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Naya [18.7K]
4 years ago
6

Plzzz help will mark the brainliest

Physics
1 answer:
ANEK [815]4 years ago
8 0
Ciara is winging....etc
The answer is : 0.60 N, toward the center of the circle


A satellite....etc
The Answer is : 7400 m/s


What is the .....etc
The Answer is : 2.60 m/s
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A very long nonconducting cylinder of diameter 10.0 cm carries charge distributed uniformly over its surface. Each meter of leng
Murrr4er [49]

Answer:

Explanation:

The concept of electric field, force acting on proton is applied and appropriate derivations were made to calculate the distance from the surface as shown in the attached file.

7 0
3 years ago
A car is moving eastward and speeding up. The momentum of the car is ______.
leva [86]
I think the answer is C. Remaining constant
7 0
3 years ago
What do you mean by velocity ratio of a wheel and axle​
IgorC [24]

Answer:

Explanation:

In wheel and axle. …with the system is the velocity ratio, or the ratio of the velocity (VF) with which the operator pulls the rope at F to the velocity at which the weight W is raised (VW). This ratio is equal to twice the radius of the large drum divided by the difference…

5 0
3 years ago
A soccer ball kicked with a force of 12.5 N accelerates at 6.2 m/s’ to the right. What is the mass of the ball? Answer in units
stepladder [879]

Answer:

m = 2.01[kg]

Explanation:

This problem can be solved using Newton's second law which tells us that the force applied on a body is equal to the product of mass by acceleration.

F =m*a

where:

F = force = 12.5 [N]

m = mass [kg]

a = acceleration = 6.2 [m/s²]

12.5=m*6.2\\m = 2.01[kg]

3 0
3 years ago
A 1.83 kg book is placed on a flat desk. Suppose the coefficient of static friction between the book and the desk is 0.442 and t
fredd [130]

Answer:7.92 N

Explanation:

Given

mass of book m=1.83\ kg

coefficient of static friction \mu _s=0.442

coefficient of kinetic friction \mu _k=0.240

To move the book, one need to overcome  the static friction  

Static friction F_s=\mu _sN

F_s=\mu _s\times 1.83\times 9.8

F_s=0.442\times 1.83\times 9.8

F_s=7.92\ N

After overcoming the Static friction , Force needed to move the block is

F_k=\mu _kN

F_k=0.240\times 1.83\times 9.8

F_k=4.30\ N

8 0
3 years ago
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