Answer:
The puck B remains at the point of collision.
Explanation:
This is an elastic collision, so both momentum and energy are conserved.
The mass of both pucks is m.
The velocity of puck B before the collision is vb.
The velocity of puck A and B after the collision is va' and vb', respectively.
Momentum before = momentum after
m vb = m vb' + m va'
vb = vb' + va'
Energy before = energy after
½ m vb² = ½ m vb'² + ½ m va'²
vb² = vb'² + va'²
Substituting:
(vb' + va')² = vb'² + va'²
vb'² + 2 va' vb' + va'² = vb'² + va'²
2 va' vb' = 0
va' vb' = 0
We know that va' isn't 0, so:
vb' = 0
The puck B remains at the point of collision.
Answer:
204 m
Explanation:
When the marble is dropped from a certain height, its gravitational potential energy converts into kinetic energy. So the kinetic energy gained is equal to the variation of gravitational potential energy:
where
m is the mass of the marble
g = 9.8 m/s^2 is the acceleration of gravity
is the change in height
In this problem, we have
m = 50 g = 0.05 kg
Solving the formula for , we find the necessary height from which the marble should be dropped:
Answer:
A lunar eclipse occurs when Earth blocks the light going to the moon. A solar eclipse occurs when the moon blocks the light coming from the sun.
Explanation:
Explanation:
decimal diamond ml dias and ka n t u t a n mo papa mo
Answer:
Explanation:
Two straight wires
Have current in opposite direction
i1=i2=i=2Amps
Distance between two wires
r=5mm=0.005m
Length of one wire is ∞
Length of second wire is 0.3m
Force between the wire,
The force between two parallel currents I1 and I2, separated by a distance r, has a magnitude per unit length given by
F/l = μoi1i2/2πr
F/l=μoi²/2πr
μo=4π×10^-7 H/m
The force is attractive if the currents are in the same direction, repulsive if they are in opposite directions.
F/l = μoi1i2/2πr
F/0.3=4π×10^-7×2²/2π•0.005
F/0.3=1.6×10^-4
Cross multiply
F=1.6×10^-4×0.3
F=4.8×10^-5N
