3. What is the charge on the leaves of an electroscope if a positively charged rod is moved near the
1 answer:
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Answer:
0.56
Explanation:
Let the coefficient of friction is μ.
m = 4.3 kg, θ = 30 degree, initial velocity, u = 0, s = 2.7 m, t = 5.8 s
By the free body diagram,
Normal reaction, N = mg Cosθ = 4.3 x 9.8 x Cos 30 = 36.49 N
Friction force, f = μ N = 36.49 μ
Net force acting on the block,
Fnet = mg Sinθ - f = 4.3 x 9.8 x Sin 30 - 36.49 μ
Fnet = 21.07 - 36.49μ
Net acceleartion, a = Fnet / m
a = (21.07 - 36.49μ) / 4.3
use second equation of motion
s = ut + 1/2 a t^2
2.7 = 0 + 1/2 x (21.07 - 36.49μ) x 5.8 x 5.8 / 4.3
By solving we get
μ = 0.56
Answer:
Acceleration of the crate is 0.362 m/s^2.
Explanation:
Given:
Mass of the box, m = 40 kg
Applied force, F = 15 N
Angle at which the force is applied, = 15°
We have to find the magnitude of the acceleration.
Let the acceleration be "a".
FBD is attached with where we can see the horizontal and vertical component of force.
⇒ and ⇒
⇒ ⇒
⇒ Applying concept of forces.
⇒
⇒
⇒ <em> ...Newtons second law Fnet = ma</em>
⇒
⇒ Plugging the values.
⇒ <em>...f is the friction which is zero here.</em>
⇒
⇒
Magnitude of the acceleration of the crate is 0.362 m/s^2.
