Answer:
A. 52 min
.A. 47 watts
Explanation:
Given that;
jim weighs 75 kg
and he walks 3.3 mph; the objective here is to determine how long must he walk to expend 300 kcal.
Using the following relation to determine the amount of calories burned per minute while walking; we have:
[tex]\dfrac{MET*weight (kg)*3.5}{200}[/tex]
here;
MET = energy cost of a physical activity for a period of time
Obtaining the data for walking with a speed of 3.3 mph From the standard chart for MET, At 3.3 mph; we have our desired value to be 4.3
However;
the calories burned in a minute = [tex]\dfrac{4.3*75 (kg)*3.5}{200}[/tex]
= 5.644
Therefore, for walking for 52 mins; Jim burns approximately 293.475 kcal which is nearest to 300 kcal.
4.
Given that:
mass m = 75 kg
intensity = 6 kcal/min
The eg ergometer work rate = ??
Applying the formula:
[tex]V_O_2 ( intensity ) = ( \dfrac{W}{m}*1.8)+7[/tex]
where ;
[tex]V_O_2 ( intensity ) = \dfrac{1 \ kcal min^{-1}*10^{-3}}{5}[/tex]
[tex]V_O_2 ( intensity ) = \dfrac{6*1 \ kcal min^{-1}*10^{-3}}{5}[/tex]
[tex]V_O_2 ( intensity ) = 0.0012[/tex]
∴[tex]0.0012 = (\dfrac{W}{75}*1.8)+7 \\ \\ W = \dfrac{0.0012-7}{1.8}*75 \\ \\ W = \dfrac{7*75}{1.8} \\ \\ W = 291.66 \ kg m /min[/tex]
Converting to watts;
Since; 6.118kg-m/min is = 1 watt
Then 291.66 kgm /min will be equal to 47.67 watts
≅ 47 watts
A student drives 105.0 mi with an average speed of 61.0 mi/h for exactly 1 hour and 30
minutes for the first part of the trip. What is the distance in miles traveled during this
time?
Answer:
91.5 miles
Explanation:
61 miles per hour so 61(x amount of hours)
so 61 x 1.5 hours is 91.5 miles
g: To open a door, you apply a force of 10 N on the door knob, directed normal to the plane of the door. The door knob is 0.9 meters from the hinge axis, and the door swings open with an angular acceleration of 5 radians per second squared. What is the moment of inertia of the door
Answer:
I =1.8 kgm^2
Explanation:
In order to calculate the moment of inertia of the door you use the following formula, which relates the torque applied to the door with its moment of inertia and angular acceleration:
[tex]\tau=I\alpha[/tex] (1)
τ: torque applied to the door
I: moment of inertia of the door
α: angular acceleration = 5 rad/s^2
The torque is also given by τ = Fd, where F is the force applied at a distance of d to the pivot of the door (hinge axis).
F = 10 N
d = 0.9 m
You replace the expression for τ, and solve for I:
[tex]Fd=I\alpha\\\\I=\frac{Fd}{\alpha}\\\\I=\frac{(10N)(0.9m)}{5rad/s^2}=1.8kgm^2[/tex]
The moment of inertia of the door is 1.8 kgm^2
Which person will most likely hear the loudest sound?
A
B
C
D
Answer:
The youngest person
Explanation:
Hearing worsens with age
Please mark brainliest
Answer:
A
Explanation:
The person closest to the origin of the sound will most likely hear the loudest sound. ^^
Two carts undergo an inelastic collision where they stick together. Cart A has an initial velocity v0, and the second cart B is initially at rest. After the collision, it is observed that the ratio of the final kinetic energy system to its initial kinetic energy is KfK0= 1/6. Determine the ratio of the carts' masses, mBmA. (Assume the track is frictionless.)
Answer:
Explanation:
Initial kinetic energy of the system = 1/2 mA v0²
If Vf be the final velocity of both the carts
applying conservation of momentum
final velocity
Vf = mAvo / ( mA +mB)
kinetic energy ( final ) = 1/2 (mA +mB)mA²vo² / ( mA +mB)²
= mA²vo² / 2( mA +mB)
Given 1/2 mA v0² / mA²vo² / 2( mA +mB) = 6
mA v0² x ( mA +mB) / mA²vo² = 6
( mA +mB) / mA = 6
mA + mB = 6 mA
5 mA = mB
mB / mA = 5 .
