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Sati [7]
3 years ago
12

A ball is projected upward at time t = 0.0 s, from a point on a roof 70 m above the ground. The ball rises, then falls and strik

es the ground. The initial velocity of the ball is What is the velocity of the ball when it is above the ground? −89 m/s − 72 m/s − 36 m/s − 107 m/s − 54 m/s
Physics
1 answer:
pentagon [3]3 years ago
8 0

Answer:

The velocity of a ball will be "-70.13 m/s".

Explanation:

The given values are:

u = 70 m

t = 0.0 s

g = a = -9.8 m/s²

s = -1 m

v = ?

As we know,

The equation of motion will be:

⇒  v^2-u^2=2as

On substituting the estimated values, we get

⇒  v^2-(70)^2=2\times (-9.8)\times (-1)

⇒  v^2-4900=19.6

⇒  v^2=19.6+4900

⇒  v^2=4919.6

⇒  v=\sqrt{4919.6}

⇒  v=70.13 \ m/s

In downward direction, it will be:

⇒  v=-70.13 \ m/s

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To solve this problem we will begin by finding the pressure through density and average depth. Later we will find the Force, by means of the relation of the pressure and the area.

P = \rho h

Here,

h = Depth average

\rho = Density

Moreover,

\text{Density of water}= \rho = 62.4lb/ft^3

Replacing,

P = (62.4lb/ft^3)(\frac{35}{2}ft)

P = 1092 lb/ft^2

Finally the force

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F = 1092lb/ft^2(101ft*52ft)

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6 0
3 years ago
A reversible reaction is in
natita [175]

Answer:

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Explanation:

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2 years ago
Compare and contrast the geocentric view of the solar system with the heliocentric view of the solar system
Mandarinka [93]
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3 years ago
Please help me with this very easy question I just don’t get it.
diamong [38]

Answer:

150m

Explanation:

The relation of speed/time and distance/time is a derivative/integral one, as in speed is the derivative of distance (the faster you go, the faster the distance changes, duh!).

So we need to compute the integral of speed over time from 0.0s to 5.0s.

The easiest way here is to compute the area under the line (it's going to be faster than computing the acceleration and using a formula of distance based on acceleration).

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Alternatively, we can use the acceleration formula:

a = (50m/s - 10m/s)/5s = 40m/s / 5s = 8m/s^2

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5 0
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b.) 1 bar = 29.53 inHg

0.92 mb(1 bar/1000 mbar)(29.53 inHg/1 bar) =<em> 0.027 inHg</em>
3 0
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