When a loose brick is resting on a wall, it has energy. When the brick is pushed off the wall and is falling down, the amount of
2 answers:
The complete sentence is as follows:
When a loose brick is resting on a wall, it has potential energy. When the brick is pushed off the wall and is falling down, the amount of potential energy is decreasing while the amount of kinetic energy is increasing.
Explanation:
- Potential energy is the energy related to the position of an object: in particular, gravitational potential energy is given by
where m is the mass of the object, g is the gravitational acceleration, and h is the height of the object above the ground.
- Kinetic energy is the energy related to the motion of an object, and it is given by
where m is the mass and v the speed of the object.
In the example, we see that:
- at the beginning, the brick is at rest on the wall, so it only has potential energy due to its height above the ground (while the kinetic energy is zero, since it is at rest, so speed v is zero)
- as it falls down, the potential energy decreases, because the height h decreases, while the kinetic energy increases, because the speed of the object increases as it falls down.
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Answer:
The acceleration of the crate is 1.8 m/s² so the answer is a.
Explanation:
The very first thing you must do when solving this problem is to draw a free body diagram. (The body diagram is attached to this answer)
So once we got the free body diagram, we can analyze it and build our sum of forces in the x and y directions. Notice that according to the diagram, there are 4 forces to this problem, Normal (N), Weight (W), kinetic friction (fk) and the 750N force.
As one may see in the free body diagram, two of the forces are vertical forces: N and W, so we can use them to build a sum of forces:
Starting with the sum of forces in the y-direction, we get:
Σ=0
We set the sum equal to zero because there is no movement in the y-direction, so the system is in vertical equilibrium.
so the sum will be:
when solving for N we get that:
N=W
where W is found by multiplying the mass of the crate by the acceleration of gravity:
N=250kg*9.8m/s²
N=2450N
Once we found the normal force, we can use it to find the kinetic friction which is given by the following formula:
μ
where μ is the kinetic friction coefficient.
So we get that the kinetic friction is:
so
With this information we can go ahead and find the sum of horizontal forces:
Σ=ma
In this case the sum is equal to mass times acceleration because the crate is moving horizontally due to the action of a force, so it will have an acceleration.
so the sum of forces look like this:
750N-=ma
so
750N-294N=(250kg)a
when solving for a we get:
so the crate's acceleration is 1.82m/s².
Answer:
Δh_air=714m
Explanation:
Given data
Solution
ΔP=P₁-P₂
=(ΔhHg)×pHg×g
=(Δh_air)× p_air ×g
Then
Δh_air=(pHg+ΔhHg)÷p_air
Δh_air=714m