Two parallel plates having charges of equal magnitude but opposite sign are separated by 21.0 cm. Each plate has a surface charge density of 39.0 nC/m2. A proton is released from rest at the positive plate. (a) Determine the magnitude of the electric field between the plates from the charge density.
Answer:
E = 3.45*10^-19 N/C
Explanation:
a) The electric field between two parallel plates id given by the following formula:
[tex]E=\frac{\sigma}{\epsilon_o}[/tex] (1)
where:
σ: surface charge density of the plates = 39.0nC/m^2
εo: dielectric permittivity of vacuum = 8.85*10^-12 C/Nm^2
You replace these values in the equation (1):
[tex]E=\frac{39.0*10^{-9}C/m^2}{8.85*10^{-12}C^2/Nm^2}\\\\E=3.45*10^{-19}\frac{N}{C}[/tex]
The electric field in between the parallel plates is 3.45*10^-19 N/C
A physics major is cooking breakfast when he notices that the frictional force between the steel spatula and the Teflon frying pan is only 0.400 N. Knowing the coefficient of kinetic friction between the two materials (0.04), he quickly calculates the normal force. What is it (in N)? N
Answer:
normal force = 10 N
Explanation:
Given data
frictional force = 0.400 N
coefficient of kinetic friction = 0.04
Solution
we get here normal force that is express as
normal force = [tex]\frac{Frictional\ force}{coefficient\ of\ friction}[/tex] ............1
put here value and we will get value
normal force = [tex]\frac{0.400}{0.04}[/tex]
solve it we get
normal force = 10 N
A driver wearing a seat beat decelerates at roughly the same rate as the car it self. Since many modern cars have a "crumble zone" built into the front of the car, let us assume that the car decelerates of a distance of 0.9 m. What is the net force acting on a 65 kg driver who is driving at 18 m/sec and comes to rest in this distance
Answer:
11,700Newton
Explanation:
According to Newton's second law, Force = mass × acceleration
Given mass = 65kg.
Acceleration if the car can be gotten using one of the equation of motion as shown.
v² = u²+2as
v is the final velocity = 18m/s
u is the initial velocity = 0m/s
a is the acceleration
s is the distance travelled = 0.9m
On substitution;
18² = 0²+2a(0.9)
18² = 1.8a
a = 324/1.8
a = 180m/²
Net force acting on the body = 65×180
Net force acting on the body = 11,700Newton
What's a line of best fit? Will give BRAINLIEST
A line of best fit expresses the relationship between the points.
Explanation:
It does not go through all the points but goes through most of them and it is like a hardrawn curve
02
Blue light has a frequency of about 7.5 x 1014 Hz. Calculate the energy, in Joules, of a single photon associated with this frequency
Answer:
49.725× 10^-24J
Explanation:
The Energy associated with a Photon us defined as;
E = hf
Where h is Planck's constant = 6.63× 10^-34m2kg/s
f is the frequency= 7.5 x 10^14 Hz
Hence
E = 6.63× 10^-34 × 7.5 x 10^14 =49.725× 10^-24J
The circular handle of a faucet is attached to a rod that opens and closes a valve when the handle is turned. If the rod has a diameter of 1cm1cm and the IMA of the machine is 66 , what is the radius of the handle
Question: The circular handle of a faucet is attached to a rod that opens and closes a valve when the handle is turned. If the rod has a diameter of 1cm and the IMA of the machine is 6, what is the radius of the handle?
Answer:
Radius of the handle = 3 cm = 0.03 m
Explanation:
Mechanical Accuracy, MA = (Radius of the handle)/(Radius of the rod).......(1)
Diameter of the rod = 1 cm
Radius of the rod = Diameter/2
Radius of the rod = 1/2
Radius of the rod = 0.5 cm
Mechanical Accuracy of the machine, MA = 6
Substitute the values into equation (1)
6 = (Radius of the handle)/0.5
Radius of the handle = 6 * 0.5
Radius of the handle = 3 cm
The Radius of the handle is = 3 cm = 0.03 m
Calculation of the radius of the handle:Since
Mechanical Accuracy, MA = (Radius of the handle)/(Radius of the rod)
Here,
Diameter of the rod = 1 cm
We know that
The radius of the rod = Diameter/2
So,
Radius of the rod = 1/2
So,
Radius of the rod = 0.5 cm
Now
Mechanical Accuracy of the machine, MA = 6
Now
6 = (Radius of the handle)/0.5
Radius of the handle = 6 * 0.5
Radius of the handle = 3 cm
Learn more about radius here: https://brainly.com/question/18648019
A fuel pump sends gasoline from a car's fuel tank to the engine at a rate of 5.37x10-2 kg/s. The density of the gasoline is 739 kg/m3, and the radius of the fuel line is 3.37x10-3 m. What is the speed at which gasoline moves through the fuel line
Answer:
Speed v = 2.04 m/s
the speed at which gasoline moves through the fuel line is 2.04 m/s
Explanation:
Given;
Mass transfer rate m = 5.37x10^-2 kg/s.
