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Alexxandr [17]
3 years ago
8

An outfielder throws a baseball with an initial speed of 81.8mi/h. Just before an infielder catches the ball at the same level,

the ball’s speed is 110 ft/s. In foot-pounds, by how much is the mechanical energy of the ball– Earth system reduced because of air drag? (The weight of a baseball is 9.0 oz.)

Physics
1 answer:
agasfer [191]3 years ago
3 0

Answer:

-20.158ft-lb

Explanation:

Check the attached files for the explanation.

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A bowling ball rolls 33\,\text m33m33, start text, m, end text with an average speed of 2.5\,\dfrac{\text m}{\text s}2.5 s m ​ 2
asambeis [7]

The ball rolled for 13.2 s

<h3>Further explanation</h3>

Speed is scalar and no direction

\tt avg~speed=\dfrac{total~distance}{elapsed~time}=\dfrac{\Delta x}{\Delta t}

A bowling ball rolls 33 m, with average speed = 2.5 m/s

So elapsed time :

\tt t=\dfrac{33~m}{2.5`m/s}=13.2~s

4 0
3 years ago
Water runs into a fountain, filling all the pipes, at a steady rate of 0.750 m3&gt;s. (a) How fast will it shoot out of a hole 4
kati45 [8]

Answer:

velocity  = 472 m/s

velocity = 52.4 m/s

Explanation:

given data

steady rate = 0.750 m³/s

diameter = 4.50 cm

solution

we use here flow rate formula that is

flow rate = Area × velocity .............1

0.750 = \frac{\pi }{4} × (4.50×10^{-2})²  × velocity

solve it we get

velocity  = 472 m/s

and

when it 3 time diameter

put valuer in equation 1

0.750 = \frac{\pi }{4} × 3 ×  (4.50×10^{-2})²  × velocity

velocity = 52.4 m/s

5 0
3 years ago
How long will bus take to travel 150 km at an average speed of 40km/hr
romanna [79]

Answer:

3.75 hours

Explanation:

By v = \frac{d}{t}

where v is the velocity (or speed in this case)

d is the distance travelled

t is the time taken

t = \frac{150}{40}=3.75
Therefore it takes 3.75 hours for the bus to travel 150 km and 40 km/hr.

7 0
2 years ago
A boxer hits punching bag and gives it a change in momentum of 12 kg multiplied by m divided by s over 7.0ms what is the magnitu
mr_godi [17]

Answer: 1700

Explanation:

8 0
2 years ago
Derive an expression for the gravitational potential energy U(r) of the object-earth system as a function of the object's distan
Drupady [299]

Answer:

U(r)=-\frac{Gm_Emr^2}{2R^3_E}

Explanation:

We are given that

Gravitational force=F_g=\frac{Gm_Emr}{R^3_E}

r=0,U(0)=0

We know that

Gravitational potential energy=-\int F_gdr

U(r)=-\int\frac{Gm_Emr}{R^3_E}dr

U(r)=-\frac{Gm_Em}{R^3_E}\times \frac{r^2}{2}+C

Substitute r=0 ,U(0)=0

0=0+C

C=0

Substitute the value

U(r)=-\frac{Gm_Emr^2}{2R^3_E}

4 0
3 years ago
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