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andreev551 [17]
2 years ago
14

a rock is tied to the end of the string and swung in a circle with a radius of 1/2 meter. if the rock goes around once in 1/10 o

f a second what is the centripetal acceleration of the rock?
Physics
1 answer:
vaieri [72.5K]2 years ago
3 0

Answer:

1974.4328 m/s²

Explanation:

r = Radius = 0.5 m

t = Time taken = 0.1 second

Rotational speed

v=\dfrac{2\pi r}{t}\\\Rightarrow v=\dfrac{2\pi 0.5}{0.1}\\\Rightarrow v=31.42\ m/s

The centripetal acceleration is given by

a_c=\dfrac{v^2}{r}\\\Rightarrow a_c=\dfrac{31.42^2}{0.5}\\\Rightarrow a_c=1974.4328\ m/s^2

The centripetal acceleration of the rock is 1974.4328 m/s²

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With time, momentum increases as it builds speed assuming their is nothing in the way to stop it. Based on the graph, you can see that example being displayed as the line on the graph gets higher
4 0
2 years ago
Salmon often jump waterfalls to reach their breeding grounds. Starting downstream, 3.18 m away from a waterfall 0.294 m in heigh
Karolina [17]

Answer:

v = 7.65 m/s

t = 0.5882 s

Explanation:

We are told that the salmon started downstream, 3.18 m away from a waterfall.

Thus, range = 3.18 m

Since the horizontal velocity component is constant, then;

Range = vcosθ × t

Thus,

vcosθ × t = 3.18 - - - (eq 1)

We are told the salmon reached a height of 0.294 m

Thus, using distance equation;

s = v_y•t + ½gt²

g will be negative since motion is against gravity.

s = v_y•t - ½gt²

Thus;

0.294 = v_y•t - ½gt²

v_y = vsinθ

Thus;

0.294 = vtsinθ - ½gt² - - - (eq 2)

From eq(1), making v the subject, we have;

v = 3.18/tcosθ

Plugging into eq 2,we have;

0.294 = (3.18/tcosθ)tsinθ - ½gt²

0.295 = 3.18tanθ - ½gt²

We are given g = 9.81 m/s² and θ = 45°

0.295 = (3.18 × tan 45) - ½(9.81) × t²

0.295 = 3.18 - 4.905t²

3.18 - 0.295 = 4.905t²

4.905t² = 2.885

t = √2.885/4.905

t = 0.5882 s

Thus;

v = 3.18/(0.5882 × cos45)

v = 7.65 m/s

8 0
2 years ago
(III) A baseball is seen to pass upward by a window with a vertical speed of If the ball was thrown by a person 18 m below on th
Ghella [55]

Answer:

<em><u>Assuming that the vertical speed of the ball is 14 m/s</u></em> we found the given values:

a) V₀ = 23.4 m/s

b) h = 27.9 m

c) t = 0.96 s

d) t = 4.8 s

 

Explanation:

a) <u>Assuming that the vertical speed is 14 m/s</u> (founded in the book) the initial speed of the ball can be calculated as follows:  

V_{f}^{2} = V_{0}^{2} - 2gh

<u>Where:</u>

V_{f}: is the final speed = 14 m/s

V_{0}: is the initial speed =?

g: is the gravity = 9.81 m/s²

h: is the height = 18 m

V_{0} = \sqrt{V_{f}^{2} + 2gh} = \sqrt{(14 m/s)^{2} + 2*9.81 m/s^{2}*18 m} = 23.4 m/s  

b) The maximum height is:

V_{f}^{2} = V_{0}^{2} - 2gh

h = \frac{V_{0}^{2}}{2g} = \frac{(23. 4 m/s)^{2}}{2*9.81 m/s^{2}} = 27.9 m

c) The time can be found using the following equation:

V_{f} = V_{0} - gt

t = \frac{V_{0} - V_{f}}{g} = \frac{23.4 m/s - 14 m/s}{9.81 m/s^{2}} = 0.96 s

d) The flight time is given by:

t_{v} = \frac{2V_{0}}{g} = \frac{2*23.4 m/s}{9.81 m/s^{2}} = 4.8 s

         

I hope it helps you!    

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In cold weather, properly designed gasoline aids in engine starting, while in hot weather, it helps prevent vapor lock. In order to meet the requirements of a modern engine, the fuel must have the volatility for which the engine's fuel system was built and an antiknock quality strong enough to prevent knock during routine operation.

During the intake phase, the air and fuel are combined before being introduced into the cylinder. The spark ignites the fuel-air mixture after the piston compresses it, resulting in combustion. During the power stroke, the piston is propelled by the expansion of the combustion gases.

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C difference in temperature
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