A satellite orbits the earth at a height equal to the radius of earth. Find its (i) orbital speed and (ii) period of revolution.
Answers
a/c to question, the satellite orbits the earth at a height equal to the radius of earth.
so, h = R = 6400 km { from data }
now, orbital speed ,
=
= 5.592 km/s
(ii) time period is given by,
here, h = R = 6400km
T = 2π(R + R)/5.592 × 10³
T = 2 × 3.14 × 2 × 6.4 × 10^6/5592
T = 14374.8 sec ≈ 4 hours
Answer:
Answer
7
3.0
Answer:
r = 0.5mm = 0.5 × 10⁻³m
= 5 × 10⁻⁴ m
T = 0.04 N/m
ρ = 0.8 gm/cc
= 0.8x 10 ⁻³/10 ⁻⁶ = 800 kg/m3
θ = 10⁰
g = 9.8 m/s²
T = rh ρ g /2 cos θ
∴ h = 2T cos θ/ r g ρ
∴ h = 2x0.04 cos10/ 5 x10⁻⁴x 800x 9.8
h = 2x 0.04x 0.9848/ 5x 800x 9.8 10⁻⁴
h = 2.01 × 10⁻² m
∴ The height of the capillary rise is 2.01 × 10⁻² m
(i) we know, orbital speed at a height h, from the surface of the earth is given by,
v_0=\sqrt{\frac{GM}{R+h}}
v_0=\sqrt{\frac{GM}{R+h}}
Physics • 19 hours ago
a/c to question, the satellite orbits the earth at a height equal to the radius of earth.
Answer:
Given :
R = 0.5cm = 0.5 × 10⁻²m
n = No. of drops formed = 10⁶
ρ = 13600 kg/m3 ,
T = 0.465 N/m
W = T∆A
P.E = mgh
Volume of single big drop
V = 4 / 3 π R ³
Volume of single small droplet = 4 /3 π r³
Volume of n droplets = n × 4/ 3 πr³
∴ 4/ 3 πR³ = n 4/ 3 π r³
∴ R³ = nr³
∴ r = R /∛n = 0.5x 10⁻² /∛ 10⁶
∴ r = 0.5 × 10⁻²m /10²
r=0.5 x 10⁻⁴m
The single drop is fallen from height h, hence its P.E. = mgh
But, P.E = Work done due to change in area ...(i)
Change in surface area ∆A = (n4πr²2 – 4πR²)
Also, W = T∆A =
T (n4πr ² – 4πR ² )
W = 4πT (nr² – R² )
∴ eq., (i) becomes,
mgh = 4πT (nr²– R² )
But, m = 4 /3 πR³ρ
∴ 4 /3 πR³ρgh = 4πT (nr² – R² )
R³ gh ρ/3 = T (nr² – R²)
∴ h = 3T (nr² – R² )/ R ³ ρ g
=3x0.465[10 ⁶ (0.5x10⁻⁴)² – (0.5x10⁻²)²]/ (0.5x10⁻²)3x13600x9.8
H=3x0.465[0.25x10⁻²)- (0.25x10⁻⁴)]/0.125x1.36x10⁴ x 9.8
On solving
=3x0.465x25x0.09/12.5x1.36x9.8
H=0.2072m
so, h = R = 6400 km { from data }
Physics • 19 hours ago
v_0=\sqrt{\frac{GM}{2R}}
v_0=\sqrt{\frac{GM}{2R}}
Explanation:
Given :
T = 435.5 dyne/cm = 0.4355 N/m,
θ = 14
0
ρ = 13600 kg/m³
d = 1 cm
∴ r = 0.5 cm
= 5 × 10⁻³ m
T = rh ρ g/ 2 cos θ
h = 2T cos θ/ r g ρ
∴ h = 2 x0.4355x cos140⁰/ 5 x10⁻³x 13600 x9.8
= 0.8710x cos( 90 + 50 )/ 5x 13.6 x9.8
=0.8710( –sin 50 ) /5x 13.6 x9.8
=–0.8710x 0.7660/ 68.0x 9.8
= – 1.001 × 10⁻³m
∴ h = – 1.001 mm
here Negative sign indicates that mercury level will be lowered by 1.001 mm.
Hence to get correct reading h = 1.001 mm has to added.
∴ h = 1.001 mm
now, orbital speed ,
\sqrt{\frac{6.67\times10^{-11}\times6\times10^{24}}{2\times6.4\times10^6}}
\sqrt{\frac{6.67\times10^{-11}\times6\times10^{24}}{2\times6.4\times10^6}}
=
Physics • 19 hours ago
= 5.592 km/s
Answer:
Given :
d1 = 1mm
∴ r 1 = d 1/2 = 1/ 2
= 0.5 mm = 0.05 cm
d2 = 1.1 mm
∴ r 2 = d2/ 2 = 1.1/ 2 = 0.55 mm = 0.55 cm
T = 75 dyne/cm
F = Tl
l = 2π (r1 + r2 )
F = Tl
= T × 2π (r1 + r2 )
∴ F = 2πT (r1 + r2 )
= 2 × 3.142 × 75 (0.05 + 0.055)
= 2 × 3.142 × 75 × 0.105
∴ F = 150 × 3.142 × 0.105
∴ F = 49.49 dynes
T=\frac{2\pi(R+h)}{v_0}
T=\frac{2\pi(R+h)}{v_0}
(ii) time period is given by,
Physics • 19 hours ago
Answer:
given :
nD = tuning fork D frequency = 340 Hz
nC = tuning fork C frequency=8 – 4 = 4 beats per second.
First nC ± nD = 8 (before filing)
nC ± nD = 4 (after filing)
From given condition nC ± nD = 8
∴ nC ± 340 = 8
∴ nC = 340 + 8 = 348 Hz
or nC = 340 – 8 = 332 Hz
when tuning fork C is filed then nC ± nD = 4
∴ nC ± 340 = 4
∴ nC = 340 + 4 = 344 Hz
or nC = 340 – 4 = 336 Hz
The frequency of tuning fork increases on filing. Hence nC ≠ 344 Hz.
If original frequency of tuning fork C is taken 332 Hz, then on filing both the value 344 Hz, and 336 Hz are greater. Also it produces 4 beats per second with tunning fork D.
∴ frequency of tuning fork C = 332 Hz
∴ nC = 332 Hz.
here, h = R = 6400km
ANSWER
T = 2π(R + R)/5.592 × 10³
Physics • 19 hours ago