All categories

3 years ago

A skateboarder is standing at the top of a tall ramp waiting to begin a trip. The skateboarder has

Physics

1 answer:

erik [133]3 years ago

3 0

Potential energy. When it starts moving, that potential energy will start transforming into kinetic energy.

You might be interested in

A converging lens is also known as a •convex •concave •double •refracted

Leni [432]

Answer:

Convex lens

Explanation:

hope this helps

4 0

3 years ago

Read 2 more answers

What is electric magnetism?

gavmur [86]

Answer:

the physical phenomenon produced by moving electric charge

Explanation:

6 0

3 years ago

Read 2 more answers

A 10-g bullet moving horizontally with a speed of 2.0 km/s strikes and passes through a 4.0-kg block moving with a speed of 4.2

SVEN [57.7K]

Answer:

$K=512J$

Explanation:

Since the surface is frictionless, momentum will be conserved. If the bullet of mass $m_1$ has an initial velocity $v_{1i}$ and a final velocity $v_{1f}$ and the block of mass $m_2$ has an initial velocity $v_{2i}$ and a final velocity $v_{2f}$ then the initial and final momentum of the system will be:

$p_i=m_1v_{1i}+m_2v_{2i}$

$p_f=m_1v_{1f}+m_2v_{2f}$

Since momentum is conserved, $p_i=p_f$ , which means:

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

We know that the block is brought to rest by the collision, which means $v_{2f}=0m/s$ and leaves us with:

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

which is the same as:

$v_{1f}=\frac{m_1v_{1i}+m_2v_{2i}}{m_1}$

Considering the direction the bullet moves initially as the positive one, and writing in S.I., this gives us:

$v_{1f}=\frac{(0.01kg)(2000m/s)+(4kg)(-4.2m/s)}{0.01kg}=320m/s$

So kinetic energy of the bullet as it emerges from the block will be:

$K=\frac{mv^2}{2}=\frac{(0.01kg)(320m/s)^2}{2}=512J$

6 0

3 years ago

Three forces that act on a roller coaster:

Andre45 [30]

Yes, all of these could be applied to a roller coaster.

6 0

3 years ago

Read 2 more answers

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