You are unloading a refrigerator from a delivery van. The ramp on the van is 5.0 mm long, and its top end is 1.4 mm above the gr
ound. As the refrigerator moves down the ramp, you are on the down side of the ramp trying to slow the motion by pushing horizontally against the refrigerator with a force of 270 NN .How much work do you do on the refrigerator during its trip down the ramp?W = ???
1 answer:
Answer:
Work done is 1.31 J.
Explanation:
Given:
Length of the ramp (L) = 5.0 mm
Height of the top end (H) = 1.4 mm
Horizontal force applied (F) = 270 N
Work done (W) = ?
We know that,
Work done is equal to the product of force and displacement caused along the line of application of force.
Here, the force acting on the refrigerator is in the horizontal direction while the refrigerator is moving down along the length of the ramp. So, we have to first find the horizontal component of the displacement caused.
Now, consider the triangle ABC representing the given situation. The point A is the top end point of the ramp, AB is the length of the ramp, AC is the vertical displacement of the refrigerator and BC is the horizontal displacement of the refrigerator.
Using Pythagoras theorem,
Now, force applied on the refrigerator is in the direction opposite to the horizontal component of displacement. So, work is negative.
Work = Force × Horizontal displacement
Therefore, work done is 1.31 J.
You might be interested in
Complete Question:
A hollow cylinder with an inner radius of 4.0mm and an outer radius of 30mm conducts a 3.0-A current flowing parallel to the axis of the cylinder. If the current density is uniform throughout the wire, what is the magnitude of the magnetic field at a point 12mm from its center?
Answer:
The magnitude of the magnetic field = 7.24 μT
Explanation:
Inner radius, a = 4.0 mm = 0.004 m
Outer radius, b = 30 mm = 0.03 m
Radius, r = 12 mm = 0.012 m
let h² = b² - a²
h² = 0.03² - 0.004²
h² = 0.000884
Let d² = r² - a²
d² = 0.012² - 0.004²
d² = 0.000128
Current I = 3A
μ = 4π * 10⁻⁷
The magnitude of the magnetic field is given by:
B = 7.24 * 10⁻⁶T
B = 7.24 μT
Answer:
The speed of the airplane relative to the air is 209.47km/hr
Explanation:
Whenever we are solving a physics problem, it's really useful to start by drawing a diagram of the problem (See picture attached). It will help us visualize the problem better.
Now, we know that the plane flew for an amount of time of 15 minutes. For our dimensions to be the same, we need to turn those 15min to hours, like this:
Once our time is rewritten as hours, we can now calculate the velocity towards north of the plane.
the plane traveled a distance to the north of 48km so the velocity is:
so
V=192km/hr j
Now, we can calculate the x and y-components of the velocity of the wind. The problem states that the wind is blowing at 42km/hr at an angle of 19° south of east, so the x and y-components of the velocity of the wind are:
and
So the velocity of the wind can be expressed as a vector as:
Once we know this, we can find the velocity of the plane with respect of the wind on x and on y:
and
So the velocity of the plane with respect to the wind can be rewritten as:
Since the problem asks us to find the speed of the plane with respect to the wind, this means that we need to find the magnitude of the velocity, since the speed is a scalar defined to be the magnitude of the velocity.
so:
speed= 209.47 km/hr
Therefore, the speed of the airplane relative to the air is 209.47km/hr
