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svetoff [14.1K]
2 years ago
6

A 1.0-kg ball has a velocity of 12 m/s downward just before it strikes the ground and bounces up with a velocity of 12 m/s upwar

d. What is the change in momentum of the ball?
Physics
1 answer:
Nezavi [6.7K]2 years ago
3 0

Answer:

The change in momentum of the ball is 24 kg-m/s  

Explanation:

It is given that,

Mass of the ball, m = 1 kg

Initial velocity of the ball, u = -12 m/s (in downwards)

Final velocity of the ball, v = +12 m/s (in upward)

We need to find the change in momentum of the ball.

Initial momentum of the ball, p_i=mu=1\ kg\times (-12\ m/s)=-12\ kg-m/s

Final momentum of the ball, p_f=mv=1\ kg\times (12\ m/s)=12\ kg-m/s

Change in momentum of the ball, \Delta p=p_f-p_i

\Delta p=12-(-12)=24\ kg-m/s

So, the change in momentum of the ball is 24 kg-m/s. Hence, this is the required solution.

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A 0.0500-kg golf ball heads off from the tee with an initial speed of 78.2 m/s and reaches to its maximum height of 37.8 m. If a
Marianna [84]

(a) 134.4 J

The total mechanical energy of the ball at the moment of launch is just equal to its initial kinetic energy:

E=K_i = \frac{1}{2}mu^2

where

m = 0.05 kg is the mass of the ball

u = 78.2 m/s is the initial speed

Substituting,

E=\frac{1}{2}(0.05)(78.2)^2=152.9 J

At the pinnacle of its trajectory, the total mechanical energy is sum of kinetic energy and potential energy:

E=K_f + U_f = K_f + mgh

where

g=9.8 m/s^2 is the acceleration of gravity

h = 37.8 m is the maximum height

Since the total energy must be conserved,

E = 152.9 J

Therefore, we can solve to find the kinetic energy of the ball at the pinnacle:

K_f = E-mgh=152.9-(0.05)(9.8)(37.8)=134.4 J

b) 74.2 m/s

When the ball is 5.60 m below the pinnacle point, the heigth of the ball is

h=37.8-5.6=32.2 m

So its potential energy is

U=mgh=(0.05)(9.8)(32.2)=15.8 J

The total energy is again the sum of potential and kinetic energy:

E = K + U

So the kinetic energy at that point is

K=E-U=152.9-15.8=137.1 J

And since the kinetic energy is

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

We can find the speed:

v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(137.1)}{0.05}}=74.2 m/s

3 0
2 years ago
A machine is currently set to a feed rate of 5.921 inches per minute (IPM). Te machinist changes this setting to 6.088 IPM. By h
lukranit [14]

Answer:

By 16.7% or 0.167 IPM

Explanation:

Substracting the final IPM (6.088) to the initial IPM (5.921) gives us the net difference, which is how much did it increase in IPM. Multiplying this number by 100 gives us the percentual increase in the feed rate.

4 0
2 years ago
Read 2 more answers
1) A fan is to accelerate quiescent air to a velocity of 8 m/s at a rate of 9 m3/s. Determine the minimum power that must be sup
azamat

Answer:

\dot{W} = 339.84 W

Explanation:

given data:

flow Q = 9 m^{3}/s

velocity = 8 m/s

density of air = 1.18 kg/m^{3}

minimum power required to supplied to the fan is equal to the POWER POTENTIAL of the kinetic energy and it is given as

\dot{W} =\dot{m}\frac{V^{2}}{2}

here \dot{m}is mass flow rate and given as

\dot{m} = \rho*Q

\dot{W} =\rho*Q\frac{V^{2}}{2}

Putting all value to get minimum power

\dot{W} =1.18*9*\frac{8^{2}}{2}

\dot{W} = 339.84 W

7 0
2 years ago
In a bedridden patient recovering from a badly fractured femur, disuse atrophy in the thigh muscles is caused by _________.
Law Incorporation [45]
In a bedridden patient recovering from a badly fractured femur, disuse atrophy in the thigh muscles is caused by <span>decreased synthesis of muscle proteins and/or increased breakdown of muscle proteins. Muscle tissue grows and heals in response to stress. Without the stress of exercise and normal daily activities, muscle tissue degenerates</span>
3 0
2 years ago
Look at the two question marks between zinc (Zn) and arsenic (As). At the time, no elements were known
marissa [1.9K]

Answer:

Mendeleev predicted the atomic mass of each element along with compounds they each should form.

Explanation:

Based on other elements in the same group he predicted the existence of eka-aluminum and eka-silicon, later to be named gallium (Ga) and germanium (Ge).

6 0
2 years ago
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