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masya89 [10]
3 years ago
6

What happens to electron flow with a conductor of the voltage source is removed?

Physics
1 answer:
9966 [12]3 years ago
4 0
The electrons stop flowing
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mary takes 6.0 seconds to run up a flight of stairs thaf is 102 meters long. if mary wieght is 87 newtons, what power has mary e
blondinia [14]
It really doesn't matter how long the flight of stairs is. What we really need to know is how much height Mary gained, straight up. We don't know that. If we knew the height then the total amount of work she did would be 87 x the height in meters. The unit is joules. Then the power she delivered would be that number divided by the six seconds and that unit is watts. By the way ... it's troubling that Mary's weight is only about 20 pounds. It's doubtful that such a tiny creature could run up a flight of stairs.
7 0
3 years ago
A dog runs after a car. The car is travelling at an average speed of 5 m/s.
n200080 [17]

Answer:

<h2>The answer is 4 s</h2>

Explanation:

The time taken can be found by using the formula

t =  \frac{d}{v}  \\

where

d is the distance

v is the velocity

From the question we have

t =  \frac{20}{5}  \\

We have the final answer as

<h3>4 s</h3>

Hope this helps you

3 0
3 years ago
Read 2 more answers
220V and 5A is supplied to the primary coil. The turns ratio is 500. What is the power on the secondary coil?
alisha [4.7K]

Answer: 1100 W

Explanation:

Input power  = 220(5) = 1100 W

The transformer will step up/down voltage, but will also step down/up current.

Neglecting hysteresis and other minor losses, the power will remain the same.

4 0
2 years ago
Both formal and informal
Tema [17]

Explanation:

is this a question????...

5 0
3 years ago
A rocket moves upward, starting from rest with an acceleration of 25.4 m/s^2 for 3.39 s. It runs out of fuel at the end of the 3
kolbaska11 [484]

Explanation:

Initial speed of the rocket, u = 0

Acceleration of the rocket, a=25.4\ m/s^2

Time taken, t = 3.39 s

Let v is the final velocity of the rocket when it runs out of fuels. Using the equation of kinematics as :

v=u+at

v=25.4\times 3.39=86.10\ m/s    

Let x is the initial position of the rocket. Using third equation of kinematics as :

v^2=u^2+2ax_o

x_o=\dfrac{v^2}{2a}

x_o=\dfrac{86.10^2}{2\times 25.4}=145.92\ m  

Let x_o is the position at the maximum height. Again using equation of motion as :

v^2-u^2=2a(x-x_o)

Now a=-g and v and u will interchange

u^2=2g(x-x_o)

x=x_o+\dfrac{u^2}{2g}

x=145.92+\dfrac{(86.10)^2}{2\times 9.8}

x = 524.14 meters

Hence, this is the required solution.

5 0
3 years ago
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