The calculation of the centripetal acceleration of an object following a circular path is based on the equation,
a = v² / r
where a is the acceleration, v is the velocity, and r is the radius.
Substituting the known values from the given above,
4.4 m/s² = (15 m/s)² / r
The value of r from the equation is 51.14 m.
Answer: 51.14 m
Answer: Mathematical Model
Explanation:
Took the test
<em>Choice-A </em>is a true statement. The string holding the ball in orbit is a contact force, whereas gravity is a non-contact force.
Choice-C is half-true. The string keeps the distance between the bodies constant, but gravity doesn't "MAKE" the distance vary.
Answer:
T= 38.38 N
Explanation:
Here
mass of can = m = 3 kg
g= 9.8 m/sec2
angle θ = 40°
From figure we see the vertical and horizontal component of tension force T
If the can is to slip - then horizontal component of tension force should become equal to force of friction.
First we find force of friction
Fs= μ R
where
μ = 0.76
R = weight of can = mg = 3 × 9.8 = 29.4 N
Now horizontal component of tension
Tx= T cos 40 = T× 0.7660 N
==>T× 0.7660 = 29.4
==> T= 38.38 N