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Alla [95]
3 years ago
15

What is required foe electricity to flow

Physics
1 answer:
lukranit [14]3 years ago
5 0
Conductivity is required for the electric current to flow.
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PLEASE HELP WITH ONE QUESTION!
nata0808 [166]
I think it is -3.99 x 102 j
6 0
3 years ago
30) A force produces power P by doing work W in a time T. What power will be produced by a force that does six times as much wor
schepotkina [342]

Answer:

A) 12P

Explanation:

The power produced by a force is given by the equation

P=\frac{W}{T}

where

W is the work done by the force

T is the time in which the work is done

At the beginning in this problem, we have:

W = work done by the force

T = time taken

So the power produced is

P=\frac{W}{T}

Later, the force does six times more work, so the work done now is

W'=6W

And this work is done in half the time, so the new time is

T'=\frac{T}{2}

Substituting into the equation of the power, we find the new power produced:

P'=\frac{W'}{T'}=\frac{6W}{T/2}=12\frac{W}{T}=12P

So, 12 times more power.

4 0
3 years ago
A boy throws a ball vertically up it returns the ground after 10 seconds find the maximum height reached by the ball
Akimi4 [234]

Answer:

Approximately 122.625\; {\rm m} (assuming that g = 9.81\; {\rm m\cdot s^{-2}}, the ball was launched from ground level, and that the drag on the ball is negligible.)

Explanation:

Let v_{0} denote the velocity at which the ball was thrown upward.

If the drag (air friction) on the ball is negligible, the ball would land with a velocity of exactly (-v_{0}). The velocity of the ball would be changed from v to (-v_{0})\! (such that \Delta v = (-v_{0}) - v_{0} = (-2\, v_{0})) within t = 10\; {\rm s}.

Also because the drag on the ball is negligible, the acceleration of the ball would be a = -g = -9.81\; {\rm m\cdot s^{-2}}. Thus:

\Delta v = a\, t = 10\; {\rm s} \times (-9.81\; {\rm m\cdot s^{-2}}) = -98.1\; {\rm m\cdot s^{-1}}.

Since \Delta v = (-2\, v_{0}):

-2\, v_{0} = \Delta v = -98.1\; {\rm m\cdot s^{-1}.

\begin{aligned}v_{0} &= \frac{-98.1\; {\rm m\cdot s^{-1}}}{-2}= 49.05\; {\rm m \cdot s^{-1}}\end{aligned}.

The ball reaches maximum height when its velocity is v_{1} = 0\; {\rm m\cdot s^{-1}}. Apply the SUVAT equation x = ({v_{1}}^{2} - {v_{0}}^{2}) / (2\, a) to find the displacement x between the original position (ground level, where v_{0} = 49.05\; {\rm m\cdot s^{-1}}) and the max-height position of the ball (where v_{1} = 0\; {\rm m\cdot s^{-1}}.)

\begin{aligned}x &= \frac{(0\; {\rm m\cdot s^{-1}})^{2} - (49.05\; {\rm m\cdot s^{-1}})^{2}}{2 \times (-9.81\; {\rm m\cdot s^{-2}})} \\ &\approx 122.625\; {\rm m\cdot s^{-1}}\end{aligned}.

7 0
2 years ago
What do you need to measure speed?
Furkat [3]

Answer:

The equation for speed is : distance divided by time.

Hoped I helped-

Sleepy~

7 0
3 years ago
A mass m0 is attached to a spring and hung vertically. The mass is raised a short distance in the vertical direction and release
iragen [17]

Answer:

The frequency of the oscillations in terms of fo will be f2=fo/3

E xplanation:

T= 2pie\frac sqrt {m}{k}

 \frac {{f2}/times {fo}}=1:3

⇒f2=fo\3

Here frequency f is inversely poportional to square root of mass m.

so the value of remainder of frequency f2 and fo is equal to 1:3.

⇒\frac{f2} {f1} = \frac sqrt{m1}[m2}

⇒\frac{f2}{fo} = 1:3

⇒f2=\frac{fo} {3}

6 0
3 years ago
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