Length of the copper rod = 2.5 meters
Speed at which sound travels through copper = 3560 meter per second
Let us assume the time taken
by sound to cover the given distance = x seconds
We already know that
Speed = Distance/ Time
Then
Time = Distance/ Speed
x = 2.5/3560 seconds
= 0.0007 seconds.
This can ve done by hitting one end of the rod and then receiving the sound at the other end and using the stop clock to measure the time taken.
Answer:
Explanation:
Given
mass of boy=36 kg
length of swing=3.5 m
Let T be the tension in the swing
At top point ![mg-T=\frac{mv^2}{r}](https://tex.z-dn.net/?f=mg-T%3D%5Cfrac%7Bmv%5E2%7D%7Br%7D)
where v=velocity needed to complete circular path
Th-resold velocity is given by ![mg-0=\frac{mv^2}{r}](https://tex.z-dn.net/?f=mg-0%3D%5Cfrac%7Bmv%5E2%7D%7Br%7D)
![v=\sqrt{gr}=\sqrt{9.8\times 3.5}=5.85 m/s](https://tex.z-dn.net/?f=v%3D%5Csqrt%7Bgr%7D%3D%5Csqrt%7B9.8%5Ctimes%203.5%7D%3D5.85%20m%2Fs)
So apparent weight of boy will be zero at top when it travels with a velocity of ![v=\sqrt{gr}](https://tex.z-dn.net/?f=v%3D%5Csqrt%7Bgr%7D)
To get the velocity at bottom conserve energy at Top and bottom
At top ![E_T=mg\times 2L+\frac{mv^2}{2}](https://tex.z-dn.net/?f=E_T%3Dmg%5Ctimes%202L%2B%5Cfrac%7Bmv%5E2%7D%7B2%7D)
Energy at Bottom ![E_b=\frac{mv_0^2}{2}](https://tex.z-dn.net/?f=E_b%3D%5Cfrac%7Bmv_0%5E2%7D%7B2%7D)
Comparing two as energy is conserved
![v_0^2=4gl+gl](https://tex.z-dn.net/?f=v_0%5E2%3D4gl%2Bgl)
![v_0^2=5gL](https://tex.z-dn.net/?f=v_0%5E2%3D5gL)
![v_0=\sqrt{5gL}=13.09 m/s](https://tex.z-dn.net/?f=v_0%3D%5Csqrt%7B5gL%7D%3D13.09%20m%2Fs%20)
Apparent weight at bottom is given by
![W=\frac{mv_0^2}{L}-mg=\frac{36\times 13.09^2}{3.5}+36\times 9.8=2115.23 N](https://tex.z-dn.net/?f=W%3D%5Cfrac%7Bmv_0%5E2%7D%7BL%7D-mg%3D%5Cfrac%7B36%5Ctimes%2013.09%5E2%7D%7B3.5%7D%2B36%5Ctimes%209.8%3D2115.23%20N)
Answer:
a)The approximate radius of the nucleus of this atom is 4.656 fermi.
b) The electrostatic force of repulsion between two protons on opposite sides of the diameter of the nucleus is 2.6527
Explanation:
![r=r_o\times A^{\frac{1}{3}}](https://tex.z-dn.net/?f=r%3Dr_o%5Ctimes%20A%5E%7B%5Cfrac%7B1%7D%7B3%7D%7D)
= Constant for all nuclei
r = Radius of the nucleus
A = Number of nucleons
a) Given atomic number of an element = 25
Atomic mass or nucleon number = 52
![r=1.25 \times 10^{-15} m\times (52)^{\frac{1}{3}}](https://tex.z-dn.net/?f=r%3D1.25%20%5Ctimes%2010%5E%7B-15%7D%20m%5Ctimes%20%2852%29%5E%7B%5Cfrac%7B1%7D%7B3%7D%7D)
![r=4.6656\times 10^{-15} m=4.6656 fm](https://tex.z-dn.net/?f=r%3D4.6656%5Ctimes%2010%5E%7B-15%7D%20m%3D4.6656%20fm)
The approximate radius of the nucleus of this atom is 4.656 fermi.
b) ![F=k\times \frac{q_1q_2}{a^2}](https://tex.z-dn.net/?f=F%3Dk%5Ctimes%20%5Cfrac%7Bq_1q_2%7D%7Ba%5E2%7D)
k=
= Coulombs constant
= charges kept at distance 'a' from each other
F = electrostatic force between charges
![q_1=+1.602\times 10^{-19} C](https://tex.z-dn.net/?f=q_1%3D%2B1.602%5Ctimes%2010%5E%7B-19%7D%20C)
![q_2=+1.602\times 10^{-19} C](https://tex.z-dn.net/?f=q_2%3D%2B1.602%5Ctimes%2010%5E%7B-19%7D%20C)
Force of repulsion between two protons on opposite sides of the diameter
![a=2\times r=2\times 4.6656\times 10^{-15} m=9.3312\times 10^{-15} m](https://tex.z-dn.net/?f=a%3D2%5Ctimes%20r%3D2%5Ctimes%204.6656%5Ctimes%2010%5E%7B-15%7D%20m%3D9.3312%5Ctimes%2010%5E%7B-15%7D%20m)
![F=9\times 10^9 N m^2/C^2\times \frac{(+1.602\times 10^{-19} C)\times (+1.602\times 10^{-19} C)}{(9.3312\times 10^{-15} m)^2}](https://tex.z-dn.net/?f=F%3D9%5Ctimes%2010%5E9%20N%20m%5E2%2FC%5E2%5Ctimes%20%5Cfrac%7B%28%2B1.602%5Ctimes%2010%5E%7B-19%7D%20C%29%5Ctimes%20%28%2B1.602%5Ctimes%2010%5E%7B-19%7D%20C%29%7D%7B%289.3312%5Ctimes%2010%5E%7B-15%7D%20m%29%5E2%7D)
![F=2.6527 N](https://tex.z-dn.net/?f=F%3D2.6527%20N)
The electrostatic force of repulsion between two protons on opposite sides of the diameter of the nucleus is 2.6527
Answer:
B. Is its acceleration constant
Explanation:
Uniform circular motion can be described as the motion of an object in a circle at a constant speed. As an object moves in a circle, it is constantly changing its direction. ... An object undergoing uniform circular motion is moving with a constant speed. Nonetheless, it is accelerating due to its change in direction.