by using v ^2 = u^2 + 2as we can find "a"
25 = 729 + 2 × a × 50
25 = 729 + 100a
a = - 7.04
so the answer is B
3,89,988 cm/min is the linear velocity
Given,
Diameter of CD = 12 cm
So, Radius of CD = 6 cm
CD is spinning at 10350 rev/min
Firstly , convert rev/min into rad/min
1 rev = 2π radians
10350 rev/min = 10350 × 2π
= 64998 rad/min
Formula used,
where,
is the Linear velocity
is the radius
is the angular velocity
= 6 cm × 64998rad/min
= 3,89,988 cm/min
Thus, linear velocity for any edge point of a 12-cm-diameter CD (compact disc) spinning at 10,350 rev/min is 389988 cm/min.
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Answer:
6.75 m/s²
Explanation:
The following data were obtained from the question:
Mass (m) of object = 40 Kg
Force applied (Fₐ) = 300 N
Force of friction (Fբ) = 30 N
Acceleration (a) =?
Next, we shall determine the net force acting on the mass. This can be obtained as follow:
Force applied (Fₐ) = 300 N
Force of friction (Fբ) = 30 N
Net force (Fₙ) =?
Fₙ = Fₐ – Fբ
Fₙ = 300 – 30
Fₙ = 270 N
Thus, the net force acting on the mass is 270 N.
Finally, we shall determine acceleration of the mass. This can be obtained as follow:
Mass (m) of object = 40 Kg
Net force (Fₙ) = 270 N
Acceleration (a) =?
Net force = mass × acceleration
Fₙ = m × a
270 = 40 × a
Divide both side by 40
a = 270 / 40
a = 6.75 m/s²
Therefore, the acceleration of the mass is 6.75 m/s²
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