It depends on where the cat is applying the force and how much the force is .. after all where and at what is distance from the axis of rotation of the door
<span>(6.0x10^-22, -1.40x10^-21, 0) kg*m/s
Momentum is a conserved quantity. The total momentum of the system before and after the interactions will not change. So, let's look at the momentum before the interaction.
(3.2x10^-21, 0, 0) kg*m/s and (0,0,0) kg*m/s
After the interaction
(2.6x10^-21, 1.40x10^-21, 0) kg*m/s
and the other proton has to have a momentum that when added to this momentum equal the original value. Since the y and z vectors were initially 0, all we need for the y and x vector values of the result is to negate them. The x vector value will be
3.2x10^-21 - 2.6x10^-21 = 0.6x10^21 = 6.0x10^-22. So the other proton will have a momentum of
(6.0x10^-22, -1.40x10^-21, 0) kg*m/s</span>
Answer:
255.4 N/m
Explanation:
We can consider the system eyeball-attached to the musculature as a mass-spring system in simple harmonic motion, whose frequency of oscillation is given by

where in this case, we know:
f = 29 Hz is the frequency of oscillation
k is the spring constant, which is unknown
m = 7.7 g = 0.0077 kg is the mass of the eyeball
Solving the equation for k, we find the spring constant of the musculature attached to the eyeball:

Answer:
4940.12 m/s
Explanation:
For the net force on the moving electron to be zero, the Magnetic force must then match the electric force.
Magnetic force = Electric force
Magnetic force = qvB
Electric force = Eq
qvB = Eq
v = (E/B)
E = Electric field = 1.65 kV/m = 1650 V/m
B = Magnetic field = 0.334 T
v = (1650/0.334)
v = 4940.12 m/s
Hope this helps!!
Answer:
50N
Explanation:
Since the floor is frictionless, the net force = mass × acceleration.
given that,
mass = 10kg
and acceleration is 5 m/s²
Thus,
Force = 10 × 5
Force = 50 N