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SpyIntel [72]
3 years ago
5

What effort force will be required to lift the 20 N object using the pulley above?

Physics
1 answer:
Umnica [9.8K]3 years ago
4 0

Answer:

I think it is 5N.

Explanation:

You might be interested in
a test tube has a diameter of 3cm . how many turns would a piece of thread of length 90.42 make round test tube​
nadya68 [22]

Answer:

4.8 turns would be made around the tube

Explanation:

You need the circumference of the tube since it's just a lifted circle

2×pi×3=18.85

90.42/18.85=4.79

And round that to 4.8

6 0
3 years ago
1. During the Middle Ages, armies often attacked castles using large siege engines such as the counterweight trebuchet at left.
fredd [130]

Answer: Some challenges that I could see would be the walls, possibly a moat, tar, and the towers.

Explanation: The wall is obliviously a main problem, trying to get over it or through it is a difficult challenge. The moat (if it has one) means that there is more than likely only one way to get in or out.  If it does have tar it means that the attackers are going to be put in a "sticky situation" And finally the towers, they have people at the top shooting arrows down at you, or throwing things at you.

8 0
3 years ago
A wire has a cross sectional area of 4.00 mm2 and is stretched by 0.100 mm by a certain force. How far will a wire of the same m
Nina [5.8K]

Answer: 0.05\ mm

Explanation:

Given

Cross-sectional area of wire A_1=4\ mm^2

Extension of wire \delta l=0.1\ mm

Extension in a wire is given by

\Rightarrow \delta l=\dfrac{FL}{AE}

where, E=\text{Youngs modulus}

\Rightarrow \delta_1=\dfrac{FL}{A_1E}\quad \ldots(i)

for same force, length and material

\Rightarrow \delta_2=\dfrac{FL}{A_2E}\quad \ldots(ii)

Divide (i) and (ii)

\Rightarrow \dfrac{0.1}{\delta_2}=\dfrac{A_2}{A_1}\\\\\Rightarrow \delta_2=0.1\times \dfrac{4}{8}\\\\\Rightarrow \delta_2=0.05\ mm

5 0
2 years ago
A police officer is called to the scene of a car accident in his accident he sketches the scene and describes it according to hi
abruzzese [7]

Answer:

The driver was not telling the truth because it is not possible for a car to hit another car from behind and generate a force to the sides that deflects it from its path.

Explanation:

First, we analyze the driver's statement.

The driver when arriving at the curve, is collided from behind by another car and deviates from his path and crashes into a tree. For the car to go to the tree there must be a force towards the tree.

The net force that causes the car to deviate must be formed by the sum of the motion vector of the first car plus the force that is directed towards the tree.

Here we verify that a car hitting from behind will not generate a force to the sides, but will generate a force in the same direction that the car moves, forward.

7 0
3 years ago
Two identical 0.400 kg masses are pressed against opposite ends of a light spring of force constant 1.75 N/cm, compressing the s
nlexa [21]

Answer:

0.853 m/s

Explanation:

Total energy stored in the spring = Total kinetic energy of the masses.

1/2ke² = 1/2m'v².................... Equation 1

Where k = spring constant of the spring, e = extension, m' = total mass, v = speed of the masses.

make v the subject of the equation,

v = e[√(k/m')].................... Equation 2

Given: e = 39 cm = 0.39 m, m' = 0.4+0.4 = 0.8 kg, k = 1.75 N/cm = 175 N/m.

Substitute into equation 2

v = 0.39[√(1.75/0.8)

v = 0.39[2.1875]

v = 0.853 m/s

Hence the speed of each mass = 0.853 m/s

7 0
3 years ago
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