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bogdanovich [222]
4 years ago
11

An object moving with a speed of 21 m/s and has a kinetic energy of 140 J, what is the mass of the object.

Physics
2 answers:
alexira [117]4 years ago
7 0

Answer:2940

Explanation:

Masteriza [31]4 years ago
7 0

Answer:

0.635 kg

Explanation:

Recall that Kinetic energy is defined as:

K.E. = (1/2) mv²,

where

K.E = Kinetic energy = given as 140J

m = mass (we are asked to find this)

v = velocity = given as 21 m/s

Substituting the above values into the equation:

K.E. = (1/2) mv²

140 = (1/2) m (21)²   (rearranging)

m = (140)(2) / (21)²

m = 0.635 kg

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A heavy flywheel is accelerated (rotationally) by a motor that provides constant torque and therefore a constant angular acceler
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Answer:

a)t_1=\frac{w_1-w_o}{\alpha}=\frac{w_1}{\alpha}sec

b)\theta_1=\frac{w_1^2}{2\alpha}rad

c)t_2=\frac{\alpha t_1}{5\alpha}=\frac{t_1}{5}sec

Explanation:

1) Basic concepts

Angular displacement is defined as the angle changed by an object. The units are rad/s.

Angular velocity is defined as the rate of change of angular displacement respect to the change of time, given by this formula:

w=\frac{\Delat \theta}{\Delta t}

Angular acceleration is the rate of change of the angular velocity respect to the time

\alpha=\frac{dw}{dt}

2) Part a

We can define some notation

w_o=0\frac{rad}{s},represent the initial angular velocity of the wheel

w_1=?\frac{rad}{s}, represent the final angular velocity of the wheel

\alpha, represent the angular acceleration of the flywheel

t_1 time taken in order to reach the final angular velocity

So we can apply this formula from kinematics:

w_1=w_o +\alpha t_1

And solving for t1 we got:

t_1=\frac{w_1-w_o}{\alpha}=\frac{w_1}{\alpha}sec

3) Part b

We can use other formula from kinematics in order to find the angular displacement, on this case the following:

\Delta \theta=wt+\frac{1}{2}\alpha t^2

Replacing the values for our case we got:

\Delta \theta=w_o t+\frac{1}{2}\alpha t_1^2

And we can replace t_1from the result for part a, like this:

\theta_1-\theta_o=w_o t+\frac{1}{2}\alpha (\frac{w_1}{\alpha})^2

Since \theta_o=0 and w_o=0 then we have:

\theta_1=\frac{1}{2}\alpha \frac{w_1^2}{\alpha^2}

And simplifying:

\theta_1=\frac{w_1^2}{2\alpha}rad

4) Part c

For this case we can assume that the angular acceleration in order to stop applied on the wheel is \alpha_1 =-5\alpha \frac{rad}{s}

We have an initial angular velocity w_1, and since at the end stops we have that w_2 =0

Assuming that t_2 represent the time in order to stop the wheel, we cna use the following formula

w_2 =w_1 +\alpha_1 t_2

Since w_2=0 if we solve for t_2 we got

t_2=\frac{0-w_1}{\alpha_1}=\frac{-w_1}{-5\alpha}

And from part a) we can see that w_1=\alpha t_1, and replacing into the last equation we got:

t_2=\frac{\alpha t_1}{5\alpha}=\frac{t_1}{5}sec

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3 years ago
The step-down transformer gives a current of 10A at 24v if the primary voltage is 240v calculate the
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You probably do this as a DC circuit which is not quite correct, but it will get you an answer. The study is a great deal more complicated.

Problem One: Secondary Power.

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Here's the thing you have to know. These transformers are 100% efficient (or are assumed so). So whatever wattage is in the secondary, it is the same as that in the primary.

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Secondary Power = 240 Watts

Primary Power = 240 Watts

W = E * I

E = 240 volts

W = 240 watts

I = W/E = 240 / 240

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Part Two

Answered above. 240 watts.

Part Three

Answered above. 240 watts.

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