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kolbaska11 [484]
3 years ago
9

a golfer hits a golf ball with a velocity of 36.0 meters/second at an angle of 28.0. if the hang time of the golf ball is 33.4 s

econds, what is the range of the golf ball
Physics
1 answer:
Kobotan [32]3 years ago
6 0
Solving this using the time, we know that range = horizontal velocity x time of flight 

since there are no horizontal forces acting on the ball, there are no horizontal accelerations and the initial horizontal velocity of 36 cos 28 will be constant throughout. If we use the correct time of flight given the launch parameters, we have 

range = 36 cos 28 x 3.44 s = 109.3 m

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A vertical block-spring system on earth has a period of 6.0 s. What is the period of this same system on the moon where the acce
Vera_Pavlovna [14]

Answer:

D) 15s

Explanation:

let Te be the period of the block-spring system on earth and Tm be the period of the same system on the moon.let g1 be the gravitational acceleration on earth and g2 be the gravitational acceleration on the moon.

the period of a pendulum is given by:

T = 2π√(L/g)

so on earth:

Te = 2π√(L/g1)

     =  6s

on the moon;

Tm = 2π√(L/g2)

since g2 = 1/6 g1 then:

Tm = 2π√(L/(1/6×g1))

      = √(6)×2π√(L/(g1))

and 2π√(L/(g1)) = Te = 6s

Tm = (√(6))×6 = 14.7s ≈ 15s

Therefore, the period of the block-spring system on the moon is 15s.

5 0
3 years ago
Why would a flare be observed in visible light, when they are so much brighter in X-ray and ultraviolet light?
raketka [301]

Answer:

Because it radiates a spectre of frequencies, this means that it radiates in a continuous range of frequencies, ones with more intensity (like x-ray and ultraviolet) and others with less intensity (like the visible light). While most of the radiation is not in the visible range, there still is a part of the radiation in the visible spectre. And while this part is not the most intense part, the radiation is so large that we can see a very bright visible light.

3 0
3 years ago
PLEASE HELP!!! 30 POINTS
Tasya [4]

The impulse and the action and reaction  law emit explain the sensation of the foot when kicking two different objects the answer in each case

a) kick the ball  the sensation is not painful

b) kick the wall the feeling is painful

Newton's second law establishes a linear relationship between force and mass

            F = m a

Where the bold letters indicate vectors, F is the force, m the mass and the acceleration

Newton's third law or action and reaction law stable that the force acts on a pairs, the kicking of an object with a force F the object responds with a force of equal magnitude but opposite direction on us.

From Newton's second law we can derive a relationship between the impulse and the variation of the momentum

            I = ∫ F dt = Δp

Where I is the impulse, t the time and Δp the variation of the momentum

The impulse is related to the sensation that the body receives due to the variation of the amount of movement, in this case we have two situations

a) Kick a soccer ball

The mass of the ball is small so the force applied by the foot creates a large acceleration without the foot having a significant decrease in foot speed. Consequently the impulse applied by the ball on the foot is small, the sensation is not painful

b) Kick a brick wall

In this case, the mass of the wall is high and when kicked it does not acquire any movement, consequently the reaction of the wall creates a high impulse on the foot that causes its speed to quickly decrease to zero, the physical sensation of this reduction in quantity movement is painful.

In conclusion, the impulse and the law of action and reaction allow us to find the answer in each case.

a) kick the ball feeling painless

b) kick the wall painful feeling

learn more about the action and reaction law and impulse here:

brainly.com/question/6677486

6 0
3 years ago
A 64.0 kg pole vaulter running at 10.2 m/s vaults over the bar. If the vaulter's horizontal component of velocity over the bar i
Lubov Fominskaja [6]

Answer:

h = 5.05 m

Explanation:

given,

mass of pole vaulter, m = 64 Kg

speed of the vaulter,V = 10.2 m/s

horizontal component of velocity in air, v = 1 m/s

height of the jump,h = ?

using energy conservation

     E_i = E_f

\dfrac{1}{2}mv_i^2 + m g h_i= \dfrac{1}{2}mv_f^2+m g h_f

initial height of the vaulter is equal to zero.

\dfrac{1}{2}v_i^2 = \dfrac{1}{2}v_f^2+gh_f

h =\dfrac{v_i^2-v_f^2}{2g}

h =\dfrac{10^2-1^2}{2\times 9.8}

 h = 5.05 m

height of the jump is equal to 5.05 m.

4 0
3 years ago
3. If the net work done on an object is negative, then the object's kinetic energy
g100num [7]

Answer:

I think the answer is B I hope this helps also Im sorry If I'm wrong

4 0
3 years ago
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