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Lady_Fox [76]
2 years ago
15

Nate the Skate was an avid physics student whose main non-physics interest in life was high-speed skateboarding. In particular,

Nate would often don a protective suit of Bounce-Tex, which he invented, and after working up a high speed on his skateboard, would collide with some object. In this way, he got a gut feel for the physical properties of collisions and succeeded in combining his two passions. On one occasion, the Skate, with a mass of 129 kg, including his armor, hurled himself against an 831 kg stationary statue of Isaac Newton in a perfectly elastic linear collision. As a result, Isaac started moving at 1.29 m/s and Nate bounced backward. What were Nate's speeds immediately before and after the collision? Ignore friction with the ground. Choose the correct option;
(a) before collision: 4.8m/sAfter Collision: 1.29m/s
(b) before collision: 2.8m/sAfter Collision: 2.29m/s
(c) before collision: 3.8m/sAfter Collision: 3.29m/s
(d) before collision: 5.8m/sAfter Collision: 1.29m/s
Physics
1 answer:
Ivenika [448]2 years ago
4 0

Answer:

v_{1i} = 4.8 m/s

v_{1f} = 3.51 m/s

Explanation:

Since the collision is perfectly elastic collision so here we can use momentum conservation

m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}

so we will have

129 v_{1i} + 0 = 129(-v_{1f}) + 831(1.29)

v_{1i} + v_{1f} = 8.31

also by elastic collision property we will have

1.29 + v_{1f} = v_{1i}

now from above two equations we have

v_{1i} = 4.8 m/s

v_{1f} = 3.51 m/s

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belka [17]

Answer:

The correct answer is:

a) remain where it is released

Explanation:

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If an object is less dense than a liquid in which it is placed, it displaces a smaller volume of the liquid than its volume, hence only some part of the object will be seen to be under the liquid, the other part will float.

If an object is denser than the liquid in which it is placed, it displaces a larger volume of the liquid than its own volume, making the object to sink and is submerged, sometimes to the bottom of the liquid, but mostly below the point at which it was released.

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8 0
2 years ago
Which of the following best represents a decomposition reaction?
Fed [463]

Answer: AB-> A+B

Explanation:

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4 0
3 years ago
Calculate the speed of a car that travels 250 miles in 4.0 hours. (remember your unit)
MrRa [10]
250/4= 62.5 mph

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6 0
2 years ago
The suspension system of a 1700 kg automobile "sags" 7.7 cm when the chassis is placed on it. Also, the oscillation amplitude de
spin [16.1K]

Answer:

the spring constant k = 5.409*10^4 \ N/m

the value for the damping constant \\ \\b = 1.518 *10^3 \ kg/s

Explanation:

From Hooke's Law

F = kx\\\\k =\frac{F}{x}\\\\where \ F = mg\\\\k = \frac{mg}{x}\\\\given \ that:\\\\mass \ of \ each \ wheel = 425 \ kg\\\\x = 7.7cm = 0.077 m\\\\g = 9.8 \ m/s^2\\\\Then;\\\\k = \frac{425 \ kg * 9.8 \ m/s^2}{0.077 \ m}\\\\k = 5.409*10^4 \ N/m

Thus; the spring constant k = 5.409*10^4 \ N/m

The amplitude is decreasing 37% during one period of the motion

e^{\frac{-bT}{2m}}= \frac{37}{100}\\\\e^{\frac{-bT}{2m}}= 0.37\\\\\frac{-bT}{2m} = In(0.37)\\\\\frac{-bT}{2m} = -0.9943\\\\b = \frac{2m(0.9943)}{T}\\\\b = \frac{2m(0.9943)}{\frac{2 \pi}{\omega}}\\\\b = \frac{m(0.9943) \ ( \omega) )}{ \pi}

b = \frac{m(0.9943)(\sqrt{\frac{k}{m})}}{\pi}\\\\b = \frac{425*(0.9943)(\sqrt{\frac{5.409*10^4}{425}) }    }{3.14}\\\\b = 1518.24 \ kg/s\\\\b = 1.518 *10^3 \ kg/s

Therefore; the value for the damping constant \\ \\b = 1.518 *10^3 \ kg/s

5 0
3 years ago
Find the solution set of x + y = 5, 2 x - y = 7
STALIN [3.7K]
X + y = 5
x = 5 - y

2x - y = 7
2(5-y) - y = 7
10 - 2y - y = 7
10 - 3y = 7
10 - 7 = 3y
3 = 3y
y = 1

x + y = 5
x + 1 = 5
x = 5 - 1
x = 4
4 0
2 years ago
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