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katovenus [111]
3 years ago
7

A golf ball is hit with a golf club. While the ball flies through the air, which forces act on the ball? Neglect air resistance.

Physics
2 answers:
34kurt3 years ago
6 0

To solve this problem it is necessary to apply the equation related to the Gravitational Force, the equation describes that

F = \frac{GMm}{r^2}

Where,

G = Gravitational Universal Constant

M = Mass of Earth (or Bigger star)

m = Mass of Object  (or smallest star)

r = Radius

From the statement we know that once the impact is made, the golf ball is subjected to the forces that are exerted in nature. Since the air resistance, which would represent the drag force, is ignored. Only the forces related to gravity remain.

The gravitational force carries 'pushes' or 'attracts' the body towards the earth, while the speed decreases as it reaches its maximum height.

When the ball has reached its maximum height only the force of gravity begins to act on it, generating the attraction to the earth in parabolic motion.

Therefore the correct answer is B.

Gnom [1K]3 years ago
4 0

Answer:

all of these

Explanation:

:)

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Answer:

D.

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6 0
2 years ago
An object, initially at rest, moves 250 m in 17 s. What is its acceleration?
mash [69]

Answer:

1.73 m/s²

Explanation:

Given:

Δx = 250 m

v₀ = 0 m/s

t = 17 s

Find: a

Δx = v₀ t + ½ at²

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2 years ago
Write down short notes on:<br>c. Power​
Fudgin [204]

Answer: In the International System of Units, the unit of power is the watt, equal to one joule per second. In older works, power is sometimes called activity. Power is a scalar quantity.

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4 0
3 years ago
Write a numerical expression for the emissive intensity (in W/m^2.sr) coming out of a tiny hole in an enclosure of surface tempe
stiks02 [169]

Answer:

6.0 × 10^{11} W/m^{2}

Explanation:

From Wien's displacement formula;

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Where: Q is the quantity of heat transferred, e is the emissivity of the surface, A is the area, and T is the temperature.

The emissive intensity = \frac{Q}{A} = eT^{4}

Given from the question that: e = 0.6 and T = 1000K, thus;

emissive intensity = 0.6 × (1000)^{4}

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                             = 6.0 × 10^{11} \frac{W}{m^{2} }

Therefore, the emissive intensity coming out of the surface is 6.0 × 10^{11} W/m^{2}.

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