We are given a box that slides up a ramp. To determine the force of friction we will use the following relationship:
Where.
To determine the Normal force we will add the forces in the direction perpendicular to the ramp, we will call this direction the y-direction as shown in the following diagram:
In the diagram we have:
Adding the forces in the y-direction we get:
Since there is no movement in the y-direction the sum of forces must be equal to zero:
Now we solve for the normal force:
To determine the y-component of the weight we will use the trigonometric function cosine:
Now we multiply both sides by "mg":
Now we substitute this value in the expression for the normal force:
Now we substitute this in the expression for the friction force:
Now we substitute the given values:
Solving the operations:
Therefore, the force of friction is 15.01 Newtons.
Answer:
26.82m/s
Explanation:
Given
Mass = m= 0.4kg
Initial Velocity = u = 0
Charge = 4.0E-5C
Distance= d = 0.5m
Object Charge = 2E-4C
First, we'll calculate the initial energy (E)
E = Potential Energy
PE = kQq / d
Where k = coulomb constant = 8.99E9Nm²/C²
Energy is then calculated by;
PE = 8.99E9 * 4E-5 * 2E-4 / 0.5
PE = 143.84J
Energy = Potential Energy = Kinetic Energy
K.E = ½mv² = 143.84J
½mv² = ½ * 0.40 * v² = 143.85
0.2v² = 143.85
v² = 143.85/0.2
v² = 719.25
v = √719.25
v = 26.81883666380777
v = 26.82m/s
Hence, the object is 26.82m/s fast when the cart moving is very far (infinity) from the fixed charge
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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Answer:
It's the duration between successive new moons. Also called a lunation or synodic month, it has a mean period of 29.53059 days (29 days 12 hours and 44 minutes).
