θ = 11.4⁰
magnetic field on the wire, B = 11. 4 x 10⁻³
length of the wire, L = 11.4 cm = 0.114 m
The magnitude of magnetic force on the wire is given by;
F = BILsinθ
F = (11.4 x 10⁻³)(11.4)(0.114)(sin 11.4°)
F = 0.00293 N
F = 2.93 x 10⁻³ N
Therefore, the correct option is "D"
A piano of mass 852 kg is lifted to a height of 3.5 m. How much gravitational potential energy is added to the piano? Acceleration due to gravity is g = 9.8 m/s2 O A. 102,282 J O B. 29,224J O C. 2982 J O D. 304.3J SUSM
Answer:
29223.6J
Explanation:
Given parameters:
Mass of Piano = 852kg
Height of lifting = 3.5m
Unknown:
Gravitational potential energy = ?
Solution:
The gravitational potential energy of a body can be expressed as the energy due to the position of a body;
G.P.E = mgh
m is the mass
g is the acceleration due to gravity
h is the height
Now insert the given parameters and solve;
G.P.E = 852 x 9.8 x 3.5 = 29223.6J
Your teacher placed a 3.5 kg block at the position marked with a “ + ” (horizontally, 0.5 m from the origin) on a large incline outlined on the graph below and let it slide, starting from rest. ***There are two images included!***
Answer:
x = 10.75 m
Explanation:
For this problem we will solve it in two parts, the first using energy and the second with kinematics
Let's use the energy work relationship to find the velocity of the block as it exits the ramp
W = [tex]Em_{f}[/tex] - Em₀
Starting point. Higher
Em₀ = U = m g h
the height from the edge of the ramp of the graph has a value
h = 9-3 = 6 m
Final point. At the bottom of the ramp
Em_{f} = K = ½ m v²
Friction force work
W = - fr d
The friction force has the formula
fr = μ N
On the ramp, we can use Newton's second law
N - W cos θ = 0
N = W cos θ
where the angle is obtained from the graph
tan θ = (9-3) / (0.5-4) = -6 / 3.5
θ = tan⁻¹ (-1,714)
θ = -59.7º
the distance d is
d = √ (Δx² + Δy²)
d = √ [(0.5-4)² + (9-3)²]
d = 6.95 m
for which the work is
W = - μ mg cos 59.7 d
we substitute
W = Em_{f} -Em₀
- μ mg cos 59.7 d = ½ m v² - m g h
In the graph o text the value of the friction coefficient is not observed, suppose that it is μvery = 0.2
- μ g cos 59.7 d = ½ v² - g h
v² = 2g (h - very d coss 59.7)
let's calculate
v² = 2 9.8 (6 - 0.2 6.95 cos 59.7)
v = √ 103.8546
v = 10.19 m / s
in the same direction as the ramp
in the second part we use projectile launch kinematics
let's look for the components of velocity
v₀ₓ = vo cos -59.7
[tex]v_{oy}[/tex] = vo sin (-59,7)
v₀ₓ = 10.19 cos (-59.7) = 5.14 m / s
v_{oy} = 10.19 if (-59.7) = -8.798 m / s
Let's find the time to get to the floor (y = o)
y = y₀ + v_{oy} t - ½ g t²
to de groph y₀=3 m
0 = 3 - 8.798 t - ½ 9.8 t²
t² - 1.796 t - 0.612 = 0
we solve the quadratic equation
t = [1.796 ±√(1.796² + 4 0.612)] / 2
t = [1,795 ± 2,382] / 2
t₁ = 2.09 s
t₂ = -0.29 s
since time must be a positive quantity the correct value is t = 2.09 s
we calculate the horizontal displacement
x = v₀ₓ t
x = 5.14 2.09
x = 10.75 m
The motion of the box, after it exits the incline is the motion and trajectory
of a projectile.
Horizontal distance from the right-hand edge of the incline to the point of
contact with the floor is approximately 1.24613 m.
