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Tcecarenko [31]
3 years ago
11

A 3.5-kg object placed on an inclined plane (angle 30? above the horizontal) is connected by a string going over a pulley to a 1

.0-kg object. The coefficient of static friction between object 1 and the surface of the inclined plane is 0.30 and equals the coefficient of kinetic friction.
What is the acceleration of the system?

Physics
1 answer:
Blababa [14]3 years ago
7 0

Answer:

a= 0.22 m/s²

Explanation:

Given that

M = 3.5 kg

θ = 30°

m = 1 kg

μ= 0.3

The force due to gravity

F₁= M g sinθ

F₁=3.5 x 10 x sin 30

F₁= 17.5 N

F₂ = m g

F₂ = 1 x 10 = 10 N

The maximum value of the friction force on the incline plane

Fr = μ M g cosθ

Fr = 0.3 x 2.5 x 10 cos30°

Fr= 6.49 N

Lets take acceleration of the system is a  m/s²

F₁ - F₂  - Fr =  (M+m) a

17.5 - 10 - 6.49 = (3.5+1)a

a= 0.22 m/s²

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Answer: 247.67 V

Explanation:

Given

Potential At A V_a=382\ V

Potential at V_c=785\ V

when particle starts from A it reaches with velocity v_b at Point while when it starts from C it reaches at point B with velocity 2v_b

Suppose m is the mass of Particle

Change in Kinetic Energy of particle moving under the Potential From A to B

q\cdot \left ( V_a-V_b\right )=0.5m\cdot (v_b)^2----1

Change in Kinetic Energy of particle moving under the Potential From C to B

q\cdot \left ( V_c-V_b\right )=0.5m\cdot (2v_b)^2-----2

Divide 1 and 2 we get

\frac{V_a-V_b}{V_c-V_b}=\frac{v_b^2}{4v_b^2}

on solving we get

V_b=\frac{4}{3}\cdot V_a-\frac{1}{3}\cdot V_c

V_b=\frac{743}{3}=247.67\ V

                     

4 0
3 years ago
Astronaut Sarah leaves Earth in a spaceship at a speed of 0.280c relative to Earth. Sarah's destination is a star-system 12.5 li
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Explanation:

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now calculating Lo

Lo = 12.5×365×24×3600×3×10^8

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now putting the values of v and Lo in the above equation we get

L=1.183\times10^{17}\sqrt{1-\frac{0.28c^2}{c^2} }

= 1.136×10^17 m

L=  = \frac{1.136\times10^{17}}{365\times24\times3600\times3\times10^8}m

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