A child pulls a wagon at a constant velocity along a level sidewalk. The child does this by applying a 22 newton force to the wa
1 answer:
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Answer:
Explanation:
The electric field inside a parallel plate capacitor is
where A is the area of one of the plates, and Q is the charge on the capacitor.
The electric force on the electron is
where q is the charge of the electron.
By definition the capacitance of the capacitor is given by
Plugging this identity into the force equation above gives
The work done by this force is equal to change in kinetic energy.
W = Fx = (30q)(0.05) = 1.5q = K
The charge of the electron is
Therefore, the kinetic energy is
Answer:
Option B, 93 cm
Explanation:
An diagram of the seed's motion is attached to this solution.
This is very close to a projectile motion question. And the quantity to be calculated, how far along the grant a seed released would travel is called the Range.
And this would be obtained from the equations of motion,
First of, the height of the plant is related to some quantities of the motion with this relation.
H = u(y) t + 0.5g(t^2)
U(y) = initial vertical component of velocity = 0 m/s, H = height at which motion began, = 20cm = 0.2 m
That means t = √(2H/g)
The horizontal distance covered, R,
R = u(x) t + 0.5g(t^2) = u(x) t (the second part of the equation goes to zero as the vertical component of the acceleration of this motion is 0)
(substituting the t = √(2H/g) derived from above
R = u(x) √(2H/g)
Where u(x) = the initial horizontal component of the bomb's velocity = maximum initial speed, that is, 4.6 m/s, H = vertical height at which the seed was released = 20 cm = 0.2 m, g = acceleration due to gravity = 9.8 m/s2
R = 4.6 √(2×0.2/9.8) = 0.929 m = 0.93 m = 93 cm. Option B.
QED!
Answer:
7350 J
Explanation:
The gravitational potential energy of the rock sitting on the edge of the cliff is given by:
where
m is the mass of the rock
g is the gravitational acceleration
h is the height of the cliff
In this problem, we have
m = 50 kg
g = 9.8 m/s^2
h = 15 m
Substituting numbers into the formula, we find: