The work done by the centripetal force during om complete revolution is 401.92 J.
<h3>What is centripetal force?</h3>
Centripetal force is a force that acts on a body undergoing a circular motion and is directed towards the center of the circle in which the body is moving.
To Calculate the work done by the centripetal force during one complete revolution, we use the formula below.
Formula:
- W = (mv²/r)2πr
- W = 2πmv²................... Equation 1
Where:
- W = Work done by the centripetal force
- m = mass of the ball
- v = velocity of the ball
- π = pie
From the question,
Given:
- m = 16 kg
- v = 2 m/s
- π = 3.14
Substitute these values into equation 1
Hence, The work done by the centripetal force during om complete revolution is 401.92 J.
Learn more about centripetal force here: brainly.com/question/20905151
Answer:
(a) the speed of the block after the bullet embeds itself in the block is 3.226 m/s
(b) the kinetic energy of the bullet plus the block before the collision is 500J
(c) the kinetic energy of the bullet plus the block after the collision is 16.13J
Explanation:
Given;
mass of bullet, m₁ = 0.1 kg
initial speed of bullet, u₁ = 100 m/s
mass of block, m₂ = 3 kg
initial speed of block, u₂ = 0
Part (A)
Applying the principle of conservation linear momentum, for inelastic collision;
m₁u₁ + m₂u₂ = v(m₁ + m₂)
where;
v is the speed of the block after the bullet embeds itself in the block
(0.1 x 100) + (3 x 0) = v (0.1 + 3)
10 = 3.1v
v = 10/3.1
v = 3.226 m/s
Part (B)
Initial Kinetic energy
Ki = ¹/₂m₁u₁² + ¹/₂m₂u₂²
Ki = ¹/₂(0.1 x 100²) + ¹/₂(3 x 0²)
Ki = 500 + 0
Ki = 500 J
Part (C)
Final kinetic energy
Kf = ¹/₂m₁v² + ¹/₂m₂v²
Kf = ¹/₂v²(m₁ + m₂)
Kf = ¹/₂ x 3.226²(0.1 + 3)
Kf = ¹/₂ x 3.226²(3.1)
Kf = 16.13 J
Answer:![x=\frac{x_m}{\sqrt{2}}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7Bx_m%7D%7B%5Csqrt%7B2%7D%7D)
Explanation:
Given
initially mass is stretched to ![x_m](https://tex.z-dn.net/?f=x_m)
Let k be the spring Constant of spring
Therefore Total Mechanical Energy is ![\frac{kx_m^2}{2}](https://tex.z-dn.net/?f=%5Cfrac%7Bkx_m%5E2%7D%7B2%7D)
Position at which kinetic Energy is equal to Elastic Potential Energy
![K=\frac{mv^2}{2}](https://tex.z-dn.net/?f=K%3D%5Cfrac%7Bmv%5E2%7D%7B2%7D)
![U=\frac{kx^2}{2}](https://tex.z-dn.net/?f=U%3D%5Cfrac%7Bkx%5E2%7D%7B2%7D)
it is given
![k=U](https://tex.z-dn.net/?f=k%3DU)
thus ![2U=\frac{kx_m^2}{2}](https://tex.z-dn.net/?f=2U%3D%5Cfrac%7Bkx_m%5E2%7D%7B2%7D)
![2\times \frac{kx^2}{2}=\frac{kx_m^2}{2}](https://tex.z-dn.net/?f=2%5Ctimes%20%5Cfrac%7Bkx%5E2%7D%7B2%7D%3D%5Cfrac%7Bkx_m%5E2%7D%7B2%7D)
![2x^2=x_m^2](https://tex.z-dn.net/?f=2x%5E2%3Dx_m%5E2)
![x=\frac{x_m}{\sqrt{2}}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7Bx_m%7D%7B%5Csqrt%7B2%7D%7D)
Answer:
![400000\ \text{N/C}](https://tex.z-dn.net/?f=400000%5C%20%5Ctext%7BN%2FC%7D)
Explanation:
= Charge at 3000 m = 40 C
= Charge at 1000 m = -40 C
= 3000 m
= 1000 m
k = Coulomb constant = ![9\times10^9\ \text{Nm}^2/\text{C}^2](https://tex.z-dn.net/?f=9%5Ctimes10%5E9%5C%20%5Ctext%7BNm%7D%5E2%2F%5Ctext%7BC%7D%5E2)
Electric field due to the charge at 3000 m
![E_1=\dfrac{k|q_1|}{r_1^2}\\\Rightarrow E_1=\dfrac{9\times 10^9\times 40}{3000^2}\\\Rightarrow E_1=40000\ \text{N/C}](https://tex.z-dn.net/?f=E_1%3D%5Cdfrac%7Bk%7Cq_1%7C%7D%7Br_1%5E2%7D%5C%5C%5CRightarrow%20E_1%3D%5Cdfrac%7B9%5Ctimes%2010%5E9%5Ctimes%2040%7D%7B3000%5E2%7D%5C%5C%5CRightarrow%20E_1%3D40000%5C%20%5Ctext%7BN%2FC%7D)
Electric field due to the charge at 1000 m
![E_2=\dfrac{k|q_2|}{r_2^2}\\\Rightarrow E_2=\dfrac{9\times 10^9\times 40}{1000^2}\\\Rightarrow E_2=360000\ \text{N/C}](https://tex.z-dn.net/?f=E_2%3D%5Cdfrac%7Bk%7Cq_2%7C%7D%7Br_2%5E2%7D%5C%5C%5CRightarrow%20E_2%3D%5Cdfrac%7B9%5Ctimes%2010%5E9%5Ctimes%2040%7D%7B1000%5E2%7D%5C%5C%5CRightarrow%20E_2%3D360000%5C%20%5Ctext%7BN%2FC%7D)
Electric field at the aircraft is
.