Answer:
v = 42.92 m/s
Explanation:
Given,
initial speed of the ball, v = 11 m/s
time taken to hit the ground = 5.5 m/s
velocity of the ball just before it hit the ground, v = ?
time taken by the ball to reach the maximum height
using equation of motion
v = u + at
final velocity = 0 m/s
0 = 11 - 9.8 t
t = 1.12 s.
time taken by the ball to reach the water from the maximum height
t' - 5.5 -1.12 = 4.38 s
using equation of motion for the calculation of speed just before it hit the water.
v = u + a t
v = 0 + 9.8 x 4.38
v = 42.92 m/s
Velocity of the ball just before it reaches the water is equal to v = 42.92 m/s
Answer:
7.1 J
Explanation:
From the question,
Work done by the mover = work done in pushing the crate + work done against friction
W = W'+Wf................. Equation 1
W = mgd+mgμd............ Equation 2
W = mgd(1+μ)................ Equation 3
Where m = mass of the crate, g = acceleration due to gravity, d = distance, μ = coefficient of friction.
Given: m = 46 kg, d = 10.5 mm = 0.0105 m, μ = 0.5
constant: g = 9.8 m/s²
Substitute these values into equation 3
W = 46×9.8×0.0105(1+0.5)
W = 7.1 J
Answer:
The change in the charge on the positve plate when the Teflon is inserted is +2.5 nC.
Explanation:
- It can be showed that the capacitance of a parallel-plate capacitor, can be expresssed as follows:
- Where ε, is the dielectric constant of the material that fills the space between plates.
- When this space is filled with air, ε= ε₀ = 8,85*10⁻¹² F/m.
- At the same time, the capacitance of a capacitor, by definition, is as follows:
- If we insert a Teflon slab, in such a way that fills completely the gap between the plates, all other parameters being equal, if ε = 2*ε₀, this means that C₂ = 2* C₁. = 50 pF
- As V₂=V₁ (due to the capacitor remains connected to the same battery) the charge must be the double, so Q₂ = 2* Q₁ = 5 nC.
- So, the change in the charge of the positive plate is +2.5 nC.
The components in a circuit don't determine the voltages in it.
The voltages are all determined by the battery or power supply
that energizes the circuit.
