A comet orbits the sun in a highly elliptical orbit. Does the comet have a constanta. linear speedb. angular speedc. angular momentum and d. kinetic energy e. potential energy andf. Total energy throughout its Orbit Neglect any mass loss of the comet when it comes very close to the Sun.
Answers
(b) No, angular speed is also not constant as as it is the rate of change of angle which is subtended between the two positions of comet and sun, since planet, speed changes so it covers unequal angles in equal interval of time so its angular speed varies.
(c) yes, angular momentum remains constant. as here there doesn't act any external torque on the system, angular momentum of system will be conserve.
(d) No, Kinetic energy is not same or conserved as the speed of comet is varying and kinetic energy depends only on the speed of particle .
(e) No, Potential energy is also not conserved or same as it depends upon separation of two bodies and as the orbit is elliptical the distance between comet and sun changes continuously so potential energy is also not conserved.
(f) yes, total energy remains constant or conserved. as you know, total energy of the system is the sum of kinetic energy and potential energy and is conserved always according to the law of conservation of energy, so even kinetic energy and potential energy varies, the total energy remains same. i.e. both are interconverted to each other.
Answer:
(a) No, Linear speed of comet is not constant. according to Kepler's 2nd law, if its speed increases when its closer to sun and speed decreases when its farther from sun.
(b) No, angular speed is also not constant as as it is the rate of change of angle which is subtended between the two positions of comet and sun, since planet, speed changes so it covers unequal angles in equal interval of time so its angular speed varies.
(c) yes, angular momentum remains constant. as here there doesn't act any external torque on the system, angular momentum of system will be conserve.
(d) No, Kinetic energy is not same or conserved as the speed of comet is varying and kinetic energy depends only on the speed of particle .
(e) No, Potential energy is also not conserved or same as it depends upon separation of two bodies and as the orbit is elliptical the distance between comet and sun changes continuously so potential energy is also not conserved.
(f) yes, total energy remains constant or conserved. as you know, total energy of the system is the sum of kinetic energy and potential energy and is conserved always according to the law of conservation of energy, so even kinetic energy and potential energy varies, the total energy remains same. i.e. both are interconverted to each other.
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
Physics • 19 hours ago
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
Physics • 19 hours ago
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
Physics • 19 hours ago
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
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.
ANSWER
Physics • 19 hours ago