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murzikaleks [220]
3 years ago
7

A generator consists of a rectangular loop with turns of wire spinning at in a uniform magnetic field. The generator output is c

onnected to a series RC circuit consisting of a and a capacitor. What is the average power delivered to the circuit?
Physics
1 answer:
iren [92.7K]3 years ago
5 0
<h3><u>Full question:</u></h3>

A generator consists of a 18-cm by 12-cm rectangular loop with 500 turns of wire spinning at 60 Hz in a 25 mT uniform magnetic field. The generator output is connected to a series RC circuit consisting of a 150 Ω and a 35 μF capacitor.What is the average power delivered to the circuit?

<h3><u>Answer:</u></h3>

The average power delivered to the circiut is 27.5 W

<h3><u>Solution:</u></h3>

Induced emf amplitude = N A B w  

N- Number of turns of the coil

B- Magnetic field

=500 \times(0.18 \times 0.12) \times 0.025 \times(2 \pi 60)

Induced emf amplitude = 101.8 Volt

\begin{aligned}&X_{c}=\frac{1}{(2 p i f C)}=\frac{1}{2 \times \pi \times 60 \times 35 \times 10^{-6}}\\&X_{c}=75.8\ \mathrm{ohm}\end{aligned}

R = 150 ohm

\begin{aligned}&z=\sqrt{\left[R^{2}+X_{c}^{2}\right]}=168 \text { ohm }\\&I_{\text {peak}}=\frac{101.8}{168}=0.606\ \mathrm{A}\\&P_{a v g}=\frac{l^{2} R}{2}=27.5\ \mathrm{W}\end{aligned}

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If an object is thrown in an upward direction from the top of a building 160 ft. High at an initial speed of 21.82 mi/h what is
viktelen [127]
To solve this problem we are going to use tow kinematic equations for falling objects.
1. Kinematic equation for final velocity: V_{f}=V_{i}+gt
where
V_{f} is the final velocity 
V_{i} is the initial velocity 
g is the acceleration due to gravity 32 \frac{ft}{s^2}
t is the time 
2. Kinematic equation for distance: d=V_{i}t+ \frac{1}{2} gt^2
where
d is the distance 
V_{i} is the initial velocity 
V_{f} is the final velocity
g is the acceleration due to gravity 32 \frac{ft}{s^2}
t is the time 

First, we are going to convert 21.82 mi/h to ft/s:
21.82 \frac{mi}{h} =31.21 \frac{ft}{s}

Next, we are going to use the first equation to find how long it takes for the rock to reach its maximum height.
We know for our problem that the object is thrown in upward direction, so its velocity at its maximum height (before falling again) will be zero; therefore: V_{f}=0. We also know that it initial speed is 31.21 ft/s, so V_{i}=31.21. Lets replace those values in our formula to find t:
V_{f}=V_{i}+gt
0=31.21+(-32)t
-32t=-31.21
t= \frac{-31.21}{-32}
t=0.98seconds

Next, we are going to use that time in our second kinematic equation to find the distance the object reach at its maximum height:
d=V_{i}t+ \frac{1}{2} gt^2
d=31.21(0.98)+ \frac{1}{2} (-32)(0.98)^2
d=15.22ft 

Now we can add the height of the building and the maximum height of the object:
d=160+15.22=175.22ft

Next, we are going to use that height (distance) in our second kinematic equation one more time to fin how long it takes for the object to fall from its maximum height to the ground:
d=V_{i}t+ \frac{1}{2} gt^2
175.22=31.21t+ \frac{1}{2} (32)t^2
16t^2+31.21t-175.22=0
t=2.47 or t=-4.43
Since time cannot be negative, t=2.47 is the time it takes the object to fall to the ground. 

Finally, we can use that time in our first kinematic equation to find the final speed of the object when it hits the ground:
V_{f}=V_{i}+gt
V_{f}=31.21+(32)(2.47)
V_{f}=110.25 ft/s

We can conclude that the speed of the object when it hits the ground is 110.25 ft/s


5 0
3 years ago
An ice sled powered by a rocket engine starts from rest on a large frozen lake and accelerates at +44 ft/s2. After some time t1,
mash [69]

Answer:

a) t₁ = 4.76 s, t₂ = 85.2 s

b) v = 209 ft/s

Explanation:

Constant acceleration equations:

x = x₀ + v₀ t + ½ at²

v = at + v₀

where x is final position,

x₀ is initial position,

v₀ is initial velocity,

a is acceleration,

and t is time.

When the engine is on and the sled is accelerating:

x₀ = 0 ft

v₀ = 0 ft/s

a = 44 ft/s²

t = t₁

So:

x = 22 t₁²

v = 44 t₁

When the engine is off and the sled is coasting:

x = 18350 ft

x₀ = 22 t₁²

v₀ = 44 t₁

a = 0 ft/s²

t = t₂

So:

18350 = 22 t₁² + (44 t₁) t₂

Given that t₁ + t₂ = 90:

18350 = 22 t₁² + (44 t₁) (90 − t₁)

Now we can solve for t₁:

18350 = 22 t₁² + 3960 t₁ − 44 t₁²

18350 = 3960 t₁ − 22 t₁²

9175 = 1980 t₁ − 11 t₁²

11 t₁² − 1980 t₁ + 9175 = 0

Using quadratic formula:

t₁ = [ 1980 ± √(1980² - 4(11)(9175)) ] / 22

t₁ = 4.76, 175

Since t₁ can't be greater than 90, t₁ = 4.76 s.

Therefore, t₂ = 85.2 s.

And v = 44 t₁ = 209 ft/s.

3 0
3 years ago
The momentum of a 0.1 kg object traveling at 2000 m/s is 20,000 kg·m/s. True or False
Alina [70]

That's false.  

The definition of momentum is (mass) x (speed), so they must be multiplied.

"20,000 kg-m/s" has the correct units resulting from multiplication, but the number could only be the result of division.

3 0
3 years ago
Which statement about two objects having the same momentum is true?
Reil [10]

Answer:

The product of mass X velocity is the same for both

Explanation:

The momentum of a body is the product of its mass and velocity.

  Momentum  = mass x velocity

Two objects having equal momentum will have the same for the product of their mass and velocity.

  • It is amount of motion a body can produce.
7 0
3 years ago
39. Draw a complete free body diagram of a 40 kg plastic crate at rest on a wooden table (us=0.7). The applied force to the righ
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In order to draw the free body diagram, first let's calculate the friction force acting on the crate:

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Since the friction force is greater than the force applied, the crate will not move, and the friction force will be equal to the force applied.

The weight force is equal to 40 * 9.8 = 392 N.

So, drawing the diagram, we have:

4 0
1 year ago
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