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cricket20 [7]
3 years ago
8

A ball is thrown straight up at 20 m. What is the balls velocity as it hits the ground?

Physics
1 answer:
deff fn [24]3 years ago
5 0

Answer:

<em>The velocity of the ball as it hit the ground = 19.799 m/s</em>

Explanation:

Velocity: Velocity of a body can be defined as the rate of change of displacement of the body. The S.I unit of velocity is m/s. velocity is expressed in one of newtons equation of motion, and is given below.

v² = u² + 2gs.......................... Equation 1

Where v = the final velocity of the ball, g = acceleration due to gravity, s = the height of the ball

<em>Given: s = 20 m, u = 0 m/s</em>

<em>Constant: g = 9.8 m/s²</em>

<em>Substituting these values into equation 1,</em>

<em>v² = 0 + 2×9.8×20</em>

<em>v² = 392</em>

<em>v = √392</em>

<em>v = 19.799 m/s.</em>

<em>Therefore the velocity of the ball as it hit the ground = 19.799 m/s</em>

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A hot air balloon competition requires a balloonist to drop a ribbon onto a target on the ground. Initially the hot air balloon
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Answer:

The answer is 3.48 seconds

Explanation:

The kinematic equation

y= y0+V0*t+1/2*a*(t*t)

-50=0+(0)t+1/2(-9.8)*(t*t)

t=3.194 seconds

During ribbons ball,

x=x0+ Vt+1/2*a*(t*t)

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x= 47.9157m

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52.084m=0+15(t)+1/2*(0)(t*t)

t=52.084/15=3.472286= 3.48seconds

3 0
3 years ago
In some recent studies it has been shown that women are men when competing in similar sports (most notably in soccer and basketb
Elza [17]

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Research has proven that women are prone to have ACL tear than men when competing in similar sports. This disparity exists due to structural differences that pose as risk factors. These includes

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3 years ago
Explain in detail what specific factors affect the momentum of a bicycle going down hill.
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Explanation:

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Two satellites, X and Y, are orbiting Earth. Satellite X is 1.2 × 106 m from Earth, and Satellite Y is 1.9 × 105 m from Earth. W
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Answer: Satellite X has a greater period and a slower tangential speed than Satellite Y

Explanation:

According to Kepler’s Third Law of Planetary motion “The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”.

T^{2}=\frac{4\pi^{2}}{GM}r^{3}    (1)

Where;

G=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}} is the Gravitational Constant

M=5.972(10)^{24}kg is the mass of the Earth

r  is the semimajor axis of the orbit each satellite describes around Earth (assuming it is a circular orbit, the semimajor axis is equal to the radius of the orbit)

So for satellite X, the orbital period T_{X} is:

T_{X}^{2}=\frac{4\pi^{2}}{GM}r_{X}^{3}    (2)

Where r_{X}=1.2(10)^{6}m

T_{X}^{2}=\frac{4\pi^{2}}{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.972(10)^{24}kg)}(1.2(10)^{6}m)^{3}    (3)

T_{X}=413.712 s    (4)

For satellite Y, the orbital period T_{Y} is:

T_{Y}^{2}=\frac{4\pi^{2}}{GM}r_{Y}^{3}    (5)

Where r_{Y}=1.9(10)^{5}m

T_{Y}^{2}=\frac{4\pi^{2}}{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.972(10)^{24}kg)}(1.9(10)^{5}m)^{3}    (6)

T_{Y}=26.064 s    (7)

This means T_{X}>T_{Y}

Now let's calculate the tangential speed for both satellites:

<u>For Satellite X:</u>

V_{X}=\sqrt{\frac{GM}{r_{X}}} (8)

V_{X}=\sqrt{\frac{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.972(10)^{24}kg)}{1.2(10)^{6}m}}

V_{X}=18224.783 m/s (9)

<u>For Satellite Y:</u>

V_{Y}=\sqrt{\frac{GM}{r_{Y}}} (10)

V_{Y}=\sqrt{\frac{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.972(10)^{24}kg)}{1.9(10)^{5}m}}

V_{Y}= 45801.13 m/s (11)

This means V_{Y}>V_{X}

Therefore:

Satellite X has a greater period and a slower tangential speed than Satellite Y

4 0
3 years ago
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