Answer:
nothing, mass is not affected by gravitational force
Explanation:
Weight is the gravitational force a planet exerts on a mass on the surface.
It is the product of the mass of an object with the gravitational acceleration that the planet produces.
The weight is the gravitational force
where,
m = Mass of the object
g = Acceleration due to gravity = 9.81 m/s²
Mass is the property that matter has which opposes the force being applied to it. It is intrinsic to the object itself and does not change according to the gravitational force. But, the weight changes.
<span>a) 10.0 m/s
b) -4.7 m/s
The formula for distance under constant acceleration is
d = 0.5AT^2
The formula for distance with a specified velocity is
d = VT
So the distance the keys travel with an initial velocity and under constant acceleration by gravity is
d = VT - 0.5AT^2
The acceleration due to gravity is 9.8 m/s^2 and the time T is 1.50 s, and finally, the distance traveled is 4.00 m. So substitute those values into the equation and solve for V
d = VT - 0.5AT^2
4.00m = 1.50s * V - 0.5 * 9.8 m/s^2 * (1.5s)^2
Do the multiplications
4.00m = 1.50s * V - 4.9m/s^2 * 2.25 s^2
Cancel the s^2 terms
4.00m = 1.50s * V - 4.9m * 2.25
Do the multiplication
4.00m = 1.50s * V - 11.025m
Add 11.025m to both sides
15.025m = 1.50s * V
Divide both sides by 1.50s
10.01667 m/s = V
Since we have 3 significant figures in the data, round results to 3 significant figures.
V = 10.0 m/s
So the keys were initially thrown upwards with a velocity of
10.0 m/s
Since it took 1.50 seconds from launch to catch, the velocity of the keys will decrease by 9.8 m/s^2 times the time. So
V = 10.0 m/s - 1.50s * 9.8 m/s^2
V = 10.0 m/s - 14.7 m/s
V = -4.7 m/s
So at the time the keys were caught, they were moving downward at a velocity of 4.7 m/s</span>
Answer:
What exactly are we supposed to be explaining? Is there another part to this?
Answer:
U/U₀ = 2
(factor of 2 i.e U = 2U₀)
Therefore, the energy stored in the capacitor is doubled when the plate separation is doubled while the capacitor has been disconnected
Explanation:
Energy stored in a capacitor can be expressed as;
U = 0.5CV^2 = Q^2/2C
And
C = ε₀ A/d
Where
C = capacitance
V = potential difference
Q = charge
A = Area of plates
d = distance between plates
So
U = Q^2/2C = dQ^2/2ε₀ A
The initial energy of the capacitor at d = d₀ is
U₀ = Q^2/2C = d₀Q^2/2ε₀ A ....1
When the plate separation is increased after the capacitor has been disconnected, the charge Q of the capacitor remain constant.
The final energy stored in the capacitor at d = 2d₀ is
U = 2d₀Q^2/2ε₀ A ...2
The factor U/U₀ can be derived by substituting equation 1 and 2
U/U₀ = (2d₀Q^2/2ε₀ A)/( d₀Q^2/2ε₀ A )
Simplifying we have;
U/U₀ = 2
U = 2U₀
Therefore, the energy stored in the capacitor is doubled when the plate separation is doubled while the capacitor has been disconnected.
Answer
given,
mass of steel ball, M = 4.3 kg
length of the chord, L = 6.5 m
mass of the block, m = 4.3 Kg
coefficient of friction, μ = 0.9
acceleration due to gravity, g = 9.81 m/s²
here the potential energy of the bob is converted into kinetic energy
v = 11.29 m/s
As the collision is elastic the velocity of the block is same as that of bob.
now,
work done by the friction force = kinetic energy of the block
d = 7.23 m
the distance traveled by the block will be equal to 7.23 m.