Reasons:
Mass of the block, m = 3.5 kg
Coefficient of kinetic friction, μ = 1.2
Location of the = 0.5 m from the origin
Required:
Horizontal distance between the block's point of contact with the floor and
the bottom right-hand edge of the incline.
Solution:
Let θ represent the angle the incline make with the horizontal.
The normal reaction of the incline on the block, [tex]F_N[/tex] = m·g·cos(θ)
Work done on friction = [tex]F_N[/tex]×μ×Length of the incline, L
Rise of the incline = 10 - 3 = 7
Run of the incline = 4
L = √(6.125² + 3.5²) = [tex]\dfrac{7 \times \sqrt{65} }{8}[/tex]
Let ΔP.E.₁ represent the potential energy transferred to kinetic energy
and work along the incline, we have;
Energy of the block at the bottom of the incline, M.E.₂, is found as follows;
K.E.₂ = mgh - m·g·μ·cos(θ)·L
[tex]K.E. =\frac{1}{2} \times 3.5 \times v^2 = 3.5 \times 9.81 \times 6.125 - 3.5 \times 9.81 \times 1.2 \times \dfrac{4}{\sqrt{65} } \times \dfrac{7 \times \sqrt{65} }{8}[/tex]
v ≈ 6.1456 m/s
The vertical component of the velocity is therefore;
[tex]v_y = v \cdot sin(\theta)[/tex]
[tex]v_y = 6.1456 \times \dfrac{7}{\sqrt{65} } \approx 5.33588[/tex]
From the equation, h = u·t + 0.5·g·t² derived from Newton's Laws of motion, we have;
ΔP.E.₁ = 3.5×9.81×7
3 = 5.33588·t + 0.5×9.81·t²
Factorizing, the above quadratic equation, we get;
The time it takes the block to reach the floor, t ≈ 0.40869 seconds
Horizontal component of the velocity is [tex]v_x \approx 6.1456 \times \dfrac{4}{\sqrt{65} } \approx 3.04908[/tex]
The horizontal distance, x = vₓ × t
∴ x = 3.04908 × 0.40869 ≈ 1.08194
Horizontal distance from the right-hand edge of the incline to the point of
contact with the floor, x ≈ 1.24613 m.
Learn more here:
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22. What happens to the volume of a 1 kg of water when it is heated from 4oC to 6oC? A. Increases B. Decreases C. Stays the same
Answer:
a
Explanation:
..................
Which of the following descriptions best describe a liquid
A
takes the shape and volume of its container
B
matter is made of atoms so tightly packed together that they cannot move around
C
has a definite volume, but takes the shape of its container
Answer:
C
Explanation:
A crossbow is fired horizontally off a cliff with an initial velocity of 15 m/s. If the arrow takes 4s to hit the ground, what is the range of the projectile?
Answer:
The range of the projectile is 60 m
Explanation:
Horizontal Motion
When an object is thrown horizontally with a speed vo from a height h, it describes a curved path ruled exclusively by gravity until it eventually hits the ground.
The horizontal component of the velocity is always constant because no acceleration acts in that direction, thus:
[tex]v_x=v_o[/tex]
The vertical component of the velocity changes in time because gravity makes the object fall at increasing speed given by:
[tex]v_y=g.t[/tex]
The horizontal distance is calculated as a constant speed motion:
[tex]x = v_x.t[/tex]
Knowing the crossbow is fired horizontally at vo=vx=15 m/s and it takes t=4 s to hit the ground, thus the range of the projectile is:
x = 15*4 = 60
The range of the projectile is 60 m
Starting from rest, a coin and a ring roll down a ramp without slipping. Which of the following are true:
A. The ring reaches the bottom first
B. The coin reaches the bottom first
C. The coin and the ring arrive at the same time
D. The one the reaches the bottom first is the one with the largest mass
E. The one that reaches the bottom first is the one with the largest diameter.
Answer:
The answer is D
Explanation:
The object with the heavy mass will reach first because the cause it is heaver so it will go faster
Squid use jet propulsion for rapid escapes. A squid pulls water into its body and then rapidly ejects the water backward to propel itself forward. A 1.5 kg squid (not including water mass) can accelerate at 20 m/s2 by ejecting 0.15 kg of water. Part A What is the magnitude of the thrust force on the squid
Answer: see attachment
Explanation:
STATION 1
Jane moved a 800kg piano to the right across the
carpet with a coefficient of friction of 0.4. What is the
magnitude of the force of friction acting on the
piano?
