Answer:
When truck is at rest while student is under motion
Explanation:
Since it is obvious that the mass of a truck is more than that of a student, we know that momentum is a product of mass and velocity
P=mv where m represent mass, v is velocity. When the student has more speed than that of truck, he exerts more momentum. The only way a student can exert more momentum is by having more speed while the truck is at rest. In such case, the momentum of truck will be zero while momentum of student will have a value
What is the force of gravity on the rocket at the planet's surface?
What is the force of gravity on the rocket at a distance of two units (twice the planet's radius from its center)?
In general, how does the force of gravity pulling on the rocket change as the distance between it and the planet increases?
cords unwind the rod rotates. Find the tension in the cords as they unwind.
Answer:
T = mg/6
Explanation:
Draw a free body diagram (see attached). There are two tension forces acting upward at the edge of the cylinder, and weight at the center acting downwards.
The center rotates about the point where the cords touch the edge. Sum the torques about that point:
∑τ = Iα
mgr = (1/2 mr² + mr²) α
mgr = 3/2 mr² α
g = 3/2 r α
α = 2g / (3r)
(Notice that you have to use parallel axis theorem to find the moment of inertia of the cylinder about the point on its edge rather than its center.)
Now, sum of the forces in the y direction:
∑F = ma
2T − mg = m (-a)
2T − mg = -ma
Since a = αr:
2T − mg = -mαr
Substituting expression for α:
2T − mg = -m (2g / (3r)) r
2T − mg = -2/3 mg
2T = 1/3 mg
T = 1/6 mg
The tension in each cord is mg/6.
As it is given that 2 kg mass is suspended by 12 cm long thread and then a horizontal force is applied on it so that it remains in equilibrium at 30 degree angle
So here we can use force balance in X and Y directions
now for X direction or horizontal direction we can use
for vertical direction similarly we can say
so here we first divide the two equations
now plug in all values in the above equation
Part b)
now in order to find the tension in the thread we can use any above equation
so tension in the thread will be 22.6 N