Density d = 739 kg/m3
radius of pipe r = 3.37x10^-3 m
We know that;
Density = mass/volume
Volume = mass/density
Volumetric flow rate V = mass transfer rate/density
V = m/d
V = 5.37x10^-2 kg/s ÷ 739 kg/m3
V = 0.00007266576454 m^3/s
V = 7.267 × 10^-5 m^3/s
V = cross sectional area × speed
V = Av
Area A = πr^2
V = πr^2 × v
v = V/πr^2
Substituting the given values;
v = 7.267 × 10^-5 m^3/s/(π×(3.37x10^-3 m)^2))
v = 0.203678639672 × 10 m/s
v = 2.04 m/s
the speed at which gasoline moves through the fuel line is 2.04 m/s
Aparticlewhosemassis2.0kgmovesinthexyplanewithaconstantspeedof3.0m/s along the direction r = i + j . What is its angular momentum (in kg · m2/s) relative to the point (0, 5.0) meters?
Answer:
[tex]\vec{L}=-30\frac{kgm^2}{s}\hat{k}[/tex]
Explanation:
In order to calculate the angular momentum of the particle you use the following formula:
[tex]\vec{L}=\vec{r}\ X\ \vec{p}[/tex] (1)
r is the position vector respect to the point (0 , 5.0), that is:
r = 0m i + 5.0m j (2)
p is the linear momentum vector and it is given by:
[tex]\vec{p}=m\vec{v}=(2.0kg)(3.0m/s)(\hat{i+\hat{j}})=6\frac{kgm}{s}(\hat{i}+\hat{j})[/tex] (3)
the direction of p comes from the fat that the particle is moving along the i + j direction.
Then, you use the results of (2) and (3) in the equation (1) and solve for L:
[tex]\vec{L}=-30\frac{kgm^2}{s}\hat{k}[/tex]
The angular momentum is -30 kgm^2/s ^k
How much displacement will a spring with a constant of 120N / m achieve if it is stretched by a force of 60N?
Answer:
Explanation:
There's a formula for this:
[tex]F = k*displacement[/tex]
F being force, k being the spring constant, and displacement being the change in x
We are given the force and the spring constant, so this is essentially isolating the Δx term. Do 60N/120N per meter. The newtons cancel out and you get a final answer of Δx = 0.5 meters
Using the equation for the distance between fringes, Δy = xλ d , complete the following. (a) Calculate the distance (in cm) between fringes for 694 nm light falling on double slits separated by 0.0850 mm, located 4.00 m from a screen. cm (b) What would be the distance between fringes (in cm) if the entire apparatus were submersed in water, whose index of refraction is 1.333? cm
Answer:
Explanation:
Distance between fringe or fringe width = xλ / d
where x is location of screen and d is slit separation
Given x = 4 m
λ = 694 nm
d = .085 x 10⁻³ m
distance between fringes
= 4 x 694 x 10⁻⁹ / .085 x 10⁻³
= 4 x 694 x 10⁻⁹ / 85 x 10⁻⁶
= 32.66 x 10⁻³ m
= 32.66 mm .
3.267 cm
b )
when submerged in water , wavelength in water becomes as follows
wavelength in water = wave length / refractive index
= 694 / 1.333 nm
= 520.63 nm
new distance between fringes
3.267 / 1.333
= 2.45 cm .
A solid exerts a force of 500 N. Calculate the pressure exerted to the surface where area
of contact is 2000 cm2.
Answer:
2500 N/m²
Explanation:
Pressure: This can be defined as the force acting normally on a surface per unit area.