If she moved it at a constant velocity what is the
applied force acting on the piano?
A rack of seven spherical bolwing balls (each 7.00 kg, radius of 9.50 cm) is positioned along a line located a distance =0.850 m from a point ,
Calculate the gravitational force
the bowling balls exert on a ping-pong ball of mass 2.70 g, centered at point .
Answer:
8.72*[tex]10^{-12}[/tex] N
Explanation:
force of attraction f = G m1m2/ r^2 = [tex]\frac{6.67*10^{-11}*7*5*2.70 }{.85*.85*1000}[/tex] = 8.72*[tex]10^{-12}[/tex] N
The gravitational force the bowling balls exert on a ping-pong is given as 8.72*[tex]10^{-12}[/tex] N.
What is force?A force is an effect that can alter an object's motion according to physics. An object with mass can change its velocity, or accelerate, as a result of a force. An obvious way to describe force is as a push or a pull. A force is a vector quantity since it has both magnitude and direction.
force of attraction
f = G [tex]m_1m_2[/tex]/r²
= [tex]6.67*10^{-11}[/tex]*7*5*2.70/.85²*1000
= 8.72*[tex]10^{-12}[/tex] N
The gravitational force the bowling balls exert on a ping-pong is given as 8.72*[tex]10^{-12}[/tex] N.
To learn more about force refer to the link:
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What is the wavelength of light falling on double slits separated by 2.00 μm if the third-order maximum is at an angle of 60.0∘?
Answer:
λ = 5.773 x 10⁻⁷ m = 577.3 nm
Explanation:
In order to solve this problem we will use the grating equation:
mλ = d Sin θ
where,
m = order = 3
λ = wavelength of light = ?
d = slit separation = 2 μm = 2 x 10⁻⁶ m
θ = angle = 60°
Therefore,
(3)λ = (2 x 10⁻⁶ m)Sin 60°
λ = 1.732 x 10⁻⁶ m/3
λ = 5.773 x 10⁻⁷ m = 577.3 nm
A parallel plate capacitor is made up of two metal squares with sides of length 8.8 cm, separated by a distance 5.0 mm. When a voltage 187 V is set up across the terminals of the capacitor, the charge stored on the positive plate is equal to __________ nC. g
Answer:
2.56 nC
Explanation:
By definition, the capacitance is expressed by the following relationship between the charge stored on one of the plates of the capacitor and the potential difference between them, as follows:[tex]C =\frac{Q}{V} (1)[/tex]
For a parallel-plate capacitor, assuming a uniform surface charge density σ, if the area of the plates is A, the charge on one of the plates can be written as follows:[tex]Q = \sigma * A (2)[/tex]
Assuming an uniform electric field E, the potential difference V can be expressed as follows:[tex]V = E*d (3)[/tex]
where d is the distance between plates.
Applying Gauss 'Law to a closed surface half within one plate, half outside it, we find that E can be written as follows:[tex]E =\frac{\sigma}{\epsilon_{0}} (4)[/tex]
Replacing (4) in (3), and (2) in (1), we can express the capacitance C as follows:[tex]C= \frac{\epsilon_{0}*A}{d} (5)[/tex]
Taking (1) and (5), as both left sides are equal each other, the right sides are also equal, so we can write the following equality:[tex]\frac{Q}{V} = \frac{\epsilon_{0}*A}{d} (6)[/tex]
Solving for Q, we get:[tex]Q = \frac{\epsilon_{0}*A*V}{d} = \frac{8.85e-12F/m*(0.088m)^{2}*187 V}{5.0e-3m} = 2.56 nC[/tex]
1. An object in motion tends to stay in
motion because it has ___?____. (inertia or
terminal velocity)
Answer: intertia
Explanation:
During a circus performance, a 72-kg humancannonball is shot out of an 18-m-long cannon. If thehuman cannonball spends 0.95 s in the cannon,determine the average net force exerted on him in thebarrel of the cannon.
