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il63 [147K]
3 years ago
14

A model airplane with mass 0.5 kg hangs from a rubber band with spring constant 45 N/m. How much is the rubber band stretched wh

en the model airplane hangs motionless?
Physics
2 answers:
e-lub [12.9K]3 years ago
8 0
Multiply 0.5 by 9.8 then divide it by 45. 0.11m
Contact [7]3 years ago
3 0

Answer:

0.109 m (10.9 cm)

Explanation:

The weight of the airplane hanging from the rubber band is given by the product between its mass (0.5 kg) and the acceleration due to gravity (g=9.8 m/s^2):

W=mg=(0.5 kg)(9.8 m/s^2)=4.9 N

This is the force that stretches the rubber band, according to Hook's law:

F=kx

where

k = 45 N/m is the spring constant

x is the stretching of the spring with respect to its equilibrium position

Substituting F = 4.9 N and re-arranging the formula, we get:

x=\frac{F}{x}=\frac{4.9 N}{45 N/m}=0.109 m=10.9 cm

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SpyIntel [72]

Answer:

Explanation:

Let after time t , Tina catches up David .

Distance travelled by them are equal ,

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s = ut + 1/2 a t²

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t = 28.57 s

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7 0
3 years ago
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Aleksandr [31]
Let's cut through the weeds and the trash
and get down to the real situation:

                  A stone is tossed straight up at  5.89 m/s .
                  Ignore air resistance.


Gravity slows down the speed of any rising object by  9.8 m/s every second.
So the stone (aka Billy-Bob-Joe) continues to rise for

                     (5.89 m/s / 9.8 m/s²)  =  0.6 seconds.

At that timer, he has run out of upward gas.  He is at the top
of his rise, he stops rising, and begins to fall.

His average speed on the way up is  (1/2) (5.89 + 0) = 2.945 m/s .

Moving for 0.6 seconds at an average speed of  2.945 m/s,
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                    (2.945 m/s) (0.6 s) =  1.767 meters above the trampoline.

With no other forces other than gravity acting on him, it takes him
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5 0
4 years ago
"What is the magnifying power of an astronomical telescope using a reflecting mirror whose radius of curvature is 5.9 m and an e
notka56 [123]

Answer:

The Magnifying power of a telescope is M = 109.26

Explanation:

Radius of curvature R = 5.9 m = 590 cm

focal length of objective f_{objective} = \frac{R}{2}

⇒ f_{objective} = \frac{590}{2}

⇒ f_{objective} = 295 cm

Focal length of eyepiece f_{eyepiece} = 2.7 cm

Magnifying power of a telescope is given by,

M = \frac{f_{objective} }{f_{eyepiece} }

M = \frac{295}{2.7}

M = 109.26

therefore the Magnifying power of a telescope is M = 109.26

4 0
3 years ago
Based on the law of conservation of energy, how can we reasonably improve a machine’s ability to do work? Move the machine to a
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Natasha2012 [34]

Answer:kinetic energy converted to heat energy

Explanation:

As the ball rolls down kinetic energy is converted to heat energy

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