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MissTica
3 years ago
6

Speed and Motion you went from the starting line to the finish line at different rates. If you repeated the activity while carry

ing weights but keeping your times the same, how would your kinetic energy be affected?
Physics
1 answer:
Phantasy [73]3 years ago
3 0

Answer:

It will cause kinetic energy to increase.

Explanation:

Given that Speed and Motion you went from the starting line to the finish line at different rates.

If you repeated the activity while carrying weights but keeping your times the same, the weight carried will add up to the mass of the body.

And since Kinetic energy K.E = 1/2mv^2

Increase in the mass of the body will definitely make the kinetic energy of the body to increase.

Since the time is the same, that means the speed V is the same.

Weight W = mg

m = W/g

The new kinetic energy will be:

K.E = 1/2(M + m)v^2

This means that there will be increase in kinetic energy.

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A wheel of radius 0.23 m, which is moving initially at 25.0 m/s, rolls to a stop in 246.0 m. The wheel's rotational inertia is 0
Digiron [165]

Answer: -1.27 m/s^2

Explanation:

a = - V^2 / 2x

a = -(25^2) / 2 x (246) = 1.27 m/ s^2

Therefore the linear acceleration of the wheel is - 1.27 m/s^2

8 0
3 years ago
A 75-hp (shaft output) motor that has an efficiency of 91% is worn out and is replaced by a high efficiency 75-hp motor that has
ZanzabumX [31]

Answer:

The heat gain of the room due to higher efficiency is 2.84 kW.

Explanation:

Given that,

Output power of shaft = 75 hp

Efficiency = 91%

High efficiency = 95.4%

We need to calculate the electric input given to motor

Using formula of efficiency

\eta=\dfrac{Work\ Output}{Work\ input}

Work\ Input =\dfrac{Work\ output}{\eta}

Work\ Input =\dfrac{75\times746}{0.91}

Work\ Input =61483.51\ W

Work\ Input = 61.48 kW

We need to calculate the electric input

For, heigh efficiency

Work\ Input_{inc} =\dfrac{75\times746}{0.954}

Work\ Input_{int} =58647.7\ W

Work\ Input_{int} = 58.64\ kW

The reduction of the heat gain of the room due to higher efficiency is

Q=Work\ Input-Work\ Input_{int}

Put the value into the formula

Q=61.48 -58.64

Q=2.84\ kW

Hence, The heat gain of the room due to higher efficiency is 2.84 kW.

3 0
3 years ago
A small, spring-loaded cannon launches a tennis ball from level ground with an initial speed vi at an angle θi with the horizont
lora16 [44]

Answer:

a. T\ =\ \dfrac{2v_isin\theta_i}{g}

b. v_icos\theta_i

c. v_1sin\theta_i

Explanation:

Given,

*initial velocity of the ball = v_i

*angle of projection = \theta_i

Horizontal component of the initial velocity of the ball = v_x\ =\ v_icos\theta_i

vertical initial component of the initial velocity of the ball = v_y\ =\ v_isin\theta_i

part a

From the kinematics,

In the y-direction motion,

total vertical displacement of the ball during the whole motion is zero.

Ball is moving under the gravitational acceleration, therefore the acceleration of the ball = -g, because gravitational acceleration always acts in the downward direction,

Let t be the time of flight of the whole motion,

y\ =\ v_yt\ -\ \dfrac{1}{2}gt^2\\\Rightarrow 0\ =\ v_isin\theta_i t\ -\ \dfrac{1}{2}gt^2\\\Rightarrow t\ =\ \dfrac{2v_isin\theta_i}{g}\\

part b.

At the peak of the path of the ball, the vertical component of the velocity of the ball becomes zero, only horizontal component of the velocity acts on the ball is equal to = v_x\ =\ v_icos\theta_i

part c.

Initial vertical component of the velocity of the ball = v_y\ =\ v_isin\theta_i

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C. Magnetism

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