A & T₂ \(\alpha\) R³ ; R_e = 6400 km
\(\frac { { T }^{ 2 } }{ { \left( 24 \right) }^{ 2 } } ={ \left( \frac { 6400 }{ 3600 } \right) }^{ 2 }\)or T = 1.7h
for the specially satellite, R is slightly greater than R_e.
So T_s>T or T_s= 2h

Sound waves can be classified in three groups according to their range of frequencies:
(1) Infrasonic waves : Sound waves having frequencies below 20. Hz are called infrasonic waves. These waves are produced during earthquakes. Human beings cannot hear these frequencies. Snakes can hear these frequencies.
(2) Audible waves : Sound waves having frequencies between 20 Hz to 20,000 Hz (20kHz) are called audible waves. Human beings can hear these frequencies.
(3) Ultrasonic waves : Sound waves having frequencies greater than 20 kHz are known as ultrasonic waves. Human beings cannot hear these frequencies. Bats can produce and hear these frequencies.

Since the process is adiabatic
P₁V₁r = P₂V_2r
\(\frac { { V }_{ 1 } }{ { V }_{ 2 } } =\frac { 2 }{ 1 } \)
\(\gamma =\frac { 7 }{ 5 } \) for hydrogen
\(\therefore\) Factor by which the pressure of the gas increases
\(\frac { { P }_{ 2 } }{ { P }_{ 1 } } =\frac { 2 }{ 1 } =(2)^{ \frac { 7 }{ 3 } }\)

mgh=\(\frac { 1 }{ 2 } m{ v }^{ 2 }\)
so h = 20.2 m

F = 36 dyne at an angle of 60°
\({ F }_{ \underline { x } }=\) F cos 60° = 18 dyne
F_x = ma_x
So \({ a }_{ x }=\frac { { F }_{ x } }{ m } =1\) cm/s²

Mass (m) = 5 kg
Co-efficient of static friction us = 0.6
Acceleration due to gravity g = 9.8 ms^-2
The value of limiting friction
F_s(max)  = us R = us mg = 0.6\(​​\times\)5\(​​\times\) 9. 8 = 29. 4
As the applied force of 3.8 N is less than the limiting friction 29.4 so the block does not move Therefore,
Force of friction = Applied force = 3.8 N.

(i) In terms of components, we can write
\(\overrightarrow { a } =\frac { { dv }_{ x } }{ dt } \hat { i } +\frac { { dv }_{ y } }{ dt } \hat { j } +\frac { dv_{ z } }{ dt } \hat { k } =\frac { d\overrightarrow { v } }{ dt } \)
Thus \({ a }_{ x }=\frac { { dv }_{ x } }{ dt } ,{ a }_{ y }=\frac { { dv }_{ y } }{ dt } ,{ a }_{ z }=\frac { { dv }_{ z } }{ dt } \) are the components of instantaneous acceleration.
(ii) Since each component of velocity is the derivative of the corresponding coordinate, we can express the components
a_x, a_y, a_z, as
\({ a }_{ x }=\frac { { d }^{ 2 }x }{ { dt }^{ 2 } } ,{ a }_{ y }=\frac { { d }^{ 2 }y }{ { dt }^{ 2 } } ,{ a }_{ z }=\frac { { d }^{ 2 }z }{ { dt }^{ 2 } } \)
(iii) Then the acceleration vector \(\overrightarrow { A } \) itself is \(\overrightarrow { a } =\frac { { d }^{ 2 }x }{ { dt }^{ 2 } } \hat { i } +\frac { { d }^{ 2 }y }{ { dt }^{ 2 } } \hat { j } +\frac { { d }^{ 2 }z }{ { dt }^{ 2 } } \hat { k } =\frac { { d }^{ 2 }\overrightarrow { r } }{ { dt }^{ 2 } } \)
Thus acceleration is the second derivative of position vector with respect to time.

distance from the earth (r) = 10000 km = 107 m
Mass of the satellite = 3000 kg
Speed v = 4500 ms^-1 = 4.5\(​​\times\)103 ms^-1
Now, the centripetal force is F = \(\frac { { mv }^{ 2 } }{ r } \)
\(F=\frac { 3000\times (4.5\times { 10 }^{ 3 }) }{ { 10 }^{ 7 } } \)
= \(\frac { 3000\times (20.25\times { 10 }^{ 6 }) }{ { 10 }^{ 7 } } \)
= \(\frac { 300\times 2.25\times { 10 }^{ 7 } }{ { 10 }^{ 7 } } \)
=300\(​​\times\)20.25\(​​\times\)10^-7x10⁷
=300\(​​\times\)20.25
The centripetal forces is, F= 6075 N

(i) Biological studies are impossible without a microscope designed using physics principles.
(ii) The invention of the electron microscope has made it possible to see even the structure of a cell. X-ray and neutron diffraction techniques have helped us to understand the structure of nucleic acids, which help to control vital life processes. X-rays are used for diagnostic purposes.
(iii) Radio-isotopes are used in radiotherapy for the cure of cancer and other diseases. In recent years, biological processes are being studied from the physics point of view.