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allsm [11]
3 years ago
12

You know your mass is 62 kg but when you stand on a bathroom scale in an elevator it says your mass is 77 kg what is the acceler

ation of the elevator
Physics
1 answer:
lbvjy [14]3 years ago
4 0
In a stationary situation, the weight of person is
W=mg=(62 kg)(9.81 m/s^2) = 608.2 N
This is the weight "felt" by the scale, which is basically the normal reaction applied by the scale on the person, and which uses the value of g (9.81) as reference to convert the weight (602.8 N) into a mass (62 kg).

When the person is in the elevator, the scale says 77 kg. The scale is still using the same value of conversion (9.81), so the apparent weight "felt" by the scale is
W' = m'g=(77 kg)(9.81 m/s^2)=755.4 N
This is the normal reaction applied by the scale on the person, and which is directed upward. Besides this force, there is still the weight W of the person, acting downward. So, if we use Newton's second law:
\sum F = ma
W-W'=ma
where a is the acceleration of the elevator. If we solve for a, we find
a= \frac{W-W'}{m}= \frac{608.2N-755.4N}{62 kg}=-2.37 m/s^2
The negative sign means the acceleration is in the opposite direction of g (which we take positive), so it means the elevator is going upward.
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2 years ago
A very small object carrying -7 μC of charge is attracted to a large, well-anchored, positively charged object. How much kinetic
miss Akunina [59]

Answer:

ΔK.E = 14 nJ

Explanation:

Solution:

- The charge that moves under the influence of an Electric Field produced between a potential difference (V) stores electric potential energy U within that is converted to kinetic energy.

- We will use conservation of energy on the system that contains the charged particle with charge q loses its electric potential energy U as it moves towards positively charged object that converts into a gain in Kinetic energy of the charged particle ΔK.E:

                                 ΔK.E = U

Where,

                                 U = V*q

                                 ΔK.E = V*q

                                 ΔK.E = (7*10^-6)*(2*10^-3)

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3 years ago
A stone is dropped into water from a bridge 52 m above the water's
dusya [7]

Answer:

Its final velocity and how much time it takes to reach the water

Explanation:

The motion of the stone is a uniformly accelerated motion, so we can use the following suvat equation to determine its final velocity:

v^2-u^2=2as

where

v is the final velocity

u = 0 is the initial velocity

a=g=9.8 m/s^2 is the acceleration of gravity

s = 52 m is the distance covered during the fall

Solving for v,

v=\sqrt{u^2+2as}=\sqrt{0^2+2(9.8)(52)}=31.9 m/s

We can also find how much time it takes to reach the water, using the equation

v=u+at

where

v = 31.9 m/s is the final velocity

u = 0 is the initial velocity

a=g=9.8 m/s^2

t is the time

And solving for t,

t=\frac{v-u}{a}=\frac{31.9-0}{9.8}=3.26 s

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3 years ago
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