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2 years ago

A ball thrown horizontally at 12.6 m/s from the roof of a building lands 20.0 m from the base of the building

Physics

1 answer:

S_A_V [24]2 years ago

7 0

Answer:

1.59 seconds

12.3 meters

but if you are wise you will read the entire answer.

Explanation:

This is a good question -- if not a bit unusual. You should try and understand the details. It will come in handy.

Time

Given

a = 0 This is the critical point. There is no horizontal acceleration.

d = 20 m

v = 12.6 m/s

Formula

d = vi * t + 1/2at^2

Solution

Since the acceleration is 0, the formula reduces to

d = vi * t

20 = 12.6 * t

t = 20 / 12.6

t = 1.59 seconds.

It takes 1.59 seconds to hit the ground

Height of the building

Givens

t = 1.59 sec

vi = 0 Another critical point. The beginning speed vertically is 0

a = 9.8 m/s^2 The acceleration is vertical.

Formula

d = vi*t + 1/2 a t^2

Solution

d = 1/2 a*t^2

d = 1/2 * 9.8 * 1.59^2

d = 12.3 meters.

The two vi's are not to be confused. The horizontal vi is a number other other 0 (in this case 12.6 m/s horizontally)

The other vi is a vertical speed. It is 0.

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An orbiting satellite can become charged by the photoelectric effect when sunlight ejects electrons from its outer surface. Sate

Rufina [12.5K]

Answer:

the longest wavelength of incident sunlight that can eject an electron from the platinum is 233 nm

Explanation:

Given data

Φ = 5.32 eV

to find out

the longest wavelength

solution

we know that

hf = k(maximum) +Ф ...............1

here we consider k(maximum ) will be zero because photon wavelength max when low photon energy

so hf = 0

and hc/ λ = +Ф

so λ = hc/Ф ................2

now put value hc = 1240 ev nm and Φ = 5.32 eV

so hc = 1240 / 5.32

hc = 233 nm

the longest wavelength of incident sunlight that can eject an electron from the platinum is 233 nm

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2 years ago

What does keplers second law of planetary monition imply

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Hitman42 [59]

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2 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:

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