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bearhunter [10]
2 years ago
5

A tennis ball is dropped from a roof. If it takes 38.9 seconds to reach the ground, how fast is the ball moving just before it h

its the ground? Answer tot he nearest tenths and no units
Physics
1 answer:
Ilya [14]2 years ago
5 0

Answer:

The velocity of the ball before it hits the ground is 381.2 m/s

Explanation:

Given;

time taken to reach the ground, t = 38.9 s

The height of fall is given by;

h = ¹/₂gt²

h = ¹/₂(9.8)(38.9)²

h = 7414.73 m

The velocity of the ball before it hits the ground is given as;

v² = u² + 2gh

where;

u is the initial velocity of the on the root = 0

v is the final velocity of the ball before it hits the ground

v² = 2gh

v = √2gh

v = √(2 x 9.8 x 7414.73 )

v = 381.2 m/s

Therefore, the velocity of the ball before it hits the ground is 381.2 m/s

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The temperature of 20 liters of helium gas is increased from 100°K to 300°K. If the pressure was kept constant, the new volume w
slavikrds [6]

Answer:

New volume (V2) = 60 liter

Explanation:

Given:

Amount of helium (V1) = 20 Liter

Temperature (T1) = 100°K

Temperature (T2) = 300°K

Find:

New volume (V2)

Computation:

According to Gas Law:

V1 / T1 = V2 / T2

20 / 100°K = V2 / 300°K

V2 = 60 liter

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8 0
3 years ago
An electrical motor spins at a constant 2662.0 rpm. If the armature radius is 6.725 cm, what is the acceleration of the edge of
Gemiola [76]

Answer:

18.73 m/s^2

Explanation:

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a = 0.06725 × 2 × 3.14 × 2662 / 60

a = 18.73 m/s^2

3 0
3 years ago
The international space station makes 15.65 revolutions per day in its orbit around the earth. assuming a circular orbit, how hi
sweet-ann [11.9K]
<span>373.2 km The formula for velocity at any point within an orbit is v = sqrt(mu(2/r - 1/a)) where v = velocity mu = standard gravitational parameter (GM) r = radius satellite currently at a = semi-major axis Since the orbit is assumed to be circular, the equation is simplified to v = sqrt(mu/r) The value of mu for earth is 3.986004419 Ă— 10^14 m^3/s^2 Now we need to figure out how many seconds one orbit of the space station takes. So 86400 / 15.65 = 5520.767 seconds And the distance the space station travels is 2 pi r, and since velocity is distance divided by time, we get the following as the station's velocity 2 pi r / 5520.767 Finally, combining all that gets us the following equality v = 2 pi r / 5520.767 v = sqrt(mu/r) mu = 3.986004419 Ă— 10^14 m^3/s^2 2 pi r / 5520.767 s = sqrt(3.986004419 * 10^14 m^3/s^2 / r) Square both sides 1.29527 * 10^-6 r^2 s^2 = 3.986004419 * 10^14 m^3/s^2 / r Multiply both sides by r 1.29527 * 10^-6 r^3 s^2 = 3.986004419 * 10^14 m^3/s^2 Divide both sides by 1.29527 * 10^-6 s^2 r^3 = 3.0773498781296 * 10^20 m^3 Take the cube root of both sides r = 6751375.945 m Since we actually want how far from the surface of the earth the space station is, we now subtract the radius of the earth from the radius of the orbit. For this problem, I'll be using the equatorial radius. So 6751375.945 m - 6378137.0 m = 373238.945 m Converting to kilometers and rounding to 4 significant figures gives 373.2 km</span>
7 0
3 years ago
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lapo4ka [179]

Explanation:

If the size and direction of the forces on the object are exactly balanced , then there is no net force acting on the object

8 0
2 years ago
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sergiy2304 [10]
You performed 0 work for the fact that work means the distance of movement made on an object not the amount of force it is exposed to. 0 work because it didn't move
7 0
3 years ago
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