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olganol [36]
3 years ago
6

I need help/this is a major grade

Physics
1 answer:
Debora [2.8K]3 years ago
8 0

Answer:

1e , 2j , 3c , 4d , 5k , 6a , 7i , 8b , 9m , 10h ,11g , 12f , 13l .

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Pete is driving down 7th Street. He
lukranit [14]

Answer:

v=10.75m/s

Explanation:

It is given that,

Pete drives 215 Meters in 20 seconds. He is driving down 7th floor. We need to find his speed.

Speed = distance/time

⇒

v=\dfrac{215\ m}{20\ s}\\\\v=10.75\ m/s

Hence, the speed of Pete is 10.75 m/s.

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What does Newton's second law describe?
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B. Newton's second law says that “when a constant force acts on a massive body, it causes it to accelerate, i.e., to change its velocity, at a constant rate.”
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What kind of animals might benefit from a locust swarm​
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A suitable model for air could be b-b's moving rapidly around and colliding with each other.
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The answer to your qestion is true
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3 years ago
Read 2 more answers
A shot putter releases the shot some distance above the level ground with a velocity of 12.0 m/s, 51.0 ∘above the horizontal. Th
alina1380 [7]

A) Zero

The motion of the shot is a projectile's motion: this means that there is only one force acting on the projectile, which is gravity. However, gravity only acts in the vertical direction: so, there are no forces acting in the horizontal direction. Therefore, the x-component of the acceleration is zero.

B) -9.8 m/s^2

The vertical acceleration is given by the only force acting in the vertical direction, which is gravity:

F=mg

where m is the projectile's mass and g is the gravitational acceleration. Therefore, the y-component of the shot's acceleration is equal to the acceleration due to gravity:

a_y = g = -9.8 m/s^2

where the negative sign means it points downward.

C) 7.6 m/s

The x-component of the shot's velocity is given by:

v_x = v_0 cos \theta

where

v_0 = 12.0 m/s is the initial velocity

\theta=51.0^{\circ} is the angle of the shot

Substituting into the equation, we find

v_x = (12.0 m/s)(cos 51^{\circ})=7.6 m/s

D) 9.3 m/s

The y-component of the shot's velocity is given by:

v_y = v_0 sin \theta

where

v_0 = 12.0 m/s is the initial velocity

\theta=51.0^{\circ} is the angle of the shot

Substituting into the equation, we find

v_y = (12.0 m/s)(sin 51^{\circ})=9.3 m/s

E) 7.6 m/s

We said at point A) that the acceleration along the x-direction is zero: therefore, the velocity along the x-direction does not change, so the x-component of the velocity at the end of the trajectory is equal to the x-velocity at the beginning:

v_x = 7.6 m/s

F) -11.1 m/s

The y-component of the velocity at time t is given by:

v_y(t) = v_y + at

where

v_y = 9.3 m/s is the initial y-velocity

a = g = -9.8 m/s^2 is the vertical acceleration

t is the time

Since the total time of the motion is t=2.08 s, we can substitute this value into the equation, and we find:

v_y(2.08 s)=9.3 m/s + (-9.8 m/s^2)(2.08 s)=-11.1 m/s

where the negative sign means the vertical velocity is now downward.

3 0
3 years ago
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