Answer:
a
b
Explanation:
From the question we are told that
The child's weight is 
The length of the sliding surface of the playground is 
The coefficient of friction is 
The angle is 
The initial speed is 
Generally the normal force acting on the child is mathematically represented as
=> 
Note 
Generally the frictional force between the slide and the child is
Generally the resultant force acting on the child due to her weight and the frictional force is mathematically represented as
Here F is the resultant force and it is represented as 
=> 
=> 
=> 
=>
So
=> 
Generally the heat energy generated by the frictional force which equivalent tot the workdone by the frictional force is mathematically represented as
=> 
=>
Generally from kinematic equation we have that
=> 
=> 
=>
Answer:
cross out the false piece in blue and write the true piece in red
Answer:
The unit you should use for work done and energy is the joule (J) which is indeed the same as the newton metre (N m).
There is another physical quantity which is the product of force and distance and that is torque or moment of a force.
The unit you should use for torque is the newton metre (Nm) and not the joule.
Naming the units of work done and torque differently helps to emphasis the fact that work done and torque refer to two different physical quantities although the definitions of both quantities have the product of force and distance in them.
work done=force→⋅displacement→ and torque→=force→×displacement→
Hope I helped
Answer: A 100-lb child stands on a scale while riding in an elevator. Then, the scale reading approaches to 100lb, while the elevator slows to stop at the lowest floor
Explanation: To find the correct answer, we need to know more about the apparent weight of a body in a lift.
<h3>What is the apparent weight of a body in a lift?</h3>
- Consider a body of mass m kept on a weighing machine in a lift.
- The readings on the machine is the force exerted by the body on the machine(action), which is equal to the force exerted by the machine on the body(reaction).
- The reaction we get as the weight recorded by the machine, and it is called the apparent weight.
<h3>How to solve the question?</h3>
- Here we have given with the actual weight of the body as 100lbs.
- This 100lb child is standing on the scale or the weighing machine, when it is riding .
- During this condition, the acceleration of the lift is towards downward, and thus, a force of ma .
- There is also<em> mg </em>downwards and a normal reaction in the upward direction.
- when we equate both the upward force and downward force, we get,
i.e. during riding the scale reads a weight less than that of actual weight.
- When the lift goes slow and stops the lowest floor, then the acceleration will be approaches to zero.
Thus, from the above explanation, it is clear that ,when the elevator moves to the lowest floor slowly and stops, then the apparent weight will become the actual weight.
Learn more about the apparent weight of the body in a lift here:
brainly.com/question/28045397
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