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Raj Aryan
Subject: Physics
, asked on 14/5/18
Ques no 8 plzz
Answer
1
Raj Aryan
Subject: Physics
, asked on 14/5/18
Fast please
7.
A spherical mass is scooped out from a solid of radius R. The distance between the cavity and centre of sphere is x. The gravitatvonal field inside the cavity is
(A) zero
(B) uniform in direction and constant in magnitude
(C) non-uniform
(D) uniform in direction and variable in magnitude
Answer
1
Raj Aryan
Subject: Physics
, asked on 14/5/18
Solve this:
4.
A system consists of n identical particles of mass m placed rigidly on the vertices of a regular polygon with each side of length a. If K
_{1}
be the kinetic energy imparted to one of the particles so that it just escapes the gravitational pull of the system and thereafter kinetic energy K
_{2}
is given to the adjacent particle to escape, then the difference
$\left({K}_{1}-{K}_{2}\right)$
is
$\left(A\right)\frac{nG{m}^{2}}{a}\left(B\right)\frac{G{m}^{2}}{na}\phantom{\rule{0ex}{0ex}}\left(C\right)\left(\frac{n}{n+1}\right)\frac{G{m}^{2}}{a}\left(D\right)\frac{G{m}^{2}}{a}$
Answer
1
Raj Aryan
Subject: Physics
, asked on 14/5/18
Solve this:
1.
A solid sphere of uniform density and radius R applies a gravitational force of attraction equal to F
_{1}
on a particle placed at a distance 2R from the centre of the sphere. A spherical cavity of radius R/2 is now made in the sphere as shown in the figure The sphere with cavity now applies a gravitational force F
_{2}
on the same particle. The ratio F
_{2}
/F
_{1}
is
(A) 5/9 (B) 7/8
(C) 3/4 (D) 7/9
Answer
1
Raj Aryan
Subject: Physics
, asked on 13/5/18
My answer is coming 4 . but answer given is 2. Plz explain!
Q. Find the ratio of the kinetic energy required to be given to the satellite to escape earth's gravitational field to the KE required to be given so that the satellite moves in a circular orbit just above earth's atmosphere.
Answer
3
Raj Aryan
Subject: Physics
, asked on 13/5/18
$Abodyisthrownfromthesurfaceoftheearthwithvelocity\sqrt{\frac{g{R}_{e}}{2}}atsomeanglewiththevertical.Ifthemaximumheightreachedbythebodyis\frac{{R}_{e}}{4},thenchoosethecorrectoption\left(s\right).({R}_{e}=Radiusofearth)\phantom{\rule{0ex}{0ex}}\left(A\right)theangleofprojectionfromtheverticalis{\mathrm{sin}}^{-1}\left(\frac{\sqrt{5}}{4}\right)\phantom{\rule{0ex}{0ex}}\left(B\right)thevelocityofthebodyatthemaximumheightisv=\sqrt{\frac{g{R}_{e}}{10}}\phantom{\rule{0ex}{0ex}}\left(C\right)theangleofprojectionfromtheverticalis{\mathrm{sin}}^{-1}\left(\frac{\sqrt{5}}{6}\right)\phantom{\rule{0ex}{0ex}}\left(D\right)thevelocityofthebodyatthemaximumheightiszero$
Answer
1
Raj Aryan
Subject: Physics
, asked on 11/5/18
Solve this:
Q.6. The ratio of KE of a planet at the point 1 and 2 is :
(A)
${\left(\frac{{r}_{1}}{{r}_{2}}\right)}^{2}$
(B)
${\left(\frac{{r}_{2}}{{r}_{1}}\right)}^{2}$
(C)
$\frac{{r}_{1}}{{r}_{2}}$
(D)
$\frac{{r}_{2}}{{r}_{1}}$
Answer
1
Raj Aryan
Subject: Physics
, asked on 11/5/18
Plz solve this fast!
13. The distance from the surface the earth at wtuch the acceleration due to gravity is the same below and above the surface of the earth.
$\left(A\right)\left(\sqrt{5}+1\right)R,\left(B\right)\left(\frac{\sqrt{5}+1}{2}\right)R,\phantom{\rule{0ex}{0ex}}\left(C\right)\frac{\left(\sqrt{5}-1\right)R}{2},\left(D\right)\left(\sqrt{5}-1\right)R,$
Answer
1
Ms Eccentric
Subject: Physics
, asked on 10/5/18
cos tetha = -1/2, tetha = ?
Answer
1
Arshak Shan
Subject: Physics
, asked on 3/5/18
A spaceship is moving directly toward a planet at a speed of 70% of the light speed.When the spaceship is 3x10^8 m from the planet, a pulse of light is emitted by the spaceship from an outside observer,how long does it take the light pulse to travel from the spaceship to the planet?
Answer
1
Laieeqa
Subject: Physics
, asked on 17/4/18
Solve this:
8. What is a period of revolution of earth satellite ? Ignore the height of satellite above the surface of earth.
Given ; (1) The value of gravitational acceleration g=10 ms
^{-2}
(2) Radius of earth R
_{E}
=6400 km. Take
$\mathrm{\pi}$
=3.14
(1) 85 minutes (2) 156 minutes
(3) 83.73 minutes (4) 90 minutes.
Answer
2
Isha Dixit
Subject: Physics
, asked on 14/4/18
two satellites are at different heights(smaller and larger)from the surface of earth.which would have greater velocity?
Answer
1
Trishu
Subject: Physics
, asked on 12/4/18
The weight of an object in the coal mine, sealevel, at the top of the mountain are W1, W2 &W3 respectively. Arrange them in increasing order of weight.
Answer
1
Rajarshi Das
Subject: Physics
, asked on 4/4/18
what is escape velocity and what is its value for Earth
Answer
1
Rajarshi Das
Subject: Physics
, asked on 4/4/18
what is orbital velocity and show that it does not depend on the mass of the body
Answer
1
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What are you looking for?

