Answer:
(a) x0 = 0m and y0 = 49.0m
(b) Vox = 15.0m/s Voy = 0m/s
(c) Vx = Vo = 15.0m/s and Vy = -gt
(d) X = 15.0t and y = 49.0 - 4.9t²
(e) t = 3.16s
(f) Vf = 34.4m/s
Explanation:
Complete Question:
Metal sphere A has a charge of − Q . −Q. An identical metal sphere B has a charge of + 2 Q . +2Q. The magnitude of the electric force on sphere B due to sphere A is F . F. The magnitude of the electric force on sphere A due to sphere B must be:
A. 2F
B. F/4
C. F/2
D. F
E. 4F
Answer:
D.
Explanation:
If both spheres can be treated as point charges, they must obey the Coulomb's law, that can be written as follows (in magnitude):
As it can be seen, this force is proportional to the product of the charges, so it must be the same for both charges.
As this force obeys also the Newton's 3rd Law, we conclude that the magnitude of the electric force on sphere A due to sphere B, must be equal to the the magnitude of the force on the sphere B due to the sphere A, i.e., just F.
Move the objects faster to get more friction.
efficiency = (useful energy transferred ÷ energy supplied) × 100
It's easy to use this formula, but we have to know both the useful energy and the energy supplied. The drawing doesn't tell us the useful energy, so we have to find a clever way to figure it out. I see two ways to do it:
<u>Way #1:</u>
We all know about the law of conservation of energy. So we know that the total energy coming out must be 250J, because that's how much energy is going in. The wasted energy is 75J, so the rest of the 250J must be the useful energy . . . (250J - 75J) = 175J useful energy.
(useful energy) / (energy supplied) = (175J) / (250J) = <em>70% efficiency</em>
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<u>Way #2: </u>
How much of the energy is wasted ? . . . 75J wasted
What percentage of the Input is that 75J ? . . . 75/250 = 30% wasted
30% of the input energy is wasted. That leaves the other <em>70%</em> to be useful energy.
Answer:
1st statement is true
Explanation:
Here statement 1 is correct
Let think about it, if you push down the bar then you are lifting your weight off the pedals.
Obviously, this question does not take into account of racing bikes with straps on pedals, where you would push on one side and pull on the other to match your legs are doing, with straps your other leg can pull pedals upward.
