Answer:
1 second
Explanation:
The impulse exerted on the boy is equal to its change of momentum:

where
F = 500 N is the push on the boy
is the contact time
m = 25 kg is the mass of the boy
is the change in velocity of the boy
Solving the formula for the contact time, we find

Answer:
Explanation:
Torque on a loop in a magnetic field
Maximum torque = M B
M is magnetic moment of loop and B is magnetic field
M = area x current
= 3.14 x .10² x i
Maximum torque = M B
1 x 10⁻³ = 3.14 x .10² x i x .65 x 10⁻⁴
i = 490 A
Current = 490 A.
Answer:
2.43J
Explanation:
Given parameters:
Mass of the arrow = 0.155kg
Velocity = 31.4m /s
Unknown:
Kinetic energy when it leaves the bow = ?
Solution:
The kinetic energy of a body is the energy in motion of the body;
it can be derived using the expression below:
K.E =
m v²
m is the mass
v is the velocity
Solve for K.E;
K.E =
x 0.155 x 31.4 = 2.43J
Answer:
final velocity will be44.72m/s
Explanation:
HEIGHT=h=100m
vi=0m/s
vf=?
g=10m/s²
by using third equation of motion for bodies under gravity
2gh=(vf)²-(vi)²
evaluating the formula
2(10m/s²)(100m)=vf²-(0m/s)²
2000m²/s²=vf²
√2000m²/s²=√vf²
44.72m/s=vf