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raketka [301]
2 years ago
15

If Steve throws the football 50 meters in 3 seconds, what is the average speed of the football? Speed = distance / time Question

2 options: A. 16.67 m/s B. 26 m/s C. 5 m/s D. 48.9 m/s
Physics
1 answer:
leva [86]2 years ago
5 0

Answer:

A. 16.67 m/s

Explanation:

Speed or velocity refers to the rate of change in distance over a change in time. That is;

Speed = Distance ÷ time

Where;

Speed is in metre/seconds

Distance is in metre

Time is in seconds.

In this question, Steve throws a football 50 meters in 3 seconds. The average speed can be calculated this:

S = D/t

Where; d = 50m, t = 3s

S = 50/3

S = 16.6666666

S = 16.67m/s

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A swimming pool is 50 ft wide and 100 ft long and its bottom is an inclined plane, the shallow end having a depth of 4 ft and th
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Explanation:

We define force as the product of mass and acceleration.

F = ma

It means that the object has zero net force when it is in rest state or it when it has no acceleration. However in the case of liquids. just like the above mentioned case, the water is at rest but it is still exerting a pressure on the walls of the swimming pool. That pressure exerted by the liquids in their rest state is known as hydro static force.

Given Data:

Width of the pool = w = 50 ft

length of the pool = l= 100 ft

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Depth of the deep end = h(d) = 10 ft.

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Solution:

a) Force on a shallow end:

F = \frac{pgwh}{2} (2x_{1}+h)

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b) Force on deep end:

F = \frac{pgwh}{2} (2x_{1}+h)

F = \frac{(62.5)(50)(10)}{2} (2(0)+10)

F = 187500 lb

c) Force on one of the sides:

As it is mentioned in the question that the bottom of the swimming pool is an inclined plane so sum of the forces on the rectangular part and triangular part will give us the force on one of the sides of the pool.

1) Force on the Rectangular part:

F = \frac{pg(l.h)}{2}(2(x_{1} )+ h)

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F = \frac{(62.5)(100)(2)}{2}(2(0)+4)

F =25000lb

2) Force on the triangular part:

F = \frac{pg(l.h)}{6} (3x_{1} +2h)

here

h = h(d) - h(s)

h = 10-4

h = 6ft

x_{1} = 4ft\\

F = \frac{62.5 (100)(6)}{6} (3(4)+2(6))

F = 150000 lb

now add both of these forces,

F = 25000lb + 150000lb

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d) Force on the bottom:

F = \frac{pgw\sqrt{l^{2} + ((h_{d}) - h(s)) } (h_{d}+h_{s})   }{2}

F = \frac{62.5(50)\sqrt{100^{2}(10-4) } (10+4) }{2}

F = 2187937.5 lb

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<h2><u>Joule</u><u>:</u></h2>

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1 Newton is the amount of force required to accelerate body of mass 1 kg by 1m/s²

So units of N is kgm/s²

So,

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1 erg is the amount of work done by a force of 1 dyne exerted for a distance of one centimetre.

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<h3><u>what you need to know for conversion</u></h3>

[1gm=0.001kg

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<h2>⇒15 J=15×10000000 Erg</h2><h2> =150000000 Erg</h2><h2> =1.5×10⁶ Erg</h2>
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