0.2 KG, just search the question and it's on yahoo answers
Answer:
Because of frictional force acting in the opposite direction.
Explanation:
The force of push on the body of Joel is 60 N. The body doesn't move. This clearly means that there must be a force acting in the opposite direction that has a magnitude equal to 60 N so that the net force acting on the body of Joel is 0 and hence the body doesn't move forward.
This opposite force is the frictional force which acts between the surface of snow and Joel. The frictional force always opposes relative motion.
So, when Joel's friend pushes him, he tends move him forward. Therefore, frictional force acts in the opposite direction to oppose the motion. If the frictional force is strong enough to stop the force of push, the body won't move.
This frictional force acting on a stationary body is a variable force and has a maximum value known as limiting friction. When the force of push exceeds the limiting friction, the body just starts to move. The body won't move till the force pf push is less than or equal to limiting friction.
Thus, Joel' friend push is less than or equal to the limiting friction and therefore, Joel is not moving.
Answer:
a) Δp = -2.0 kgm / s, b) Δp = -4 kg m / s
Explanation:
In this exercise the change in moment of a ball is asked in two different cases
a) clay ball, in this case the ball sticks to the door and we have an inelastic collision where the final velocity of the ball is zero
Δp = p_f - p₀
Δp = 0 - m v₀
Δp = - 0.100 20
Δp = -2.0 kgm / s
b) in this case we have a bouncing ball, this is an elastic collision, as the gate is fixed it can be considered an object of infinite mass, therefore the final speed of the ball has the same modulus of the initial velocity, but address would count
v_f = - v₀
Δp = p_f -p₀
Δp = m v_f - m v₀
Δp = m (v_f -v₀)
Δp = 0.100 (-20 - 20)
Δp = -4 kg m / s
Answer:
a)32.34 N/m
b)10cm
c)1.6 Hz
Explanation:
Let 'k' represent spring constant
'm' mass of the object= 330g =>0.33kg
a) in order to find spring constant 'k', we apply Newton's second law to the equilibrium position 10cm below the release point.
ΣF=kx-mg=0
k=mg / x
k= (0.33 x 9.8)/ 0.1
k= 32.34 N/m
b) The amplitude, A, is the distance from the equilibrium (or center) point of motion to either its lowest or highest point (end points). The amplitude, therefore, is half of the total distance covered by the oscillating object.
Therefore, amplitude of the oscillation is 10cm
c)frequency of the oscillation can be determined by,
f= 1/2π 
f= 1/2π 
f= 1.57
f≈ 1.6 Hz
Therefore, the frequency of the oscillation is 1.6 Hz
Answer:
Explanation:
Given that,
The distance between two spheres, r = 25 cm = 0.25 m
The capacitance, C = 26 pF = 26×10⁻¹² F
Charge, Q = 12 nC = 12 × 10⁻⁹ C
We need to find the work done in moving the charge. We know that, work done is given by :
Put all the values,
So, the work done is 
.