The expression for pressure is give as
P = F/A...................... Equation 1
Where P = pressure (N/m²), F = force (N), A = Contact area (m²)
Given: F = 500 N, A = 2000 cm² = (2000/10000) m = 0.2 m.
Substitute into equation 1
P = 500/0.2
P = 2500 N/m²
Hence the pressure exerted to the surface is 2500 N/m²
Astrophysicist Neil deGrasse Tyson steps into an elevator on the 29th floor of a skyscraper. For some odd reason, there is a scale on the floor of the elevator. Neil, whose mass is about 115 kg, decides to step on the scale and presses the button for a lower floor. The elevator starts traveling downwards with a constant acceleration of 1.5 m/s2 for 6.0 seconds, and then travels at a constant velocity for 6.0 seconds. Finally, the elevator has an upward acceleration of 1.5 m/s2 for 6.0 seconds as it comes to a stop.
A. If each floor is approximately 4 m tall, which floor does the elevator stop at?
B. If the mass of the elevator is 1,200 kg, what is the maximum tension of the elevator cable?
Answer:
A. Final Floor = 15.5 = 15 (Considering downward portion of elevator)
B. T = 14859.5 N = 14.89 KN
Explanation:
A.
First we calculate distance covered by the elevator during downward motion. Downward motion consists of two parts. First one is uniformly accelerated. For that part we use 2nd equation of motion:
s₁ = Vi t + (0.5)at²
where,
s₁ = distance covered during accelerated downward motion = ?
Vi = initial speed = 0 m/s (since elevator is initially at rest)
t = time taken = 6 s
a = acceleration = 1.5 m/s²
Therefore,
s₁ = (0 m/s)(6 s) + (0.5)(1.5 m/s²)(6 s)²
s₁ = 4.5 m
also we find the final velocity using 1st equation of motion:
Vf = Vi + at
Vf = 0 m/s + (1.5 m/s²)(6 s)
Vf = 9 m/s
Now, the second part of downward motion is with constant velocity. So:
s₂ = vt
where,
s₂ = distance covered during constant speed downward motion = ?
v = Vf = 9 m/s
t = 6 s
Therefore,
s₂ = (9 m/s)(6 s)
s₂ = 54 m
Now for distance covered during upward motion is given by the 2nd equation of motion. Since the values of acceleration and time are same. Therefore, it will be equal in magnitude to s₁:
s₃ = s₁ = 4.5 m
Therefore, the total distance covered by elevator is given by following equation:
s = s₁ + s₂ - s₃ (Downward motion taken positive)
s = 4.5 m + 54 m - 4.5 m
s = 54 m
Therefore, net motion of the elevator was 54 m downwards.
So the final floor will be:
Final Floor = Initial Floor - Distance Covered/Length of a floor
Final Floor = 29 - 54 m/4m
Final Floor = 15.5 = 15 (Considering the downward portion or floor of elevator)
B.
The maximum tension will occur during the upward accelerated motion. It is given by the formula:
T = m(g + a)
where,
T = Max. Tension in Cable = ?
m = total mass of person and elevator = 115 kg + 1200 kg = 1315 kg
g = 9.8 m/s²
a = acceleration = 1.5 m/s²
Therefore,
T = (1315 kg)(9.8 m/s² + 1.5 m/s²)
T = 14859.5 N = 14.89 KN
Why do bears activity increase as certain points during the day
Because they are well rested and have to work to get food in their system.
If radio waves are used to communicate with an alien spaceship approaching Earth at 10% of the speed of light c, the aliens would receive our signals at speed of:_______.
a. 0.99c
b. 1.10c
c. 1.00c
d. 0.90c
e. 0.10c
Answer:
3×10^7 m/s or 0.10c (e)
Explanation: If the actual value of the speed of light were to be put into consideration.
Given that the speed of light is c = 3.0×10^8m/s
The alien spaceship is approaching at the rate of 10% of the speed of light.
10% of 3.0×10^8m/s
10/100 × 3.0×10^8m/s
0.1 ×3.0×10^8m/s
3×10^7 m/s. Which is the same thing as 0.1 of c = 0.1×c
Answer: 1.00c
Explanation: I got it correct on the homework
Two radio antennas A and B radiate in phase. Antenna B is a distance of 100 m to the right of antenna A. Consider point Q along the extension of the line connecting the antennas, a horizontal distance of 50.0 m to the right of antenna B. The frequency, and hence the wavelength, of the emitted waves can be varied.
