An 56 kg sled is being pulled across the snow, at constant speed,by a horizontal force of 15 N, find the coefficient of kinetic
1 answer:
Answer:
The coefficient of kinetic friction = 0.026
Explanation:
An 56 kg sled is being pulled across the snow, at constant speed,by a horizontal force of 15 N.
Here we have to note that the weight is pulled at a constant speed . This means that the net force acting on the weight is zero.
The external force acting on the body is in the forward direction and the friction acts in the backward direction.
Friction increases as the mass of the body increases.
Friction =
We now equate this to the external force of 15 N.
15 =
=
= 0.026
The coefficient of kinetic friction = 0.026
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(a) The momentum of each vehicle prior to collision is 80000 kgm/s for truck and 13500 kgm/s for car.
(b) The size of momentum is 93500 kgm/s and it will be directed towards South.
Explanation:
The mass of the truck moving due south is given as 4000 kg and the speed is 20 m/s. Similarly, the mass of the car moving due north is 1350 kg and the speed is 10 m/s.
(a) Then the momentum of each vehicle can be obtained by the product of mass with their respective speed.
Similarly, the momentum of car will be
So, the momentum of each vehicle prior to collision is 80000 kgm/s for truck and 13500 kgm/s for car.
(b) Since, after collision, the vehicles stick together, the momentum after collision will be equal to the total momentum of both the vehicles before collision. This is because, it will obey conservation of momentum.
Momentum of vehicles after collision = total momentum before collision
Momentum after collision = 80000+13500 = 93500 kgm/s.
The direction of the vehicles after collision will be towards south as the mass and speed of the truck is greater than car. So the impact or force exerted on the car by the truck will be greater and thus both the vehicles will be directed towards south after collision.
Thus, the size of momentum is 93500 kgm/s and it will be directed towards South.
Explanation:
It is given that,
Mass of lithium,
It is accelerated through a potential difference, V = 224 V
Uniform magnetic field, B = 0.724 T
Applying the conservation of energy as :
q is the charge on an electron
v = 78608.58 m/s
To find the radius of the ion's path in the magnetic field. The centripetal force is balanced by the magnetic force as :
r = 0.0078 meters
So, the radius of the path of the ion is 0.0078 meters. Hence, this is the required solution.