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Ivanshal [37]
4 years ago
5

Two forces are acting on an object, but the net force on the object in 0n. For the net force to be 0n, all the forces acting on

the object must cancel each other out. What must must be true for the two forces acting on the object to cancel each other out?
Physics
1 answer:
s344n2d4d5 [400]4 years ago
8 0

The forces acting on the book (Fn and Fg) are equal. Therefore, the netforce is zero and it is an unbalanced force.

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A metal object is four feet away from a magnet. If you move the object two feet closer toward the
Temka [501]
The flux will be 4 times stronger ( due to the inverse square law )
8 0
4 years ago
A block of 250-mm length and 48 × 40-mm cross section is to support a centric compressive load P. The material to be used is a b
Marrrta [24]

Answer:

153.6 kN

Explanation:

The elastic constant k of the block is

k = E * A/l

k = 95*10^9 * 0.048*0.04/0.25 = 729.6 MN/m

0.12% of the original length is:

0.0012 * 0.25 m = 0.0003  m

Hooke's law:

F = x * k

Where x is the change in length

F = 0.0003 * 729.6*10^6 = 218.88 kN (maximum force admissible by deformation)

The compressive load will generate a stress of

σ = F / A

F = σ * A

F = 80*10^6 * 0.048 * 0.04 = 153.6 kN

The smallest admisible load is 153.6 kN

8 0
3 years ago
John runs 120 meters in 10 seconds and then runs back to where he started in another 10 seconds. Which statement is true? Rememb
atroni [7]

Answer:

B

Explanation:

<em>A. His speed is 0 m/s </em>

<em>B. His velocity is 12 m/s </em>

<em>C. His velocity is 0 m/s </em>

<em>D. His acceleration is 12 m/s</em>

Total distance traveled by John = 120 + 120 = 240 meters

Total time taken by John to cover the distance = 10 + 10 = 20 s

<em>Average speed of John = total distance traveled/total time taken</em>

      = 240/20 = 12 m/s

Hence, the average speed/velocity of John throughout the journey is 12 m/s.

The correct option is B.

4 0
3 years ago
The moon's gravity is one-sixth that of the earth. what is the period of a 2.00m long pendulum on the moon
erik [133]

Answer:

6.96 s

Explanation:

The period of a simple pendulum is given by:

T=2 \pi \sqrt{\frac{L}{g}}

where

L is the length of the pendulum and g the acceleration due to gravity.

In this problem, we have a pendulum with length L = 2.00 m, while the acceleration due to gravity is 1/6 that of the earth:

g' = \frac{g}{6}=\frac{9.8 m/s^2}{6}=1.63 m/s^2

So, the period of the pendulum on the moon is

T=2 \pi \sqrt{\frac{2.00 m}{1.63 m/s^2}}=6.96 s

3 0
3 years ago
Can someone who knows slot about engineering explain to me by using an example on how to properly read a triangular cone shaped
Vlad1618 [11]

srry cant help u there

4 0
3 years ago
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