Required:
a. What is the longest wavelength for which there will be destructive interference at point Q?
b. What is the longest wavelength for which there will be constructive interference at point Q?
Answer:
a. 200 m
b. 100 m
Explanation:
Solution:-
- We will first draw three points marked A,B and Q from left most to right most.
- We are told that the antennas at A and B radiate in phase. This means the radio-waves emitted by each antenna are synchronous in terms of ( frequency and wavelength ).
- We will denote the common wavelength of coherent sources of radio-waves ( A and B ) with λ.
- The relation between the wavelength ( λ ) and the path difference between the source and observation point ( Q ) for the case of destructive interference is:
AQ - BQ = n*λ/2
Where,
n: The order of wavelength
AQ: The distance between antenna A and point Q
BQ: The distance between antenna B and point Q
- The point Q is positioned ( 100 + 50 ) m away from antenna A and 50 m from antenna B. Hence,
150 - 50 = n*λ/2
- To determine the longest wavelength ( λ ) to meet destructively at point Q with the given path difference. The order of wavelength ( n ) must be minimum ( 1 ). Therefore,
100 = λ/2
λ = 200 m .... Answer
- The relation between the wavelength ( λ ) and the path difference between the source and observation point ( Q ) for the case of constructive interference is:
AQ - BQ = n*λ
Where,
n: The order of wavelength
AQ: The distance between antenna A and point Q
BQ: The distance between antenna B and point Q
- The point Q is positioned ( 100 + 50 ) m away from antenna A and 50 m from antenna B. Hence,
150 - 50 = n*λ
- To determine the longest wavelength ( λ ) to meet constructively at point Q with the given path difference. The order of wavelength ( n ) must be minimum ( 1 ). Therefore,
100 = λ
λ = 100 m .... Answer
An aluminum wing on a passenger jet is 30 m long when its temperature is 27 C. At what temperature would the wing be 0.03 shorter?
Answer:2000
Explanation:
To open a door, you apply a force of 10 N on the door knob, directed normal to the plane of the door. The door knob is 0.9 meters from the hinge axis, and the door swings open with an angular acceleration of 5 radians per second squared. What is the moment of inertia of the door
Answer:
The moment of inertia is [tex]I = 1.8 \ kg m^2[/tex]
Explanation:
From the question we are told that
The force applied is [tex]F = 10 \ N[/tex]
The distance of the knob to the hinge is [tex]d = 0.9 \ m[/tex]
The angular acceleration is [tex]a = 5 \ rad/s[/tex]
The moment is mathematically represented as
[tex]I = \frac{d Fsin(\theta)}{a}[/tex]
Here [tex]\theta = 90^o[/tex] This is because the force direction is perpendicular to the plane of the door
substituting values
[tex]I = \frac{0.9 * 10 * sin (90)}{5}[/tex]
[tex]I = 1.8 \ kg m^2[/tex]
determine the smallest mass of lead that when tied using a string to a wooden boat on a pond will be enough to sink the toy boat. assuming specific gravity of wood is 0.5 and density of water is 1000kg per cubic metre?
A 2 kg car moving towards the right at 4 m/s collides head on with an 8 kg car moving towards the left at 2 m/s, and they stick together. After the collision, the velocity of the combined bodies is:_____________.
a) 2.4 m/s towards the left.
b) 2.4 m/s towards the right.
c) 0.8 m/s towards the left.
d) 0
e) 0.8 m/s towards the right.
Answer:
correct answer is c
v = -0.8 m / s
Explanation:
This is a problem of quantity of movement, for this we must define a system formed by the two cars, so that the forces during the collision are internal and therefore the quantity of movement is conserved
initial
p₀ = m₁ v₁ - m₂ v₂
final
= (m₁ + m₂) v
We have taken the direction to the right as positive
p₀ =p_{f}
m₁ v₁ - m₂ v₂ = (m₁ + m₂) v
v = (m₁ v₁ - m₂ v₂) / (m₁ + m₂)
we calculate
v = (2 4 - 8 2) / (2 + 8)
v = (8 -16) / 10
v = -0.8 m / s
the negative sign indicates that the set is moving to the left
correct answer is c
To study the properties of various particles, you can accelerate the particles with electric fields. A positron is a particle with the same mass as an electron but the opposite charge ( e). If a positron is accelerated by a constant electric field of magnitude 286 N/C, find the following.
