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vovangra [49]
3 years ago
12

Q 6: A body throws a ball vertically up. It returns to the ground after 5 seconds. Find

Physics
1 answer:
Rasek [7]3 years ago
8 0

Answer:

Taking gravity to be 9.8m/s2, The velocity is 24.5m/s2.

Taking gravity to be 10m/s2, The velocity is 25m/s2.

Explanation:

According the first formula of motion under the influence of gravity for upward motion, v=u-gt, where v=final velocity, u=initial velocity, and t= time taken.

Here the time taken for the ball to reach the maximum point is half of 5, which is 2.5 seconds.

And v is 0, since at the maximum point gravity slows down the velocity to 0.

Finding the initial velocity,

v=u-gt

0=u-10(2.5)

u=10(2.5)

u=25m/s

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3 years ago
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
Nina [5.8K]

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

Depth of the shallow end = h(s) = 4 ft

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

weight density = ρg = 62.5 lb/ft

Solution:

a) Force on a shallow end:

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

F = \frac{(62.5)(50)(4)}{2}(2(0)+4)

F = 25000 lb

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)

x_{1} = 0\\h_{s} = 4ft

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

F = 175000lb

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

7 0
3 years ago
A truck heading east has an initial velocity of 6 m/s. It accelerates at 2 m/s2 for 12 seconds. What distance does the truck tra
gavmur [86]
I believe it would be 144 m!
7 0
3 years ago
Read 2 more answers
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