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taurus [48]
3 years ago
7

A toaster uses 400 W of power. How much does it use in 5 seconds?

Physics
1 answer:
gavmur [86]3 years ago
8 0

Answer:

This one I think is either division or multiplication

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Estimate how much power is saved if the voltage is stepped up from 150 VV to 1500 VV and then down again, rather than simply tra
gtnhenbr [62]

The question is incomplete! The complete question along with answer and explanation is provided below.

Question:

Suppose 79 kW is to arrive at a town over two 0.115 Ω lines. Estimate how much power is saved if the voltage is stepped up from 150 V to 1500 V and then down again, rather than simply transmitting at 150 V. Assume the transformers are each 99%% efficient. Express your ans up to two decimal points.

Answer:

Power saved = 30772.58 W or 30.77 kW

Explanation:

Since the power loss depends upon current, we need to find the current at each voltage level.

Current at 150 V:

I = P/V = 79,000/150 = 526.67 A

Current at 1500 V:

Considering transformer efficiency of 99%

79,000/0.99 = 79797.97 W

I = P/V = 79,797.97/1500 = 53.19 A

Power loss at 150 V:

P = I²R = (526.67)²*0.115 = 31898.84 W

Power loss at 1500 V:

Here we have to consider both of the losses, due to the transmission line and due to transformer itself

P = I²R = (53.19)²*0.115 = 325.35 W

So the total power at the sending end would be

Receiving end power + losses

79,000 + 325.35 = 79,325 W

Power at the input of the transformer is

79,325/0.99 = 80126.26

Power lost in the transformer is

80,126.26 - 79,000 = 1126.26 W

So the power that can be saved by stepping up the voltage from 150 to 1500 V is

Power saved = 31898.84 - 1126.26

Power saved = 30772.58 W

Power saved = 30.77 kW

3 0
3 years ago
Interactive LearningWare 8.1 reviews the approach that is necessary for solving problems such as this one. A motorcyclist is tra
STALIN [3.7K]

Answer:

The angular displacement of each wheel is 269.92 rad

Explanation:

Given:

Angular acceleration \alpha = 6.10 \frac{rad}{s^{2} }

Time to pass cyclist t = 4.36 s

Angular velocity \omega _{f} = 75.2 \frac{rad}{s}

According to the equation of kinematics,

  \omega _{f} = \omega _{i} + \alpha   t

   \omega _{i} = \omega _{f} - \alpha   t

   \omega _{i} = 75.2 - 6.10 \times 4.36

  \omega _{f} = 48.60 \frac{rad}{s}

For finding angular displacement,

    \omega _{f} ^{2}  - \omega _{i} ^{2}  = 2 \alpha  \theta

Where \theta = angular displacement,

  \theta  = \frac{\omega _{f}^{2} - \omega _{i} ^{2}  }{2\alpha }

  \theta  = \frac{5655.04 - 2361.96  }{2\times 6.10 }

  \theta = 269.92 rad

Therefore, the angular displacement of each wheel is 269.92 rad

8 0
2 years ago
Can someone please help, ty!!<br> (Will mark brainliest)
Alexxandr [17]
28 is balanced and unbalanced
3 0
3 years ago
When unpolarized light is incident on a sheet of polarizing material with a transmission axis oriented vertically, what percenta
yulyashka [42]

Answer:

If the light were incident upon two polarizers at right angles, no light would get thru - thus each polarizer must block 50% of the light.

One polarizer would allow 50% of the light to pass.

4 0
2 years ago
What is the acceleration of a proton moving with a speed of 6.5 m/s at right angles to a magnetic field of 1.8 t ?
defon
The magnetic force acting on the proton is 
F=qvB \sin \theta
where
q is the proton charge
v is its speed
B is the intensity of the magnetic field
\theta is the angle between the direction of v and B; since the proton is moving perpendicular to the magnetic field, \theta=90^{\circ} and \sin \theta=1, so the force becomes
F=qvB

this force provides the centripetal force that keeps the proton in circular motion:
m \frac{v^2}{r} = q v B
where the term on the left is the centripetal force, with
m being the mass of the proton
r the radius of its orbit

Re-arranging the previous equation, we can find the radius of the proton's orbit:
r= \frac{mv}{qB}= \frac{(1.67 \cdot 10^{-27} kg)(6.5 m/s)}{(1.6 \cdot 10^{-19} C)(1.8 T)}=3.77 \cdot 10^{-8}m

And now we can calculate the centripetal acceleration of the proton, which is given by
a_c =  \frac{v^2}{r}= \frac{(6.5 m/s)^2}{3.77\cdot 10^{-8}m}=1.12 \cdot 10^9 m/s^2

7 0
3 years ago
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