(a) Find the acceleration of the positron. m/s2
(b) Find the positron's speed after 8.70 × 10-9 s. Assume that the positron started from rest. m/s
Answer:
a) a = 5.03x10¹³ m/s²
b) [tex]V_{f} = 4.4 \cdot 10^{5} m/s [/tex]
Explanation:
a) The acceleration of the positron can be found as follows:
[tex] F = q*E [/tex] (1)
Also,
[tex] F = ma [/tex] (2)
By entering equation (1) into (2), we have:
[tex] a = \frac{F}{m} = \frac{qE}{m} [/tex]
Where:
F: is the electric force
m: is the particle's mass = 9.1x10⁻³¹ kg
q: is the charge of the positron = 1.6x10⁻¹⁹ C
E: is the electric field = 286 N/C
[tex] a = \frac{qE}{m} = \frac{1.6 \cdot 10^{-19} C*286 N/C}{9.1 \cdot 10^{-31} kg} = 5.03 \cdot 10^{13} m/s^{2} [/tex]
b) The positron's speed can be calculated using the following equation:
[tex] V_{f} = V_{0} + at [/tex]
Where:
[tex]V_{f}[/tex]: is the final speed =?
[tex]V_{0}[/tex]: is the initial speed =0
t: is the time = 8.70x10⁻⁹ s
[tex] V_{f} = V_{0} + at = 0 + 5.03 \cdot 10^{13} m/s^{2}*8.70 \cdot 10^{-9} s = 4.4 \cdot 10^{5} m/s [/tex]
I hope it helps you!
Suppose the demand for air travel decreases (as illustrated in the graph below). A decrease in demand _____ the equilibrium price for air travel and _____ the equilibrium quantity for air travel. decreases, decreases increases, increases decreases, increases
Answer:
decreases, decreases
Explanation:
A decrease in the demand will create a fall in equilibrium prices and the quantity supplied will also decrease. As the equilibrium prices in the market are the price in which the quantity demanded equals to quantity supplied. If the demand for the air decreases then the quantity of the air travel will also decrease and thus when the supply and demand change so do the changes associated with the equilibrium prices.The Great Lakes are all part of what? The Mississippi River The St. Lawrence Seaway A large body of salt lakes The Missouri River
Answer:
St Lawrence Sea way
Explanation:
The great lake connects the middle of North America which is at the Canada-United states border connecting to the Atlantic Ocean through the St Lawrence River.
Coherent light that contains two wavelengths, 660 nm and 470 nm , passes through two narrow slits with a separation of 0.280 mm and an interference pattern is observed on a screen which is a distance 5.50 m from the slits.
Required:
What is the disatnce on the screen between the first order bright fringe for each wavelength?
Answer:
λ1 = 0.0129m = 1.29cm
λ2 = 0.00923m = 0.92 cm
Explanation:
To find the distance between the first order bright fringe and the central peak, can be calculated by using the following formula:
[tex]y_m=\frac{m\lambda D}{d}[/tex] (1)
m: order of the bright fringe = 1
λ: wavelength of the light = 660 nm, 470 nm
D: distance from the screen = 5.50 m
d: distance between slits = 0.280mm = 0.280 *10^⁻3 m
ym: height of the m-th fringe
You replace the values of the variables in the equation (1) for each wavelength:
For λ = 660 nm = 660*10^-9 m
[tex]y_1=\frac{(1)(660*10^{-9}m)(5.50m)}{0.280*10^{-3}m}=0.0129m=1.29cm[/tex]
For λ = 470 nm = 470*10^-9 m
[tex]y_1=\frac{(1)(470*10^{-9}m)(5.50m)}{0.280*10^{-3}m}=0.00923m=0.92cm[/tex]
Water, in a 100-mm-diameter jet with speed of 30 m/s to the right, is deflected by a cone that moves to the left at 14 m/s. Determine (a) the thickness of the jet sheet at a radius of 230 mm. and (b) the external horizontal force needed to m
Answer:
Explanation:
The velocity at the inlet and exit of the control volume are same [tex]V_i=V_e=V[/tex]
