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4 years ago

11

ick jumps off his roof which is 10 meters tall onto his trampoline which has a height of 1 meter and acts like a spring. Nick ha

s a mass of 70 kg and sinks 0.4 meters into the trampoline before coming to a complete stop. What is the spring constant of the trampoline

1 answer:

AysviL [449]4 years ago

3 0

Answer: The spring constant of the trampoline is 87500N/m

Explanation:

From the formula for elastic potential energy,

Work done on the spring = 1/2 Ke²

Work done = Potential energy of the body = mass(kg) * g * height(meters)

mgh = 1/2 Ke²,

where m = 70Kg, g = 10 m/s², h = 10m, K = spring constant, e = extension or compression = 0.4m

70 * 10 * 10 = 1/2 K * (0.4)²

7000 = 0.08 * K

K = 87500N/m

The spring constant of the trampoline is 87500N/m

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The Answer would be True

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3 years ago

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A driver enters a one-lane tunnel at 34.4 m/s. The driver then observes a slow-moving van 154 m ahead travelling (in the same di

navik [9.2K]

Answer:

Ans. B) 22 m/s (the closest to what I have which was 20.16 m/s)

Explanation:

Hi, well, first, we have to find the equations for both, the driver and the van. The first one is moving with constant acceleration (a=-2m/s^2) and the van has no acceletation. Let´s write down both formulas so we can solve this problem.

$X(van)=5.65t+154$

$X(driver)=34.4t+\frac{(-2)t^{2} }{2}$

or by rearanging the drivers equation.

$X(driver)=34.4t+t^{2}$

Now that we have this, let´s equal both equations so we can tell the moment in which both cars crashed.

$X(van)=X(driver)$

$5.65t+154=34.4t-t^{2}$

$0=t^{2} -(34.4-5.65)t+154$ $0=t^{2} -28.75t+154$

To solve this equation we use the following formulas

$t=\frac{-b +\sqrt{b^{2}-4ac } }{2a}$

$t=\frac{-b +\sqrt{b^{2}-4ac } }{2a}$

Where a=1; b=-28.75; c=154

So we get:

$t=\frac{28.75 +\sqrt{(-28.75)^{2}-4(1)(154) } }{2(1)}=21.63s$ $t=\frac{28.75 -\sqrt{(-28.75)^{2}-4(1)(154) } }{2(1)}=7.12s$

At this point, both answers could seem possible, but let´s find the speed of the driver and see if one of them seems ilogic.

$V(driver)=V_{0} +at$ }

$V(driver)=34.4\frac{m}{s} -2\frac{m}{s^{2} } *(7.12s)=20.16\frac{m}{s}$ $V(driver)=34.4\frac{m}{s} -2\frac{m}{s^{2} } *(21.63s)=-8.86\frac{m}{s}$

This means that 21.63s will outcome into a negative speed, for that reason we will not use the value of 21.63s, we use 7.12s and if so, the speed of the driver when he/she hits the van is 20.16m/s, which is closer to answer A).

Best of luck

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3 years ago

Explain how the gravitational potential energy of an object can be changed.

inna [77]

Answer:

Change in Potential energy, pe = Final potential energy - Initial potential energy

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3 years ago

A person, with his ear to the ground, sees a huge stone strike the concrete pavement. A moment later two sounds are heard from t

marishachu [46]

Answer:

The impact occured at a distance of 2478.585 meters from the person.

Explanation:

(After some research on web, we conclude that problem is not incomplete) The element "Part A" may lead to the false idea that question is incomplete. Correct form is presented below:

<em>A person, with his ear to the ground, sees a huge stone strike the concrete pavement. A moment later two sounds are heard from the impact: one travels in the air and the other in the concrete, and they are 6.4 seconds apart. How far away did the impact occur? (Sound speed in the air: 343 meters per second, sound speed in concrete: 3000 meters per second)</em>

Sound is a manifestation of mechanical waves, which needs a medium to propagate themselves. Depending on the material, sound will take more or less time to travel a given distance. From statement, we know this time difference between air and concrete ( $\Delta t$ ), in seconds:

$\Delta t = t_{A}-t_{C}$ (1)

Where:

$t_{C}$ - Time spent by the sound in concrete, in seconds.

$t_{A}$ - Time spent by the sound in the air, in seconds.

By suposing that sound travels the same distance and at constant speed in both materials, we have the following expression:

$\Delta t = \frac{x}{v_{A}}-\frac{x}{v_{C}}$

$\Delta t = x\cdot \left(\frac{1}{v_{A}}-\frac{1}{v_{C}} \right)$

$x = \frac{\Delta t}{\frac{1}{v_{A}}-\frac{1}{v_{C}} }$ (2)

Where:

$v_{C}$ - Speed of the sound in concrete, in meters per second.

$v_{A}$ - Speed of the sound in the air, in meters per second.

$x$ - Distance traveled by the sound, in meters.

If we know that $\Delta t = 6.4\,s$ , $v_{C} = 3000\,\frac{m}{s}$ and $v_{A} = 343\,\frac{m}{s}$ , then the distance travelled by the sound is:

$x = \frac{\Delta t}{\frac{1}{v_{A}}-\frac{1}{v_{C}} }$

$x = 2478.585\,m$

The impact occured at a distance of 2478.585 meters from the person.

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3 years ago