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bekas [8.4K]
3 years ago
11

Suppose you toss a tennis ball upward.a) Does the kinetic energy of the ball increase or decrease as it moves higher?b) What hap

pens to the potential energy of the ball as it moves higher?c) If the same amount of energy were imparted to a ball the same size as a tennis ball but of twicethe mass, how high would it go in comparison to the tennis ball?
Physics
1 answer:
Lilit [14]3 years ago
7 0

Answer:

a) No. The kinetic energy of the ball decreases.

b) The potential energy of the ball increases.

c) The ball would go half of the original distance.

Explanation:

a) The kinetic energy would be converted to potential energy as the ball goes higher. Since the total mechanical energy is conserved, the kinetic energy would decrease.

b) The potential energy of the ball would increase. Since the total mechanical energy of the ball is conserved, the ball would lose speed, and therefore kinetic energy. In order to compensate the loss of kinetic energy, the ball would gain potential energy as it goes higher.

c) The relation of the energy and mass is as follows:

K = \frac{1}{2}mv^2\\U = mgh

According to the energy conservation

K_1 + U_1 = K_2 + U_2\\\frac{1}{2}mv^2 + 0 = 0 + mgh\\\frac{1}{2}v^2 = gh

The maximum height that the ball reaches is proportional to the initial velocity. If the ball would be imparted with the same amount of energy, its final potential energy would be the same. However, in order to have the same potential energy (mgh), its height would be half of the original case.

mgh = (2m)g(\frac{h}{2})

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In this problem, you will practice applying this formula to several situations involving angular acceleration. In all of these s
riadik2000 [5.3K]

Answer:

Part a)

\alpha = \frac{2(m_1 - m_2)g}{(m_1 + m_2)L}

Part b)

\alpha = \frac{6(m1 - m_2)g}{3(m_1 + m_2)L + m_{bar}L}

Explanation:

As we know that the see saw bar is massless so here torque due to two masses is given as

\tau = I\alpha

here we will have

\tau = (m_1g - m_2g)(\frac{L}{2})

now we will have inertia of two masses given as

I = (m_1 + m_2)(\frac{L}{2})^2

now we have

I = (m_1 + m_2)\frac{L^2}{4}

now the angular acceleration is given as

\alpha = \frac{\tau}{I}

so we have

\alpha = \frac{2(m_1 - m_2)g}{(m_1 + m_2)L}

Part b)

Now if the rod is not massles then we will have total inertia given as

I = (m_1 + m_2)(\frac{L}{2})^2 + \frac{m_{bar}L^2}{12}

so we will have

I = (m_1 + m_2)\frac{L^2}{4} + \frac{m_{bar}L^2}{12}

now the acceleration is given as

\alpha = \frac{\tau}{I}

\alpha = \frac{6(m1 - m_2)g}{3(m_1 + m_2)L + m_{bar}L}

7 0
3 years ago
A wheelbarrow is pushed with a force of 40 N. If 6,000 J of work is
stepladder [879]

Answer:

Distance = 150 meters

Explanation:

Given the following data;

Work done = 6,000 Joules

Force = 40 Newton

To find the total distance covered by the wheelbarrow;

Workdone = force * distance

Substituting into the formula, we have;

6000 = 40 * distance

Distance = 6000/40

Distance = 150 meters

Therefore, the total distance the wheelbarrow was pushed is 150 meters.

5 0
3 years ago
An important diagnostic tool for heart disease is the pressure difference between blood pressure in the heart and in the aorta l
butalik [34]

Answer:

a)   f ’’ = f₀ \frac{1 + \frac{v}{c} }{1- \frac{v}{c} } , b)   Δf = 2 f₀ \frac{v}{c}

Explanation:

a) This is a Doppler effect exercise, which we must solve in two parts in the first the emitter is fixed and in the second when the sound is reflected the emitter is mobile.

Let's look for the frequency (f ’) that the mobile aorta receives, the blood is leaving the aorta or is moving towards the source

                    f ’= fo\frac{c+v}{c}

This sound wave is reflected by the blood that becomes the emitter, mobile and the receiver is fixed.

                   f ’’ = f’ \frac{c}{ c-v}

where c represents the sound velocity in stationary blood

therefore the received frequency is

                 f ’’ = f₀   \frac{c}{c-v}

let's simplify the expression

                f ’’ = f₀ \frac{c+v}{c-v}

                f ’’ = f₀ \frac{1 + \frac{v}{c} }{1- \frac{v}{c} }

         

b) At the low speed limit v <c, we can expand the quantity

                 (1 -x)ⁿ = 1 - x + n (n-1) x² + ...

                 ( 1- \frac{v}{c} ) ^{-1} = 1 + \frac{v}{c}

 

                f ’’ = fo ( 1+ \frac{v}{c}) ( 1 + \frac{v}{c} )

                f ’’ = fo ( 1 + 2 \frac{v}{c} + \frac{v^2}{ c^2} )

leave the linear term

               f ’’ = f₀ + f₀ 2\frac{v}{c}

the sound difference

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The complete observation about adding bulb 3 is the brightness of the bulbs has to do with power which considers both the voltage and the current: less voltage x less current = dimmer bulbs.  In circuit A, the voltage is divided across the resistors and the current decreases as resistance increases.  In circuit B, the voltage is the same in each parallel section of the circuit and the current through that section of the circuit only depends on the resistor in that section.

<h3>What is power of the circuit?</h3>

The power of the bulb or any resistor is equal to the product of voltage and  current flowing through it.

P = VI

Circuit A has bulbs in series while the circuit B has bulbs in parallel.

When bulb 3 added to circuit A,  the brightness of all the bulbs dimmed but when bulb 3 (R3) added to circuit B, nothing changed in the brightness of the bulb.

The brightness is depended on the power of the circuit. When both the voltage and current are less, the bulb will be dimmed. In circuit A, series resistors divide the voltage across them. In circuit B, voltage is equal for all the resistors.

Thus, the last option is correct.

Learn more about power.

brainly.com/question/2933971

#SPJ1

4 0
2 years ago
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