Answer:
A
Explanation:
they are atoms of the same element that have the same number of protons but different number of neutrons.
Answer
given,
mass of steel ball, M = 4.3 kg
length of the chord, L = 6.5 m
mass of the block, m = 4.3 Kg
coefficient of friction, μ = 0.9
acceleration due to gravity, g = 9.81 m/s²
here the potential energy of the bob is converted into kinetic energy
![m g L = \dfrac{1}{2} mv^2](https://tex.z-dn.net/?f=m%20g%20L%20%3D%20%5Cdfrac%7B1%7D%7B2%7D%20mv%5E2)
![v= \sqrt{2gL}](https://tex.z-dn.net/?f=v%3D%20%5Csqrt%7B2gL%7D)
![v= \sqrt{2\times 9.8\times 6.5}](https://tex.z-dn.net/?f=v%3D%20%5Csqrt%7B2%5Ctimes%209.8%5Ctimes%206.5%7D)
v = 11.29 m/s
As the collision is elastic the velocity of the block is same as that of bob.
now,
work done by the friction force = kinetic energy of the block
![f . d = \dfrac{1}{2} mv^2](https://tex.z-dn.net/?f=f%20.%20d%20%3D%20%5Cdfrac%7B1%7D%7B2%7D%20mv%5E2)
![\mu m g. d = \dfrac{1}{2} mv^2](https://tex.z-dn.net/?f=%5Cmu%20m%20g.%20d%20%3D%20%5Cdfrac%7B1%7D%7B2%7D%20mv%5E2)
![d=\dfrac{v^2}{2\mu g}](https://tex.z-dn.net/?f=d%3D%5Cdfrac%7Bv%5E2%7D%7B2%5Cmu%20g%7D)
![d=\dfrac{11.29^2}{2\times 0.9 \times 9.8}](https://tex.z-dn.net/?f=d%3D%5Cdfrac%7B11.29%5E2%7D%7B2%5Ctimes%200.9%20%5Ctimes%209.8%7D)
d = 7.23 m
the distance traveled by the block will be equal to 7.23 m.
Answer:
kick 1 has travelled 15 + 15 = 30 yards before hitting the ground
so kick 2 travels 25 + 25 = 50 yards before hitting the ground
first kick reached 8 yards and 2nd kick reached 20 yards
Explanation:
1st kick travelled 15 yards to reach maximum height of 8 yards
so, it has travelled 15 + 15 = 30 yards before hitting the ground
2nd kick is given by the equation
y (x) = -0.032x(x - 50)
![Y = 1.6 X - 0.032x^2](https://tex.z-dn.net/?f=Y%20%3D%201.6%20X%20-%200.032x%5E2)
we know that maximum height occurs is given as
![x = -\frac{b}{2a}](https://tex.z-dn.net/?f=x%20%3D%20-%5Cfrac%7Bb%7D%7B2a%7D)
![y =- \frac{1.6}{2(-0.032)} = 25](https://tex.z-dn.net/?f=y%20%3D-%20%5Cfrac%7B1.6%7D%7B2%28-0.032%29%7D%20%3D%2025)
and maximum height is
![y = 1.6\times 25 - 0.032\times 25^2](https://tex.z-dn.net/?f=y%20%3D%201.6%5Ctimes%2025%20-%200.032%5Ctimes%2025%5E2)
y = 20
so kick 2 travels 25 + 25 = 50 yards before hitting the ground
first kick reached 8 yards and 2nd kick reached 20 yards
Answer:
Part a)
![EMF = 14 \times 10^{-3} V](https://tex.z-dn.net/?f=EMF%20%3D%2014%20%5Ctimes%2010%5E%7B-3%7D%20V)
Part b)
![EMF = 15.67 \times 10^{-3} V](https://tex.z-dn.net/?f=EMF%20%3D%2015.67%20%5Ctimes%2010%5E%7B-3%7D%20V)
Explanation:
As we know that magnetic flux through the loop is given as
![\phi = B.A](https://tex.z-dn.net/?f=%5Cphi%20%3D%20B.A)
now we have
![\phi = B\pi r^2](https://tex.z-dn.net/?f=%5Cphi%20%3D%20B%5Cpi%20r%5E2)
now rate of change in flux is given as
![\frac{d\phi}{dt} = B(2\pi r)\frac{dr}{dt}](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5Cphi%7D%7Bdt%7D%20%3D%20B%282%5Cpi%20r%29%5Cfrac%7Bdr%7D%7Bdt%7D)
now we know that
![A = \pi r^2](https://tex.z-dn.net/?f=A%20%3D%20%5Cpi%20r%5E2)
![0.285 = \pi r^2](https://tex.z-dn.net/?f=0.285%20%3D%20%5Cpi%20r%5E2)
![r = 0.30 m](https://tex.z-dn.net/?f=r%20%3D%200.30%20m)
Now plug in all data
![EMF = (0.20)\times 2\pi\times (0.30) \times (0.037)](https://tex.z-dn.net/?f=EMF%20%3D%20%280.20%29%5Ctimes%202%5Cpi%5Ctimes%20%280.30%29%20%5Ctimes%20%280.037%29)
![EMF = 14 \times 10^{-3} V](https://tex.z-dn.net/?f=EMF%20%3D%2014%20%5Ctimes%2010%5E%7B-3%7D%20V)
Part b)
Now the radius of the loop after t = 1 s
![r_1 = r_0 + \frac{dr}{dt}](https://tex.z-dn.net/?f=r_1%20%3D%20r_0%20%2B%20%5Cfrac%7Bdr%7D%7Bdt%7D)
![r_1 = 0.30 + 0.037](https://tex.z-dn.net/?f=r_1%20%3D%200.30%20%2B%200.037)
![r_1 = 0.337 m](https://tex.z-dn.net/?f=r_1%20%3D%200.337%20m)
Now plug in data in above equation
![EMF = (0.20)\times 2\pi\times (0.337) \times (0.037)](https://tex.z-dn.net/?f=EMF%20%3D%20%280.20%29%5Ctimes%202%5Cpi%5Ctimes%20%280.337%29%20%5Ctimes%20%280.037%29)
![EMF = 15.67 \times 10^{-3} V](https://tex.z-dn.net/?f=EMF%20%3D%2015.67%20%5Ctimes%2010%5E%7B-3%7D%20V)