When an object does not move even on pushing , static frictional force acts on in opposite direction of the applied force to stop the object from moving. static frictional force is a self adjusting force and it adjust its value according to the applied force if the applied force is smaller than the maximum value of static frictional force. The object starts moving once the applied force on it becomes greater than the maximum static frictional force. hence the statement is true.
Answer:
P = 5.22 Kg.m/s
Explanation:
given,
mass of the projectile = 1.8 Kg
speed of the target = 4.8 m/s
angle of deflection = 60°
Speed after collision = 2.9 m/s
magnitude of momentum after collision = ?
initial momentum of the body = m x v
= 1.8 x 4.8 = 8.64 kg.m/s
final momentum after collision
momentum along x-direction
P_x = m v cos θ
P_x = 1.8 x 2.9 x cos 60°
P_x = 2.61 kg.m/s
momentum along y-direction
P_y = m v sin θ
P_y = 1.8 x 2.9 x sin 60°
P_y = 4.52 kg.m/s
net momentum of the body


P = 5.22 Kg.m/s
momentum magnitude after collision is equal to P = 5.22 Kg.m/s
Answer:
f = 19,877 cm and P = 5D
Explanation:
This is a lens focal length exercise, which must be solved with the optical constructor equation
1 / f = 1 / p + 1 / q
where f is the focal length, p is the distance to the object and q is the distance to the image.
In this case the object is placed p = 25 cm from the eye, to be able to see it clearly the image must be at q = 97 cm from the eye
let's calculate
1 / f = 1/97 + 1/25
1 / f = 0.05
f = 19,877 cm
the power of a lens is defined by the inverse of the focal length in meters
P = 1 / f
P = 1 / 19,877 10-2
P = 5D
The work done to pull the sled up to the hill is given by

where
F is the intensity of the force
d is the distance where the force is applied.
In our problem, the work done is

and the distance through which the force is applied is

, so we can calculate the average force by re-arranging the previous equation and by using these data:
Answer:
True
Explanation:
Given that Power whose unit is Watt equates to one joule of work per second. It implies that Power is directly proportional to the work done and inversely proportional to the time to do the work.
Therefore, in this case, the right answer to the question is that it is TRUE that the power is inversely proportional with time