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Galina-37 [17]
3 years ago
11

2. A woman prevents a 3kg brick from falling by pressing it against a vertical wall. The coefficient of friction

Physics
1 answer:
Anettt [7]3 years ago
6 0
<h3>Answer:</h3>

49 N

<h3>Explanation:</h3>

<u>We are given;</u>

  • Mass of the brick as 3 kg
  • The coefficient of friction as 0.6

We are required to determine the force that must be applied by the woman so the brick does not fall.

  • We need to importantly note that;
  • For the brick not to fall the, the force due to gravity is equal to the friction force acting on the brick.
  • That is; Friction force = Mg

But; Friction force = μ F

Therefore;

μ F = mg

0.6 F = 3 × 9.8

0.6 F = 29.4

      F = 49 N

Therefore, she must use a force of 49 N

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sammy [17]

Answer:

The correct answer is in the x-axis direction, where the distance travelled is 1.0 cm and the surface is 8 cm^{2}.

Explanation:

The electrical resistance is lower when the contact surface is larger and the length is shorter.

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R = ρ * \frac{l}{S}

Where:

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l: Length.

S: Surface of the conductor.

Thus, the direction in which the electrical resistance is lower is when the distance is minimum and the area is greater.

Have a nice day!

8 0
3 years ago
What happens to the car if it is traveling in a circular path suddenly encounters ice?
pychu [463]

the car's friction is reduced to zero

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the car continues in a straight path from the point at which it encountered the ice

8 0
3 years ago
Read 2 more answers
A moving curling stone, A, collides head on with stationary stone, B. Stone B has a larger mass than stone A. If friction is neg
Kitty [74]

Answer:

The correct answer is option 'c': Smaller stone rebounds while as larger stone remains stationary.

Explanation:

Let the velocity and the mass of the smaller stone be 'm' and 'v' respectively

and the mass of big rock be 'M'

Initial momentum of the system equals

p_i=mv+0=mv

Now let after the collision the small stone move with a velocity v' and the big roch move with a velocity V'

Thus the final momentum of the system is

p_f=mv'+MV'

Equating initial and the final momenta we get

mv=mv'+MV'\\\\m(v-v')=MV'.....i

Now since the surface is frictionless thus the energy is also conserved thus

E_i=\frac{1}{2}mv^2

Similarly the final energy becomes

E_f=\frac{1}{2}mv'^2+\frac{1}{2}MV'^2\

Equating initial and final energies we get

\frac{1}{2}mv^2=\frac{1}{2}mv'^2+\frac{1}{2}MV'^2\\\\mv^2=mv'^2+MV'^2\\\\m(v^2-v'^2)=MV'^2\\\\m(v-v')(v+v')=MV'^2......(ii)

Solving i and ii we get

v+v'=V'

Using this in equation i we get

v'=\frac{v(m-M)}{(M-m)}=-v

Thus putting v = -v' in equation i  we get V' = 0

This implies Smaller stone rebounds while as larger stone remains stationary.

4 0
3 years ago
Newton began his academic career in 1667 for how long was he a working scientists was he a very productive scientist
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6 0
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The work done when a force moves a body through a distance of 15m is 1800j. What is the value of the force applied
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Answer:

120

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