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Answer:
The coefficient of kinetic friction between the sled and the snow is 0.0134
Explanation:
Given that:
M = mass of person = 52 kg
m = mass of sled = 15.2 kg
U = initial velocity of person = 3.63 m/s
u = initial velocity of sled = 0 m/s
After collision, the person and the sled would move with the same velocity V.
a) According to law of momentum conservation:
Total momentum before collision = Total momentum after collision
MU + mu = (M + m)V
Substituting values:
The velocity of the sled and person as they move away is 2.81 m/s
b) acceleration due to gravity (g) = 9.8 m/s²
d = 30 m
Using the formula:
The coefficient of kinetic friction between the sled and the snow is 0.0134
gravitational potential is directly proportional to the height of the object relative to a reference line and is given as
PE = mgh
where m = mass of object , g = acceleration due to
gravity and h = height of the object above the reference line .
as the skydiver falls , its height above the ground decrease and hence the gravitational potential energy of the skydiver decrease.
as per conservation of energy , total energy of the skydiver must remain constant all the time . hence the decrease in potential energy appears as increase in kinetic energy by same amount to keep the total energy constant
KE + PE = Total energy
so as the skydiver falls , it gains speed and hence the kinetic energy of skydiver increase since kinetic energy is directly proportional to the square of the speed.
when the parachute opens, the skydiver experience force in upward which tries to balance the weight of the skydiver. hence the speed of the skydiver decrease until upward force becomes equal to the downward force. hence the kinetic energy decrease just after the parachute opens
Answer:
1.925 μC
Explanation:
Charge: This can be defined as the product of the capacitance of a capacitor and the voltage. The S.I unit of charge is Coulombs (C)
The formula for the charge stored in a capacitor is given as,
Q = CV ................... Equation 1
Where Q = charge, C = Capacitor, V = Voltage.
Note: 1 μF = 10⁻⁶ F
Given: C = 0.55 μF = 0.55×10⁻⁶ F, V = 3.5 V.
Substitute into equation 1
Q = 0.55×10⁻⁶×3.5
Q = 1.925×10⁻⁶ C.
Q = 1.925 μC
Hence the charge on the plate = 1.925 μC
