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3 years ago

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You are riding in an elevator that is accelerating upward. Suppose you stand on a scale. The reading on the scale is __________.

You are riding in an elevator that is accelerating upward. Suppose you stand on a scale. The reading on the scale is __________.
equal to your true weight
less than your true weight
greater than your true weight

1 answer:

olchik [2.2K]3 years ago

5 0

Answer:

greater than your true weight

Explanation:

When going up in an elevator the acceleration of the elevator is added to the acceleration due to gravity. This will increase the reading on the scale.

The expression of the resultant weight will be

$N=m(a+g)$

where,

m = Mass of the person

g = Acceleration due to gravity = 9.81 m/s²

a = Acceleration of the elevator.

Hence, the reading on the scale is <u>greater than your true weight.</u>

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Answer:

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b) $\theta_1=80.5 \textdegree$

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From the question we are told that:

Initial speed of 1st ball $u_{1}=0 m/s$

Mass of 1st ball $m_1=0.585kg$

Mass of 2nd ball $m_2=0.420kg$

Initial speed of 2nd ball $u_{2}=0.270 m/s$

Final speed of 2nd ball $v_{2}=0.220 m/s$

Angle of collision $\angle=36.9 \textdegree$

a)

Generally the equation for law of conservation is mathematically given by

$m_1u_1+m_2u_2=m_1v_1^2+m_2v_2^2$

The final velocity $v_1$ is given as

$0+(0.420)(0.270)=(0.585)(v_1)^2+(0.420)(0.220)^2$

$(v_1)^2=\frac{(0.420)(0.270)-(0.420)(0.220)^2}{0.585}$

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$(v_1)=0.3989m/s$

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Generally the equation for law of conservation is mathematically given by

$m_1u_1+m_2u_2=m_1v_1cos\theta_1+m_2\theta_2$

$0+(0.420)(0.270)=(0.585)(1.511)cos\theta_1+(0.420)(0.220)cos36 \textdegree$

$cos\theta_1= \frac{(0.420)(0.270)-(0.420)(0.220)cos36 \textdegree}{(0.585)(0.3989)}$

$cos\theta_1=0.1656$

$\theta_1=80.5 \textdegree$

c)

Generally the equation for kinetic energy is mathematically given by

$K.E=\frac{1}{2} mv^2$

1st Ball

$K.E=\frac{1}{2} (0.585)(0.3989)^2$

$K.E=0.0465J$

2nd ball

$K.E=\frac{1}{2} (0.420)(0.220)^2$

$K.E=0.101J$

Therefore the change in the total kinetic energy of the two balls as a result of the collision is

$0.101-0.0465$

$K.E=0.036J$

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