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professor190 [17]
3 years ago
10

What is the weight of a 4.5 kg mass on Earth?

Physics
1 answer:
swat323 years ago
4 0

Answer:

7.535×10^25 earth mass

Explanation:

for an approximate result,divide the mass value by 9.223e+18

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A particle is being accelerated through space by a 10 N force. Suddenly the particle encounters a second of 10 N force in the op
Lyrx [107]

Answer:

The particle will be rendered still

Explanation:

The particle will be rendered still because of the Newton’s 1st law of motion states that a body will continue to move on forever until contacted by an equal and opposite force. The same case has been produced here and it is justifying the law. If the Force has another direction and magnitude than 10 N then the body will continue to move on in the direction of the greater force.

7 0
4 years ago
A tennis ball is dropped from 1.16 m above the
sweet-ann [11.9K]

Answer:

Vf = 4.77 m/s

Explanation:

During the downward motion we can easily find the final velocity or the velocity with which the ball hits the ground, by using third equation of motion. The third equation of motion is given as follows:

2gh = Vf² - Vi²

where,

g = acceleration due to gravity = 9.8 m/s²

h = height = 1.16 m

Vf = Final Velocity of Ball = ?

Vi = Initial Velocity of Ball = 0 m/s (Since, ball was initially at rest)

Therefore, using these values in the equation, we get:

(2)(9.8 m/s²)(1.16 m) = Vf² - (0 m/s)²

Vf = √(22.736 m²/s²)

<u>Vf = 4.77 m/s</u>

6 0
3 years ago
Jess kicks a soccer ball, and it rolls across the ground. The force diagram shows all the forces acting on the ball at the momen
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Arrow at the left side pointing towards right side represents the frictional force as it always acts opposite to motion
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When a sound source is moving away from an observer, the observer will hear a lower pitch. This is due to
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Lower Pitch Due to the change to a lower frequency
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You're driving down the highway late one night at 21 m/s when a deer steps onto the road 35 m in front of you. Your reaction tim
andreev551 [17]

Speed with which initially car is moving is 21 m/s

Reaction time = 0.50 s

distance traveled in the reaction time d = v t

d = 21 * 0.50 = 10.5 m

deceleration after this time = -10 m/s^2

now the distance traveled by the car after applying bakes

v_f^2 - v_i^2 = 2a d

0 - 21^2 = 2(-10)d

d = 22.05 m

so total distance moved before it stop

d = 22.05 + 10.5 = 32.55 m

so the distance from deer is 35 - 32.55 = 2.45 m

now to find the maximum speed with we can move we will assume that we will just touch the deer when we stop

so our distance after brakes are applied is d = 35 - 10.5 = 24.5 m

again by kinematics

v_f^2 - v_i^2 = 2 ad

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v = 22.1 m/s

so maximum speed would be 22.1 m/s

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