Can u show the problem ?
Acceleration= the rate and change of an object also cars go fast so I think it would be positive acceleration
Hope I helped
Answer:
Explanation:
1) either all the way to the ground, or, if the cliff is high enough,
h = ½(9.8)1.4² = 9.6 m
2) If the cliff is high enough and we ignore air resistance
v = 9.8(2.7) = 26 m/s
In our Solar System, Jupiter is the largest planet we have. it has the surface area of 23.71 billion mi^2. it beats all the other planets in both mass and volume.
Answer:
Explanation:
Given
zero vertical displacement
suppose \theta is the inclination angle and u is the initial velocity
considering horizontal motion
range is given by
where t is the time of flight
Now consider vertical motion
a=acceleration
for zero displacement y=0
substitute the value of t in Range
for maximum vale of must be equal to 1
i.e.
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.