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Anna35 [415]
3 years ago
8

A customer sits in an amusement park ride in which the compartment is to be pulled downwards in the negative direction of y axis

with accelaration magnitude of 1.24 g(g=9.8 m/sec ^2). A 0.567 gms coin is placed on customers knee. once motion begins,
find coins accelaration relative to
a)ground
b)customer
c) how long does the coin take to reach the compartment ceiling 2.2 m above the knee.
d)the actual force on the coin
e)the apparent force according to customer's measure of coins accelaration
Physics
1 answer:
Brums [2.3K]3 years ago
3 0
This movement of the coin relative to the knee is not under an inertial frame, because knee is accelerated relative to the coin.

a) coin acceleration relative to the ground: g = 9.81 m/s^2. This is the acceleration due to the atraction of the earth, given that the coin is not tied to the car or to the knee.

b) coin acceleration relative to the knee: acceleration of the coin relative to the ground - acceleration of the knee relative to the ground = - g - (-1.24g) = 0.24 g = 0.24 * 9.81 m^2^2 = 2.354 m/s^2

c) time to the coin move 2.2 m upward, relative to the car

d =  a*t^2 / 2 , where a is the acceleration relative to the car (same of the knee)

=> t = √(2d/a) = √(2*2.2m/2.354 m/s^2) = 1.37 s

d) actual force on the coin

Use the acceleration relative to the ground, a=g = 9.81 m/s^2

F = m*a = 0.567 g * 10^-3 kg/g * 9.81 m/s^2 = 5.56*10^-3 N


e) apparent force

Use apparent acceleration, a = 2.354 m/s^2

F = m*a = 0.567*10^-3 kg * 2.354 m/s^2 = 1.33*10^-3 N
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Anna drew a diagram to compare the strong and weak force. Which labels belong in the areas marked X, Y, and Z? X: infinite range
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Answer:

For areas marked X, Y, Z, X is attractive only, Y has a very small range, and Z is attractive and repulsive

Explanation:

Solution

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From the question stated, Anna drew a diagram to compare forces that are strong and weak.

Now,

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calculate the diameter of a silver wire of length 75cm , which is extended by 1.85mm when a 10kg mass is suspended from it's end
sdas [7]

Answer:0.8\ mm

Explanation:

Given

length of wire l=75\ cm

change in length \Delta l=1.85\ mm

mass of wire m=10\ kg

Young's modulus for silver E=7.9\times 10^{10}\ N/m^2

load on wire F=mg

F=10\times 9.8=98\ kg

change in length is given by

\Delta l=\dfrac{Pl}{AE}

Where A=area of cross-section

A=\dfrac{Pl}{\Delta lE}

A=\dfrac{98\times 0.75}{1.85\times 10^{-3}\times 7.9\times 10^{10}}

A=\dfrac{73.5}{14.615\times 10^{7}}

A=5.029\times 10^{-7}\ m^2

also wire is the shape of cylinder so cross-section is given by

A=\dfrac{\pi d^2}{4}=5.029\times 10^{-7}\ m^2

\Rightarrow d^2=\dfrac{5.029\times 10^{-7}\times 4}{\pi }

\Rightarrow d^2=64.02\times 10^{-8}

\Rightarrow d=8\times 10^{-4}\ m

\Rightarrow d=0.8\ mm

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3 years ago
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