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ICE Princess25 [194]
4 years ago
15

Suppose a person whose mass is m is being held up against the wall with a constant tangential velocity v greater than the minimu

m necessary. find the magnitude of the frictional force between the person and the wall.
Physics
2 answers:
Naddika [18.5K]4 years ago
7 0
In the magnitud is in good
Kamila [148]4 years ago
7 0

Answer:

friction force must be equal to the magnitude of weight of the object

Explanation:

When the object is stuck against the wall

then we will have

F_n = \frac{mv^2}{r}

also we know that

mg = F_f

so if the block is at rest and not sliding then in vertical direction the force must be balanced

so here we can say

F_g = F_f

so friction force must be equal to the magnitude of weight of the object

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2. If you exert a force of 10.0 N to lift a box a distance of 0.9 m, how much work have you done?
KonstantinChe [14]
<span> force of 10.0 N
</span>
<span>distance of 0.9 m
w=f*d
w=10*0.9
=9.0 j</span>
6 0
3 years ago
Read 2 more answers
Object A has of mass 7.20 kilograms, and object B has a mass of 5.75 kilograms. The two objects move along a straight line towar
Vera_Pavlovna [14]
If the collision is elastic, there is no loss in kinetic energies, which means that the total energies before and after impact are the same.  So no need to worry about final velocities.

Final energy
= initial energy
= (1/2) (7.20*2.00^2+5.75*(-1.30)^2)
=19.26 joules

Answer: the total kinetic energy is 19.3 J. after collision.

8 0
4 years ago
Three crates with various contents are pulled by a force F pull = 3615 N across a horizontal, frictionless roller‑conveyor syste
Dmitry_Shevchenko [17]

Answer:

m1=914.9kg

m2=604.9kg

m3=864.75kg

Explanation

I think we are suppose to find the mass of the crate.

The effective force that moves the body in positive x direction is 3615N

ΣFx = Σma

Then Fx=3615N

Then the masses be m1, m2 and m3

Then,

ΣF = Σ(ma)

3615=(m1+m2+m3)a

Given that a=1.516

The masses are

m1+m2+m3=, 2384.56. Equation 1

Between mass 1 and mass 2 is, F12=1387.

The effective force that pull mass 1 is 1387.

F12=m1 ×a

Therefore,

m1=F12/a

m1=1387/1.516

m1=914.9kg.

The effective force that pulls crate 1 and crate 2 is F23

F23=(m1+m2)a

Therefore

2304=(m1+m2)a

Therefore, since a=1.516

m1+m2=2304/1.516

m1+m2=1519.8kg

Since m1=914.9kg

So, m2=1519.8-m1

m2=1519.8-914.9

m2=604.9kg

Also from equation 1

m1+m2+m3=2384.56

Since m1=914.9kg and m2=604.9kg

Then, m3=2384.56-604.9-914.9

m3=864.75kg

3 0
3 years ago
Which of the following quantities would be acceptable representations of weight? Check all that apply.
Oxana [17]

Only 12 lb and 1600 kN .

7 0
3 years ago
Joanna's house is 8000 feet due west of her school. If her house is assigned the position of zero and her school is assigned the
olchik [2.2K]

Answer:

Joanna's position would be -100.

Explanation:

Joanna's position would be 100 feets west to her house. Her position is negative because Joanna walked west of her house. The position would have been positive if she had moved east of his house, that is, to the school.

8 0
3 years ago
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