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cestrela7 [59]
2 years ago
7

An accessory such as a coupling that performs a mechanical rather than an electrical function is called?

Physics
1 answer:
Assoli18 [71]2 years ago
5 0
Fitting is the answer
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All are examples of electromagnetic energy except ______.
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All are examples of electromagnetic energy except <span>circles forming when a rock drops into a pool. The correct option among all the options that are given in the question is the third option or option "C". The other choices can be negated. I hope that this is the answer that has actually come to your help.</span>
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What are the major weather fronts? How do they differ from each other?
dmitriy555 [2]

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they differ each other bc it is a lot of with earth quakes and everything.

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2 years ago
A 4 kg car with frictionless wheels is on a ramp that makes a 25° angle with the horizontal. The car is connected to a 1 kg mass
stellarik [79]
Refer to the diagram shown below.

W₁ = (4 kg)*(9.8 m/s²) = 39.2 N
W₂ = (1 kg)*(9.8 m/s²) = 9.8 N

The normal reaction on the 4-kg mass is
N = (39.2 N)*cos(25°) = 35.5273 N
The force acting down the inclined plane due to the weight is
F = (39.2 N)*sin(25°) = 16.5666 N

The net force that accelerates the 4-kg mass at a m/²s down the plane is
F - W₂ = (4 kg)*(a m/s²)
4a = 16.5666 - 9.8
a = 1.6917 m/s²

Answer: 1.69 m/s²  (nearest hundredth)

7 0
3 years ago
A negative test charge experiences a force to the right as a result of an electric field. Which is the best conclusion to draw
Alona [7]

Answer:a

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4 0
2 years ago
A 25.0-kg child is standing at the edge of a horizontal merry-go-round with a radius of 2.40 m and a moment of inertia of 356 kg
motikmotik

Answer:

\omega_{f}=1.634\ rad/s  

Explanation:

given,  

diameter of merry - go - round = 2.40 m  

moment of inertia = I = 356 kg∙m²

speed of the merry- go-round = 1.80 rad/s

mass of child = 25 kg  

initial angular momentum of the system  

L_i = I\omega_i  

L_i =356\times 1.80  

L_i =640.8\ kg.m^2/s  

final angular momentum of the system  

L_f = (I_{disk}+mR^2)\omega_{f}  

L_f = (356 + 25\times 1.2^2)\omega_{f}  

L_f= (392)\omega_{f}  

from conservation of angular momentum  

L_i = L_f  

640.8= (392)\omega_{f}  

\omega_{f}=1.634\ rad/s  

8 0
2 years ago
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