The coefficient of static friction between the puck and the surface.
In fact, that coefficient describes exactly how "hard" it is to cause the puck to start moving, if it starts from an idle condition.
Answer:
Explanation:
Force of gravity = GMm/r^2 = ma
a being the acceleration due to gravity at some distance r from the center of the Earth. I'll use 6400 km for the radius of the Earth so r = 6734 km or 6734000 meters
a = GM/r^2
plugging in G = 6.67 x 10^-11
M is the mass of the Earth = 6 x 10^24
and r is from above
a = 8.825 m/s^2 = 0.9g
so 90% the acceleration of gravity on the surface.
Answer:
0.05 m
Explanation:
From the question given above, the following data were obtained:
Mass of first object (M1) = 9900 kg
Gravitational force (F) = 12 N
Mass of second object (M2) = 52000 kg
Distance apart (r) =?
Gravitational constant (G) = 6.67×10¯¹¹ Nm²/Kg²
Thus, we can obtain the distance between the two objects as shown below:
F = GM1M2/r²
12 = 6.67×10¯¹¹ × 9900 × 52000 /r²
Cross multiply
12 × r² = 6.67×10¯¹¹ × 9900 × 52000
Divide both side by 12
r² = (6.67×10¯¹¹ × 9900 × 52000)/12
Take the square root of both side
r = √[(6.67×10¯¹¹ × 9900 × 52000)/12]
r = 0.05 m
Therefore, the distance between the two objects is 0.05 m
Answer:
(A) –14m/s
(B) –42.0m
Explanation:
The complete solution can be found in the attachment below.
This involves the knowledge of motion under the action of gravity.
Check below for the full solution to the problem.
Incomplete question.The complete one is here
A runner taking part in the 200m dash must run around the end of a track that has a circular arc with a radius curvature of 30m. The runner starts the race at a constant speed. If she completes the 200m dash in 23s and runs at constant speed throughout the race, what is her centripetal acceleration as she runs the curved portion of the track?
Answer:
Explanation:
Given data
Required
Centripetal acceleration
Solution
According to the motions equation the velocity given by:
The centripetal acceleration is given by:
