Answer: F=0.075N
Explanation:
T*cos(θ) = mg=0.005X9.8
T*cos(θ)=0.049N
T = 0.049/cosθ
F = k*q1*q2 / r2
F = 8.9875 * 10^9 * q1*q2 / r^2 q1
F = 8.9875 * 10^9 * (5 x 10^-6)^2 / r^2
F=0.2246875/r^2
r= 2Lsinθ = 2 x 1 x sinθ =2sinθ
Kindly check the attached for further solution
80 joule is momentarily stored in each spring
<em><u>Solution:</u></em>
Given that,
The coil springs on a car's suspension have a value of k = 64000 N/m
When the car strikes a bump the springs briefly compress by 5.0 cm (.05 m)
By compressing the spring, we apply a force over a distance
As a result we have done work on the spring
Doing work means that we have transferred energy to spring in form of elastic potential
Therefore,
k = 64000 N/m
x = 0.05m
<em><u>The elastic potential energy is given as:</u></em>
Where, "k" is the spring constant and "x" is the displacement
Thus 80 joule is momentarily stored in each spring
The car A has a mass of 1200 kg.
The car B has the mass of 1900 kg.
It is given that velocity of car A is given as 22 Km/hr
The car B has the velocity of 25 Km/hr.
Let the mass of two bodies are denoted as
Let the velocity of cars A and B are denoted as
The momentum before collision is-
[Here p stand for momentum.]
We are asked to calculate the final momentum of the system after collision.
The answer of the question is based law of conservation of linear momentum.
As per law of conservation of linear momentum the sum total linear momentum for an isolated system is always constant.Hence irrespective of the type of collision[elastic and inelastic],the momentum of the system is always constant which is a universal truth.
Let after the collision the velocity of A and B are
Hence the final momentum of the system is-
As per the law of conservation of linear momentum, the initial and final momentum must be equal i.e
Hence the option A is right.