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bearhunter [10]
3 years ago
6

An object is held at an unknown height above Earth’s surface, where the acceleration due to gravity of the object is considered

to be constant. After the object is released from rest, a student must determine the object’s speed the instant the object makes contact with the ground.
Which of the following equations could the student use to determine the object’s speed by using the fewest measuring tools if the student does not have access to a motion sensor? Select two answers.

A) vₓ =vₓ₀ +aₓ t
B) x= x₀ + vₓ₀ t + (1/2)aₓt²
C) v²x=v² x₀+2aₓ (x−x₀ )
D) v¯=(x−x₀) /t
Physics
1 answer:
Aleks04 [339]3 years ago
7 0

Answer:

Options A and B.

Explanation:

Gravitational acceleration, initial height, intial speed and time are required to determine final speed. The option D is incorrect, since speed varies in time. Option C is dimentionally wrong.

The correct strategy is calculating the initial height from option B. Later, substituting time in equation A to derive an expression of the final velocity in terms of position. Hence, the required equations are options A and B.

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I believe the answer is A
5 0
3 years ago
Two asteroids identical to those above collide at right angles and stick together; i.e, their initial velocities were perpendicu
11111nata11111 [884]

Answer:

velocity = 62.89 m/s  in 58 degree measured from the x-axis

Explanation:

Relevant information:

Before the collision, asteroid A of mass 1,000 kg moved at 100 m/s, and asteroid B of mass 2,000 kg moved at 80 m/s.

Two asteroids moving with velocities collide at right angles and stick together. Asteroid A initially moving to right direction and asteroid B initially move in the upward direction.

Before collision Momentum of A = 1000 x 100 = $ 10^5$ kg - m/s in the right direction.

Before collision Momentum of B = 2000 x 80 = 1.6 x $ 10^5$  kg - m/s in upward direction.

Mass of System of after collision = 1000 + 2000 = 3000 kg

Now applying the Momentum Conservation, we get

Initial momentum in right direction = final momentum in right direction = $ 10^5$

And, Initial momentum in upward direction = Final momentum in upward direction = 1.6 x $ 10^5$

So, $ V_x = \frac{10^5}{3000} $  = $ \frac{100}{3} $  m/s

and $ V_y=\frac{160}{3}$  m/s

Therefore, velocity is = $ \sqrt{V_x^2 + V_y^2} $

                                   = $ \sqrt{(\frac{100}{3})^2 + (\frac{160}{3})^2} $

                                   = 62.89 m/s

And direction is

tan θ = $ \frac{V_y}{V_x}$     = 1.6

therefore, $ \theta = \tan^{-1}1.6 $

                   = $ 58 ^{\circ}$  from x-axis

4 0
3 years ago
What is the total kinetic energy of a 0.15 kg hockey puck sliding at 0.5 m/s and rotating about its center at 8.4 rad/s
AlexFokin [52]

Answer:

K=0.023J

Explanation:

From the question we are told that:

Mass m=0.15

Velocity v=0.5m/s

Angular Velocity \omega=8.4rad/s

Generally the equation for Kinetic Energy is mathematically given by

K=\frac{1}{2}M(v^2+\frac{1}{2}R^2\omega^2)

K=\frac{1}{2}0.15(0.5^2+\frac{1}{2}(0.038)^2.(8.4rad/s^2))

K=0.023J

8 0
3 years ago
a goalkeeper catches a 491 g soccer ball traveling horizontally at 29.4 m/s. if it took 2,218 n of force to stop the ball, how m
yarga [219]

The ball will take 2.551 seconds to reach its peak position.

<h3>How much time will the ball take to land?</h3>

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<h3>How quickly does a ball drop?</h3>

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To know more about balls visit:-

brainly.com/question/19930452

#SPJ4

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vlabodo [156]

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