Calculate the inlet and exit velocity of water jet
[tex]V=V_j+V_e\\\\V=30+14\\\\V=44m/s[/tex]
The conservation of mass equation of steady flow
[tex]\sum ^e_i\bar V. \bar A=0\\\\(-V_iA_i+V_eA_e)=0[/tex]
[tex]A_i\ \texttt {is the inlet area of the jet}[/tex]
[tex]A_e\ \texttt {is the exit area of the jet}[/tex]
since inlet and exit velocity of water jet are equal so the inlet and exit cross section area of the jet is equal
The expression for thickness of the jet
[tex]A_i=A_e\\\\\frac{\pi}{4} D_j^2=2\pi Rt\\\\t=\frac{D^2_j}{8R}[/tex]
R is the radius
t is the thickness of the jet
D_j is the diameter of the inlet jet
[tex]t=\frac{(100\times10^{-3})^2}{8(230\times10^{-3}} \\\\=5.434mm[/tex]
(b)
[tex]R-x=\rho(AV_r)[-(V_i)+(V_c)\cos 60^o]\\\\=\rho(V_j+V_c)A[-(V_i+V_c)+(V_i+V_c)\cos 60^o]\\\\=\rho(V_j+V_c)(\frac{\pi}{4}D_j^2 )[V_i+V_c](\cos60^o-1)][/tex]
[tex]1000kg/m^3=\rho\\\\44m/s=(V_j+V+c)\\\\100\times10^{-3}m=D_j[/tex]
[tex]R_x=[1000\times(44)\frac{\pi}{4} (10\times10^{-3})^2[(44)(\cos60^o-1)]]\\\\=-7603N[/tex]
The negative sign indicate that the direction of the force will be in opposite direction of our assumption
Therefore, the horizontal force is -7603N
The current through an inductor of inductance L is given by I(t) = Imax sin(ωt).
(a) Derive an expression for the induced emf in the inductor as a function of time.
(b) At t = 0, is the current through the inductor increasing or decreasing?
(c) At t = 0, is the induced emf opposing or aiding the flow of the charge carriers? (Remember that the direction of a positive induced emf is the same as the current direction and the direction of a negative induced emf is opposite the current direction.)
(d) How are the answers to parts b and c consistent with the behavior of inductors discussed in the text?
Answer:
(a) [tex]emf_L=-LI_{max}\omega cos(\omega t)[/tex]
(b) neither increasing or decreasing
(c) opposite to the flow of charge carriers
Explanation:
The current through an inductor of inductance L is given by:
[tex]I(t)=I_{max}sin(\omega t)[/tex] (1)
(a) The induced emf is given by the following formula
[tex]emf_L=-L\frac{dI}{dt}[/tex] (2)
You derivative the expression (1) in the expression (2):
[tex]emf_L=-L\frac{d}{dt}(I_{max}sin(\omega t))\\\\emf_L=-LI_{max}\omega cos(\omega t)[/tex]
(b) At t=0 the current is zero
(c) At t = 0 the emf is:
[tex]emf_L=-\omega LI_{max}[/tex]
w, L and Imax have positive values, then the emf is negative. Hence, the induced emf is opposite to the flow of the charge carriers.
(d) read the text carefully
At t zero, the current through the inductor neither increasing nor decreasing because current is zero.
The current through an inductor of inductance L can be calculated by
[tex]\bold {I_t = I_m_a_x sin (\omega t)}[/tex].........1
(a) The induced emf can be calculated by
[tex]\bold {emf_L = - L \dfrac {dI}{dt}}[/tex]............2
Derivative the equation (1) in the equation (2)
[tex]\bold {emf _L= -L \dfrac {d (I _m_a_x sin (\omega t)} {dt}}\\\\\bold {emf _L= -L (I _m_a_x \omega cos( \omega t) }[/tex]
(b) At t=0 the current is zero,
(c) At t = 0 the emf is:
[tex]\bold {emf_L = -\omega LI _m_a_x}[/tex]
Therefore, at t zero, the current through the inductor neither increasing nor decreasing because current is zero.
To know more about inductance,
https://brainly.com/question/17431388
If a cart of 8 kg mass has a force of 16 newtons exerted on it, what is its acceleration?
Answer:
Explanation:
From Newton's 2nd Law,
F = m×a
Where F is Force
m is mass
a is acceleration
Hence a= F/m
a= 16/8= 2m/s2