Answer:
Explanation:
The question relates to motion on a circular path .
Let the radius of the circular path be R .
The centripetal force for circular motion is provided by frictional force
frictional force is equal to μmg , where μ is coefficient of friction and mg is weight
Equating cenrtipetal force and frictionl force in the case of car A
mv² / R = μmg
R = v² /μg
= 26.8 x 26.8 / .335 x 9.8
= 218.77 m
In case of moton of car B
mv² / R = μmg
v² = μRg
= .683 x 218.77x 9.8
= 1464.35
v = 38.26 m /s .
Answer:
0.004 m away from the film
Explanation:
u = Object distance
v = Image distance
f = Focal length = 50 mm

The image distance is 0.051 m
When u = 50 cm

The image distance is 0.055 m
The lens has moved 0.055-0.051 = 0.004 m away from the film
Answer:
f = 15 N
Explanation:
It is given that, when you push with a horizontal 15-N force on a book that slides at constant velocity, a frictional force also acts on it. Frictional force is an opposing force. The magnitude of applied force and frictional forces are same. So, the force of friction on the book is equal to 15 N.
Answer:
Kinetic energy.
The explanation never existed
Answer:
Explanation:
The time period of the pendulum containing simple pendulum is given by

where, L is the length of the pendulum and g is the value of acceleration due to gravity.
The time period of the clock using the spring mechanism is given by

where, m is the mass of the block attached to the spring and k is the spring constant.
here we observe the time period of the pendulum depends on the value of acceleration due to gravity. The value of acceleration due to gravity decreases as we go on the heights that means when the clock is taken to the mountain, the value of g decreases and thus, the value of time period increases and the clock runs slow.
So, the clock containing the spring system gives the accurate reading rather than the clock containing simple pendulum.