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tamaranim1 [39]
2 years ago
10

For elevator questions, does the weight of a person increase when the elevator is going up or if it's going down-

Physics
1 answer:
TiliK225 [7]2 years ago
6 0

The weight of a person increase when the elevator is going up.

<h3>Weight of the person in the elevator</h3>

The weight of the person in the elevator is calculated as follows;

<h3>When the person is going up</h3>

F = ma + mg

F = m(a + g)

where;

  • a is acceleration of the person
  • g is acceleration due to gravity

<h3>When the person is going down</h3>

F = mg - ma

F = m(g - a)

Thus, the weight of a person increase when the elevator is going up.

Learn more about weight here: brainly.com/question/2337612

#SPJ1

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Answer:

Adding a reducing agent

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Homodimeric proteins are formed by non-covalent interactions between two identical proteins (monomers).

Some of these intermolecular interactions are mediated by disulphide bonds which can be broken by a  reducing agent like beta-mercaptoethanol (2-ME).

By adding the reducing agent to the gel, it is probable that the dimer is broken and the band observed will correspond to the two identical 22.5-kDa monomers.

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4 years ago
An force of 50 N is applied to a shovel, which acts as a lever to
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Answer:

the answer is 18

Explanation:

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A truck is hauling a 300-kg log out of a ditch using a winch attached to the back of the truck. Knowing the winch applies a cons
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Answer:

0.128 s

Explanation:

We have to start by calculating the net force acting on the log. We have two forces:

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- The frictional force, backward

The frictional force is given by

F_f = \mu mg

where

\mu=0.45 is the coefficient of friction

m = 300 kg is the mass of the log

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

Substituting,

F_f = (0.45)(300)(9.8)=1323 N

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Now, we can find the acceleration of the log by using Newton's second law

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a=\frac{F}{m}=\frac{1177}{300}=3.92 m/s^2

And finally we can find the time it takes for the log to reach a speed of

v = 0.5 m/s

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4 years ago
A force of 1.200×103 N pushes a man on a bicycle forward. Air resistance pushes against him with a force of 615 N. If he starts
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Answer:

15.45 m/s

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Net force, Fnet = 1.2 x 1000 - 615 = 1200 - 615 = 585 N

acceleration = Fnet / mass = 585 / 122.45 = 4.78 m/s^2

Use third equation of motion

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