7.A spherical mass is scooped out from a solid of radius R. The distance between the cavity and centre of sphere is x. The gravitatvonal field inside the cavity is (A) zero(B) uniform in direction and constant in magnitude

(C) non-uniform

(D) uniform in direction and variable in magnitude

4.A system consists of n identical particles of mass m placed rigidly on the vertices of a regular polygon with each side of length a. If K_{1}be the kinetic energy imparted to one of the particles so that it just escapes the gravitational pull of the system and thereafter kinetic energy K_{2}is given to the adjacent particle to escape, then the difference $\left({K}_{1}-{K}_{2}\right)$ is$\left(A\right)\frac{nG{m}^{2}}{a}\left(B\right)\frac{G{m}^{2}}{na}\phantom{\rule{0ex}{0ex}}\left(C\right)\left(\frac{n}{n+1}\right)\frac{G{m}^{2}}{a}\left(D\right)\frac{G{m}^{2}}{a}$

1.A solid sphere of uniform density and radius R applies a gravitational force of attraction equal to F_{1}on a particle placed at a distance 2R from the centre of the sphere. A spherical cavity of radius R/2 is now made in the sphere as shown in the figure The sphere with cavity now applies a gravitational force F_{2}on the same particle. The ratio F_{2}/F_{1}is(A) 5/9 (B) 7/8

(C) 3/4 (D) 7/9

Q. Find the ratio of the kinetic energy required to be given to the satellite to escape earth's gravitational field to the KE required to be given so that the satellite moves in a circular orbit just above earth's atmosphere.

Q.6. The ratio of KE of a planet at the point 1 and 2 is :

(A) ${\left(\frac{{r}_{1}}{{r}_{2}}\right)}^{2}$

(B)${\left(\frac{{r}_{2}}{{r}_{1}}\right)}^{2}$

(C) $\frac{{r}_{1}}{{r}_{2}}$

(D) $\frac{{r}_{2}}{{r}_{1}}$

13. The distance from the surface the earth at wtuch the acceleration due to gravity is the same below and above the surface of the earth.

$\left(A\right)\left(\sqrt{5}+1\right)R,\left(B\right)\left(\frac{\sqrt{5}+1}{2}\right)R,\phantom{\rule{0ex}{0ex}}\left(C\right)\frac{\left(\sqrt{5}-1\right)R}{2},\left(D\right)\left(\sqrt{5}-1\right)R,$

8. What is a period of revolution of earth satellite ? Ignore the height of satellite above the surface of earth.

Given ; (1) The value of gravitational acceleration g=10 ms

^{-2}(2) Radius of earth R

_{E}=6400 km. Take $\mathrm{\pi}$=3.14(1) 85 minutes (2) 156 minutes

(3) 83.73 minutes (4) 90 minutes.