Calculate the amount of work done to draw a current of 8A from a point at 100V to a point at 120V in 2 seconds?

A disk rotates around an axis through its center that is perpendicular to the plane of the disk. The disk has a line drawn on it

natka813 [3]

Answer:

$t = \frac{\sqrt{\omega0^2+2*\theta*\alpha} -\omega0}{\alpha}$

Explanation:

The rotated angle is given by:

$\theta=\omega0*t+1/2*\alpha*t^2$

Since this is a quadratic equation it can be solved using:

$x=\frac{-b \± \sqrt{b^2-4*a*c} }{2*a}$

Rewriting our equation:

$1/2*\alpha*t^2+\omega0*t-\theta=0$

$t = \frac{\±\sqrt{\omega0^2+2*\theta*\alpha} -\omega0}{\alpha}$

Since $\sqrt{\omega0^2+2*\theta*\alpha} >\omega0$ we discard the negative solution.

$t = \frac{\sqrt{\omega0^2+2*\theta*\alpha} -\omega0}{\alpha}$

8 0

2 years ago

A gymnast dismounts the uneven parallel bars with some angular momentum about her transverse axis. Just after release, she is in

Luda [366]

Answer:

a. Her moment of inertia increases and she rotates slower.

Explanation:

As we know that initially when she starts her motion she is in piked position due to which her whole mass is concentrated near the axis of rotation

So here the rotational Inertia of her body will be smaller

Now when is comes closer to the position of landing she extends into layout position due to which her mass will move away from the axis of rotation

Due to this the rotational inertia of her body will increase

now we know that there is no external torque on the system

so here angular momentum must be conserved

So we will have

$I\omega = constant$

so if rotational inertia is increasing then angular speed must be slower

so correct answer will be

a. Her moment of inertia increases and she rotates slower.

7 0

2 years ago

Every action force has an opposite and equal reaction force. Determine which of these are action/reaction pairs. Check all that

denis23 [38]

A girl leans against a wall and the wall pushes on the girl.

A baseball hits a glove and the glove pushes on the ball.

A cat rubs up against a tree and the tree pushes against the cat

3 0

2 years ago

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During the 440, a runner changes his speed as he comes out of the curve onto the home stretch from 18 ft/sec to 38 ft/sec over a

Sloan [31]

Answer:

$6.67ft/s^2$

Explanation:

We are given that

Initial velocity=u=18ft/s

Final velocity,v=38ft/s

Time=t=3 s

We have to find the average acceleration over that 3 s period.

We know that

Average acceleration,a= $\frac{v-u}{t}{t}$

Using the formula

Average acceleration,a= $\frac{38-18}{3}ft/s^2$

Average acceleration,a= $\frac{20}{3}ft/s^2$

Average acceleration,a= $6.67ft/s^2$

Hence, the average acceleration= $6.67ft/s^2$

5 0

2 years ago

Physics Help!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

djverab [1.8K]

Good work on solving part a).
b) may look complicated, but it's not too bad.

It says that the body is 25% efficient in converting fat to mechanical energy.
In other words, only 25% of the energy we get from our stored fat shows up
in the physical, mechanical moving around that we do. (The rest becomes
heat, which dissipates into the environment as we keep our bodies warm,
breathe hot air out,and perspire.)

You already know how much mechanical energy the climber needed to lift
himself to the top of the mountain... 2.4x10⁶ joules.
That's 25% of what he needs to convert in order to accomplish the climb.
He needs to pull 4 times as much energy out of fat.

-- Fat energy required = 4 x (2.4 x 10⁶) = 9.6 x 10⁶ joules.

-- Amount stored in 1kg of fat = 3.8 x 10⁷ joules

-- Portion of a kilogram he needs to use = (9.6 x 10⁶) / (3.8 x 10⁷)

Note:
That much of a kilogram weighs about 8.9 ounces ... which shows why it's so
hard to lose weight with physical exercise alone. It also helps you appreciate
that fat is much more efficient at storing energy than batteries are ... that one
kilogram of fat stores the amount of energy used by a 100-watt light bulb, to
burn for 105 hours (more than 4-1/2 days ! ! !)

5 0

2 years ago

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