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borishaifa [10]
3 years ago
15

A rock is thrown from the upper edge of a tall cliff at some angle above the horizontal. It reaches its highest point and starts

falling down. Which of the following statements about the rock's motion are true just before it hits the ground?
a. Its speed is the same as it was just as it was launched

b. Its horizontal velocity component is the same as it was just as it was launched

c. Its velocity is vertical.

d. Its vertical velocity component is the same as it was just as it was launched.

e. Its horizontal velocity component is zero.
Physics
1 answer:
lesya [120]3 years ago
3 0

Answer:

b. Its horizontal velocity component is the same as it was just as it was launched

Explanation:

When the rock is thrown from the upper edge of the cliff, at some angle with the horizontal. There is no any acceleration in horizontal direction, therefore, the horizontal velocity remains the same, whereas there is a constant downward acceleration of the magnitude g= 9.81 m/s^2. This causes the increase in vertical velocity of the rock in downward direction.

therefore  true statement just before it hits the ground

b. Its horizontal velocity component is the same as it was just as it was launched.

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Explain how balanced and unbalanced forces effect an objects motion differently
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Balanced Forces acting on an object will not change the object's motion. Unbalanced Forces acting on an object will change the change the object's motion.
5 0
3 years ago
What is Marie's instantaneous speed at 20 minutes in miles/min?
AVprozaik [17]

Answer:

0.25miles/min

Explanation:

Instantaneous speed of a person or an object is its speed at a particular moment usually at a period of time.

The speedometer of a car reports the instantaneous speed.

 It can be mathematically expressed as;

        Instantaneous speed  = \frac{distance}{time}

At 20min the distance covered is 5miles;

    Instantaneous speed  = \frac{5 miles }{20mins}   = 0.25miles/min

8 0
3 years ago
a 5.0 kg ball is dropped from a 2.5 m high window. what is the velocity of the ball just before it hits the ground?
julsineya [31]

Answer:

Approximately 7.0\; \rm m \cdot s^{-1}.

Assumption: the ball dropped with no initial velocity, and that the air resistance on this ball is negligible.

Explanation:

Assume the air resistance on the ball is negligible. Because of gravity, the ball should accelerate downwards at a constant g = 9.81 \; \rm m \cdot s^{-2} near the surface of the earth.

For an object that is accelerating constantly,

v^2 - u^2 = 2\, a \, x,

where

  • u is the initial velocity of the object,
  • v is the final velocity of the object.
  • a is its acceleration, and
  • x is its displacement.

In this case, x is the same as the change in the ball's height: x = 2.5\; \rm m. By assumption, this ball was dropped with no initial velocity. As a result, u = 0. Since the ball is accelerating due to gravity, a = 9.81\; \rm m \cdot s^{-2}.

v^2 - u^2 = 2\, g \cdot h.

In this case, v would be the velocity of the ball just before it hits the ground. Solve for

v^2 = 2\, a\, x + u^2.

\begin{aligned}v &= \sqrt{2\, a\, x + u^2} \\ &= \sqrt{2\times 9.81 \times 2.5 + 0} \\ &\approx 7.0\; \rm m\cdot s^{-1}\end{aligned}.

3 0
3 years ago
A student drops a ball from the top of a tall building; it takes 2.9 s for the ball to reach the ground.
AlexFokin [52]

<h2><u>We have</u>,</h2>

  • Initial velocity (u) = 0 m/s
  • Time taken (t) = 2.9s
  • Acceleration due to gravity (g) = + 10 m/s² [Down]

<h2><u>To calculate</u>,</h2>

  • Final velocity (v)
  • Height (h)

<h2><u>Solution</u><u>,</u></h2>

→ v = u + gt

→ v = 0 + 10(2.9)

→ v = 29 m/s \qquad … ( Ans )

And,

→ h = ut + ½gt²

→ h = 0(2.9) + ½ × 10 × (2.9)²

→ h = 5 × 8.41

→ h = 42.05 m \qquad … ( Ans )

4 0
3 years ago
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