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3 years ago

A uniform thin circular rubber band of mass M and spring constant k has an original radius R?

Physics

1 answer:

Gemiola [76]3 years ago

6 0

Answer: Yes

Explanation:

A ruber band can be described by his mas, M, the spring constant K, and his rest radius, wich is the radius of the circular rubber band when there is no force applied to it, so yes, a uniform thin circular rubber band of mass M and spring constant k has an original radius R

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Choose the correct statement regarding energy production in the interior of a star and energy production in a power plant.

sergey [27]

' C ' is the only correct statement here.

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3 years ago

Imagine a landing craft approaching the surface of Callisto, one of Jupiter's moons. If the engine provides an upward force (thr

nordsb [41]

Answer:

$W=3456 N$

Explanation:

Force 1 $F_1=3456$

Force 2 $F_2=2333N$

Acceleration at stage 2 $a_2=0.39$

Generally the weight of the Craft W is given as

W= upward force(thrust)

Therefore

$W=3456 N$

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3 years ago

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BigorU [14]

Answer:

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3 years ago

As the boat in which he is riding approaches a dock at 3.0 m/s, Jasper stands up in the boat and jumps toward the dock. Jasper a

Annette [7]

Im pretty sure it’s a because it makes more sense you know?.

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3 years ago

A student that jumps a vertical height of 50 cm during the hang time activity.

muminat

The hang time of the student is 0.64 seconds, and he must leave the ground with a speed of 3.13 m/s

Why?

To solve the problem, we must consider the vertical height reached by the student as max height.

We can use the following equations to solve the problem:

Initial speed calculations:

$v_{f}^{2}=v_{o}^{2}+2*a*d$

At max height, the speed tends to zero.

So, calculating, we have:

 $v_{f}^{2}=v_{o}^{2}+2*a*d\\\\0=v_{o}^{2}+2*(-9.81\frac{m}{s^{2}})*0.5m\\\\v_{o}^{2}=9.81\frac{m^{2} }{s^{2}}\\\\v_{o}=\sqrt{9.81\frac{m^{2} }{s^{2}}}=3.13\frac{m}{s}$ 

Hang time calculations:

We must remember that the total hang time is equal to the time going up plus the time going down, and both of them are equal,so, calculating the time going down, we have have:

$y-yo=vo.t+\frac{1}{2}*9.81\frac{m}{s^{2}}*t^{2} \\\\0.5m-0=0*t+4.95*t^{2}\\\\t^{2}=\frac{0.5}{4.95}\\\\t=\sqrt{0.101}=0.318s$

Then, for the total hang time, we have:

$TotalHangTime=2*0.318seconds=0.64seconds$

Have a nice day!

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4 years ago