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nasty-shy [4]
3 years ago
14

A 35 kg box is pulled with a rope from above at an angle of 20. The box begins to move when F 200N. Calculate the frictional for

ce acting on the box. Calculate the normal force acting on the box.

Physics
1 answer:
Otrada [13]3 years ago
3 0

Answer:

b) 188 N

c) 275 N

Explanation:

Draw a free body diagram of the box.  There are four forces:

Applied force F at an angle θ

Weight mg pulling down

Normal force Fn pushing up

Friction force Ff pushing left

(b)

Sum of the forces in the x direction:

∑F = ma

F cos θ − Ff = 0

Ff = F cos θ

Ff = 200 cos 20°

Ff = 188 N

(c)

Sum of the forces in the y direction:

∑F = ma

F sin θ + Fn − mg = 0

Fn = mg − F sin θ

Fn = (35)(9.8) − 200 sin 20°

Fn = 275 N

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KonstantinChe [14]

Explanation:

The net force of each square is the combination of the forces in each direction. The direction is the... direction the square would go in due to the net force. The magnitude of the net force is how large it is. So if you had a force pushing 2N to the left and 1N to the right, then the net force would be 1N to the left; because the two oppose eachother. If they were going in the same direction, then they'd add to each other. And perpendicular net forces (like one pushing up and another pushing left) can create net forces in diagonal directions.

I'm not going to do all of these for you because they're basically all the same thing and it's good practice for you anyway. But I'll do the first three just so you can get the idea:

1. The net force's magnitude is 4N and it's direction is to the right.

2. The net force's magnitude is 4N and it's direction is to the left.

3. The net force's magnitude is 0N and it has no direction because they are equal forces acting in opposite directions.

7 0
3 years ago
Two resistors, R1 = 25.0 ohms and
Leni [432]

Answer: 0.078

Explanation:

Given variables:

R1 = 25.0 , R2 = 64.0 , I1 = 0.200

Step 1: find V

0.200 = v/25

0.200 * 25 = v

5 = v

Step 2: Solve for I2

I = V/R

I = 5/64

I = 0.078125

Hope this helps!

3 0
3 years ago
A ray of monochromatic light (f=5.09x10^14 hz) passes through air and a rectangular transparent block calculate the absolute ind
sasho [114]
The right answer for the question that is being asked and shown above is that: "The angle of refraction of the light ray as it enters the transparent block from air = 20 degrees." A ray of monochromatic light (f=5.09x10^14 hz) passes through air and a rectangular transparent block calculate the absolute index of refraction for the <span>medium of the transparent block </span>

3 0
4 years ago
Speedy Sue, driving at 34.0 m/s, enters a one-lane tunnel. She then observes a slow-moving van 160 m ahead traveling at 5.20 m/s
Zielflug [23.3K]

Answer:

there will be collision

Explanation:

v_{s} =  speed of sue = 34 m/s

v_{v} = speed of van = 5.20 m/s

v_{sv} = speed of sue relative to van  = v_{s} - v_{v} = 34 - 5.20 = 28.8 m/s

d_{s} = stopping distance after brakes are applied

D = distance between sue and van = 160 m

v_{f} = final speed of sue = 0 m/s

a = acceleration = - 1.80 m/s²

Using the kinematics equation

v_{f}^{2} = v_{o}^{2} + 2 a d_{s}

0^{2} = 28.8^{2} + 2 (1.80) d_{s}

d_{s} = 230.4 m

Since  d_{s} < D

hence there will be collision

7 0
4 years ago
Auto companies frequently test the safety of automobiles by putting themthrough crash tests to observe the integrity of the pass
e-lub [12.9K]
Answer:

The average force that brings the car to rest is 175000N

Explanation:

The mass of the car, m = 1000 kg

Initial speed, u = 14 m/s

Final speed, v = 0 m/s (Since the car comes to a stop)

The time taken, t = 8 x 10^-2 s

The average force is calculated as:

\begin{gathered} F=\frac{m(u-v)}{t} \\ F=\frac{1000(14-0)}{8\times10^{-2}} \\ F=\frac{14000}{0.08} \\ F=175000N \end{gathered}

The average force that brings the car to rest is 175000N

4 0
1 year ago
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