A meter stick balances horizontally on a knife-edge at the 50.0cm mark. With two 5.0g coins stacked over the 12.0cm mark l, the
1 answer:
Answer:
mass of the scale is 7.44 g
Explanation:
As we know that the meter scale is balanced at 45.5 cm mark when two coins are placed at 12 cm mark
now we can say that the torque due to weight of the coin is balanced by the weight of scale above the knife edge because the scale remains horizontal
so we will have
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Answer:
Explanation:
An object is at rest along a slope if the net force acting on it is zero. The equation of the forces along the direction parallel to the slope is:
(1)
where
is the component of the weight parallel to the slope, with m being the mass of the object, g the acceleration of gravity,
the angle of the slope
is the frictional force, with
being the coefficient of friction and R the normal reaction of the incline
The equation of the forces along the direction perpendicular to the slope is
where
R is the normal reaction
is the component of the weight perpendicular to the slope
Solving for R,
And substituting into (1)
Re-arranging the equation,
This the condition at which the equilibrium holds: when the tangent of the angle becomes larger than the value of , the force of friction is no longer able to balance the component of the weight parallel to the slope, and so the object starts sliding down.
Answer:
Explanation:
Since the front and back of the rocket simultaneously line up with forward and backward end of the platform respectively .
Then length of the platform = length of the train rocket .
A )
Time to cross a particular point on the platform
= length of rocket train / .96 x 3 x 10⁸
= 90 / .96 x 3 x 10⁸
= 31.25 x 10⁻⁸ s
B) Rest length of the rocket = length of platform = 90 m
C ) length of platform as viewed by moving observer =
=
= 321 m
D ) For the observer on platform time taken = 31.25 x 10⁻⁸ s
for the observer in the rocket , time will be dilated so time recorded by observer in motion ,
8.75 x 10⁻⁸ s .