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erastovalidia [21]
4 years ago
8

The force pressing two stacked objects together is equal to the weight of the object on top. The friction between a piece of woo

d on a stone surface is 22.5 newtons. The same piece of wood is placed on another surface, and the friction between the wood and the surface is now 46.5 newtons.
What could be the second surface?

A) metal
B) concrete
C) brick
D) skin
Physics
1 answer:
34kurt4 years ago
3 0

Answer: B. Concrete

Explanation:

Let N = reacting force pressing the bodies in context together (units in Newtons),

The question stated that the force pressing the two mounted/stacked objects together is equal to the weight of the object on top.

We need to start by finding the weight of the piece of wood.

friction is given by

f = μN

The value of f is 22.5,

and from the chart reference the coefficient of friction between wood and stone, μ is 0.30.

22.5 = 75. 0.30

Putting the values into the equation: 22.5 = 0.30N.

Divide both sides by 0.30 to find the value of N:

N= 22.5/0.3 = 75

Now that the piece of wood will be placed on another surface, its weight of 75 Newton is the force pressing the two bodies together.

To determine the new surface, you should find the new coefficient of friction by using the new value of the force of friction given 46.5:

46.5 = µ(75).

Divide both sides by 75 to isolate μ.

The refer chart also indicates that the coefficient of friction equals 0.62 between wood and concrete, so the new surface corresponding to 0.62 is the concrete, which is (B).

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Inessa [10]

Answer:

W=2219pounds

Explanation:

If the weight  is a linear function of the amount of fuel, the following correlation is fulfilled :

\frac{2153pounds-1999pounds}{46gallons-18gallons} = \frac{W-1999pounds}{58gallons-18galons}

we solve the equation:

W=2219pounds

6 0
3 years ago
Earning Goal: To be able to calculate work done by a constant force directed at different angles relative to displacement
lana [24]

Answer:

the work done by the 30N force is 4156.92 J.

For this problem, they don´t ask you to determine the work of the total force applied in the block. They only want the work done for the force of 30N, with an angle of 30º respectively of the displacement and a traveled distance of 160m. So:

W=F·s·cos(α)=30N·160m·cos(30º)=4156.92J

8 0
3 years ago
Part A
7nadin3 [17]

Answer:

2.5 m/s²

Explanation:

Using the formula, v = u + at ( v = Final velocity; u = Initial velocity; t = Time; a = Acceleration)

25 = 0 + 10a

a = 25/10 = 2.5 m/s²

8 0
3 years ago
At a race car driving event, a staff member notices that the skid marks left by the race car are 9.06 m long. The very experienc
vaieri [72.5K]

Answer:

27.1 m/s

Explanation:

Given that at a race car driving event, a staff member notices that the skid marks left by the race car are 9.06 m long. The very experienced staff member knows that the deceleration of a car when skidding is -40.52 m/s2.

Using third equation of motion,

V^2 = U^2 + 2aS

Since the car is decelerating, the final velocity V = 0

Substitute all the parameter into the equation above,

0 = U^2 - 2 * 40.52 * 9.06

U^2 = 734.22

U = \sqrt{734.22}

U = 27.096

U = 27.1 m/s  approximately

Therefore, the staff member can estimate for the original speed of the race car to be 27.1 m/s if it came to a stop during the skid

5 0
3 years ago
A 25kg chair initially at rest on a horizontal floor requires a 165 N horizontal force to set it in motion. Once the chair is in
SCORPION-xisa [38]

Answer:

\mu_k=0.51  

Explanation:

Given that

Mass , m = 25 kg

We know that when body is in rest condition then static friction force act on the body and when body is in motion the kinetic friction force act on the body .That is why these two forces are given as follows

Static friction force ,fs= 165 N

Kinetic friction force ,fk = 127 N

If the body is moving with constant velocity ,it means that acceleration of that body is zero and all the forces are balanced.

Lets take coefficient of kinetic friction  = μk

The kinetic friction is given as follows

fk = μk  m g

Now by putting the values

127 = μk x 25 x 9.81

\mu_k=\dfrac{127}{25\times 9.81}

\mu_k=0.51

Therefore the value of coefficient of kinetic friction will be 0.51

4 0
3 years ago
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