Potential energy can be converted into kinetic energy, a good example of this is when a pole-vaulter bends the pole during a lea
2 answers:
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Answer:
Fnet = 0
Explanation:
- Since the block slides across the floor at constant speed, this means that it's not accelerated.
- According Newton's 2nd Law, if the acceleration is zero, the net force on the sliding mass must be zero.
- This means that there must be a friction force opposing to the horizontal component of the applied force, equal in magnitude to it:
- In the vertical direction, the block is not accelerated either, so the sum of the normal force and the vertical component of the applied force, must be equal in magnitude to the force of gravity on the block:
⇒ 169 N + Fn = Fg = 216 N (3)
- This means that there must be a normal force equal to the difference between Fappy and Fg, as follows:
- Fn = 216 N - 169 N = 47 N (4)
Answer:
Explanation:
A parallel-plate capacitors consist of two parallel plates charged with opposite charge.
Since the distance between the plates (1 cm) is very small compared to the side of the plates (19 cm), we can consider these two plates as two infinite sheets of charge.
The electric field between two infinite sheets with opposite charge is:
where
is the surface charge density, where
Q is the charge on the plate
A is the area of the plate
is the vacuum permittivity
In this problem:
- The side of one plate is
L = 19 cm = 0.19 m
So the area is
Here we want to find the maximum charge that can be stored on the plates such that the value of the electric field does not overcome:
Substituting this value into the previous formula and re-arranging it for Q, we find the charge:
I can't answer this question without a figure. I've found a similar problem as shown in the first picture attached. When adding vectors, you don't have to add the magnitudes only, because vectors also have to factor in the directions. To find the resultant vector C, connect the end tails of the individual vectors. <em>The red line (second picture) represents the vector C.</em>
