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Inessa [10]
3 years ago
12

Three identical 6.0-kg cubes are placed on a horizontal frictionless surface in contact with one another. The cubes are lined up

from left to right and a force is applied to the left side of the left cube causing all three cubes to accelerate to the right at 2.0 m/s2. What is the magnitude of the force exerted on the middle cube by the left cube in this case
Physics
1 answer:
Aliun [14]3 years ago
6 0

Answer:

24 Newtons

Explanation:

The force exerted in the middle cube needs to be enough to move the middle cube and the right cube with an acceleration of 2 m/s2.

The mass of those two cubes combined is 6 + 6 = 12 kg

So, using the following equation, we can find the force:

Force = mass * acceleration

Force = 12 * 2

Force = 24 Newtons

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A cat of mass of 1kg runs at a speed of 2.5 m/s what is the momentum of the cat?
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A famous physics tale is about a rich man who was found dead. He was
anyanavicka [17]
<h2>Answer: C) He could have thrown the bag of money sideways, creating a  horizontal reaction force on himself.</h2>

Explanation:

According to Newton's third law of motion, when two bodies interact between them, appear equal forces and opposite senses in each of them.

To understand it better:

Each time a body or object exerts a force on a second body or object, it (the second body) will exert a force of equal magnitude and direction but in the opposite direction on the first.

So, if the rich man had pushed the bag of money horizontally opposite of where he was, he could have saved himself.

7 0
3 years ago
A stretched string is 2.11 m long and has a mass of 19.5 g. When the string oscillates at 440 Hz , which is the frequency of the
blsea [12.9K]

Answer:

The tension of the string is 41.876 N

Explanation:

Given;

length of the string, L = 2.11 m

mass of the string, m = 19.5 g = 0.0195 kg

frequency of the wave, f = 440 Hz

wavelength, λ = 15.3 cm = 0.153 m

The velocity of the wave is given by;

v = fλ

v = 440 x 0.153

v = 67.32 m/s

Also the velocity of the wave is given by

v = \sqrt{\frac{T}{\mu} }

where;

μ is mass per unit length = 0.0195 / 2.11 = 0.00924 kg/m

T is the tension of the string

T = v²μ

T = (67.32)²(0.00924)

T = 41.876 N

Therefore, the tension of the string is 41.876 N

5 0
4 years ago
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