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9966 [12]
1 year ago
7

Q3. You throw a ball into the air, it reaches a certain height and then comes back to you.

Physics
1 answer:
Grace [21]1 year ago
5 0

The If a car is going round a curve , there is an acceleration because the direction of the velocity changes.

<h3>What is the direction of the velocity?</h3>

Now we know that if you throw the ball upwards, the motion is in opposite direction to gravity thus the ball is experiencing deceleration and the speed decreases. The velocity decreases and the acceleration is negative.

If the ball is coming down, then the ball is accelerated thus it speeds up and the direction of the acceleration is positive.

If a car is going round a curve, the vehicle is accelerating because the direction of the velocity changes even if its amount remains constant.

When a board is moving down a hill at 2 ms-1, it is experiencing an acceleration because the motion is in the same direction as gravity.

If a car is coming to a stop at a point, it experiences a deceleration and not an acceleration since the change of velocity with time is negative as the car comes to rest.

Learn more about acceleration:brainly.com/question/12550364

#SPJ1

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The potential energy an object has due to it position Is called _______ potential energy
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In a controlled experiment, which variable does the investigator change?
Bas_tet [7]

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The manipulated variable is also known as the independent variable

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3 years ago
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A vibrating object produces periodic waves with a wavelength of 53 cm and a frequency of 15 Hz. How fast do these waves move awa
DerKrebs [107]

Answer:

v = 7.95 m/s

Explanation:

Given that,

Wavelength of a wave, \lambda=53\ cm=0.53\ m

Frequency of a wave, f = 15 Hz

We need to find the speed of the wave. The speed of a wave is given by :

v=f\lambda\\\\v=15\ Hz\times 0.53\ m\\\\v=7.95\ m/s

So, the wave move with a speed of 7.95 m/s.

5 0
2 years ago
Question 2
Delvig [45]

Answer:

Approximately 73\; {\rm N}, assuming that the acceleration of this ball is constant during the descent.

Explanation:

Assume that the acceleration of this ball, a, is constant during the entire descent.

Let x denote the displacement of this ball and let t denote the duration of the descent. The SUVAT equation x = (1/2)\, a\, t^{2} would apply.

Rearrange this equation to find an expression for the acceleration, a, of this ball:

\begin{aligned} a &= \frac{2\, x}{t^{2}}\end{aligned}.

Note that x = 11\; {\rm m} and t = 1.5\; {\rm s} in this question. Thus:

\begin{aligned} a &= \frac{2\, x}{t^{2}} \\ &= \frac{2 \times 11\; {\rm m}}{(1.5\; {\rm s})^{2}} \\ &\approx 9.78\; {\rm m \cdot s^{-2}}\end{aligned}.

Let m denote the mass of this ball. By Newton's Second Law of Motion, if the acceleration of this ball is a, the net external force on this ball would be m\, a.

Since m = 7.5\; {\rm kg} and a \approx 9.78\; {\rm m\cdot s^{-2}}, the net external force on this ball would be:

\begin{aligned} (\text{net force}) &= m\, a \\ &\approx 7.5\; {\rm kg} \times 9.78\; {\rm m\cdot s^{-2}} \\ &\approx 73\; {\rm kg \cdot m \cdot s^{-2} \\ &= 73\; {\rm N} && (1\; {\rm N} = 1\; {\rm kg \cdot m\cdot s^{-2}}) \end{aligned}.

4 0
2 years ago
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