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Viktor [21]
2 years ago
13

if a person with a mass of 70kg is standing on scale in an elevator when the cable snaps, what will the reading on the scale be

during free-fall?
Physics
1 answer:
dybincka [34]2 years ago
3 0

Answer:

The reading will be the same.

Explanation:

Mass does not depend upon anything and it remains the same anywhere. What changes is the weight of the body because it depends upon gravity and is different at different places.

Giving me the brainest will be helpful.

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a stone with a mass of 2.40 kg is moving with velocity (6.60î − 2.40ĵ) m/s. find the net work (in j) on the stone if its velocit
ch4aika [34]

By the work energy theorem, the total work done on the stone is given by its change in kinetic energy,

W = \Delta K = \dfrac m2 ({v_2}^2 - {v_1}^2)

We have

\vec v_1 = (6.60\,\vec\imath - 2.40\,\vec\jmath)\dfrac{\rm m}{\rm s} \implies {v_1}^2 = \|\vec v_1\|^2 = 49.32 \dfrac{\rm m^2}{\rm s^2}

\vec v_2 = (8.00\,\vec\imath + 4.00\,\vec\jmath) \dfrac{\rm m}{\rm s} \implies {v_2}^2 = \|\vec v_2\|^2 = 80.0\dfrac{\mathrm m^2}{\mathrm s^2}

Then the total work is

W = \dfrac{2.40\,\rm kg}2 \left(80.0\dfrac{\rm m^2}{\rm s^2} - 49.32\dfrac{\rm m^2}{\rm s^2}\right)  \approx \boxed{36.8\,\rm J}

5 0
1 year ago
Joe runs 10 m north, 20 m south, 9m south, and then 15 m north. What is Joe's<br> displacement?
tresset_1 [31]

Answer:

Joes displacement is 54m

4 0
1 year ago
X + 10 times X = 50 what is the answer
tia_tia [17]

Answer:

4.54

Explanation:

X+10X=50

11X=50

X=4.54#

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3 0
3 years ago
If a cell phone is dropped from a very tall building, how far has the phone fallen after 2.7 seconds, neglecting air resistance?
Zielflug [23.3K]
The free fall of the phone is an uniformly accelerated motion toward the ground, with constant acceleration equal to
g=9.81 m/s^2

So, assuming the downward direction as positive direction of the motion, since the phone starts from rest the distance covered by the phone after a time t is given by
y(t) =  \frac{1}{2}gt^2
And if we substitute t=2.7 s, we find the distance covered:
y(t)=  \frac{1}{2}(9.81 m/s^2)(2.7 s)^2=35.8 m
6 0
2 years ago
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