Answer:
2872.8 N
Explanation:
We have the following information
m =n72kg
Δy = 18m
t = 0.95s.
From here we use the equation
Δy=1/2at2 in order to solve for the acceleration.
So a
=( 2x 18m)/(0.95s²)
= 36/0.9025
= 39.9m/s2.
From there we use the equation
F = ma
F=(72kg) x (39.9)
= 2872.8N.
2872.8N is the average net force exerted on him in the barrel of the cannon.
Thank you!
Use the above picture to fill in the blanks for the following statement.
One of the element carbon combines with one of the element oxygen to form one of the compound carbon dioxide.
Answer:
C + 2O ------ CO2
Explanation:
"One" element of Carbon combines with "Two" elements of Oxygen, to form "One" compound of Carbon dioxide.
I didn't really get what you meant but this is my guess of what you meant
Block A is also connected to a horizontally-mounted spring with a spring constant of 281 J/m2. What is the angular frequency (in rad/s) of simple harmonic oscillations of this system?
Answer:
This question is incomplete
Explanation:
This question is incomplete. However, the formula to be used here is
ω = 2π/T
Where ω is the angular frequency (in rad/s)
T is the period - the time taken for Block A to complete one oscillation and return to it's original position.
To solve for this period T, the formula below should be used
T = 2π√m/k
where m is the mass of the object (Block A) and k is the spring constant (281 J/m²)
1. Opposite charges
O repel
attract
Answer:
attract
Explanation:
that is the answer
Answer:
Attract.
Explanation:
I took the quiz.
what transition metal has 5 more protons than the halogen found in period 3?
Answer: Titanium (Ti)
Explanation:
First, each element has a unique atomic number Z, that is equal to the number of protons in the nucleus.
The halogen in period 3 is chlorine (Ch)
Chlorine's atomic number is Z = 17, this means that it has 17 protons.
Now we want to find a transition metal that has 5 more protons, then this transition metal has Z = 17 + 5 = 22
Now we can look at the periodic table and find the element with Z = 22, and if this is in the d-block, then this will be a transition metal.
The element with Z = 22 is titanium (Ti)
The propeller of an aircraft accelerates from rest with an angular acceleration α = 7t + 8, where α is in rad/s2 and t is in seconds. What is the angle in radians through which the propeller rotates from t = 1.00 s to t = 6.10 s?
Answer:
The value is [tex]\theta =407.3 \ radian[/tex]
Explanation:
From the question we are told that
The angular acceleration is [tex]\alpha = (7t + 8) \ rad/ s^2[/tex]
The first time is [tex]t_1 = 1.00 \ s[/tex]
The second time [tex]t_2 = 6.10 \ s[/tex]
Generally the angular velocity is mathematically represented as
[tex]w = \int\limits {\alpha } \, dt[/tex]
=> [tex]w = \int\limits {7t + 8 } \, dt[/tex]
=> [tex]w =\frac{ 7t^2}{2} + 8 t[/tex]
Generally the angular displacement is mathematically represented as
[tex]\theta = \int\limits^{t_2}_{t_1} { w} \, dt[/tex]
=> [tex]\theta = \int\limits^{t_2}_{t_1} { \frac{7t^2}{2} + 8t } \, dt[/tex]
=> [tex]\theta = { \frac{7t^3}{6} + \frac{8t^2}{2} } | \left \ t_2} \atop {t_1}} \right.[/tex]
=> [tex]\theta = { \frac{7t^3}{6} + 4t^2} } | \left \ 6.10} \atop {1}} \right.[/tex]
=> [tex]\theta =[ { \frac{7}{6}[6.10 ]^3 + 4[6.10]^2} } ] -[ { \frac{7}{6}[1 ]^3 + 4[1]^2} } ][/tex]
=> [tex]\theta =407.3 \ radian[/tex]