Ask question

Ask question

All categories

nataly862011 [7]

4 years ago

11

1. why does the sun have the greatest amount of gravity in our solar system?

2 answers:

gogolik [260]4 years ago

6 0

The lighter object orbits the heavier one, and the Sun is, by far, the heaviest object in the solar system. ... Heavier objects (really, more massive ones) produce a bigger gravitational pull than lighter ones, so as the heavyweight in our solar system, the Sun exerts the strongest gravitational pull.

melomori [17]4 years ago

3 0

Answer:

it is the biggest object that is the closest to us

Explanation:

You might be interested in

In which situation is the speed of the car constant while its velocity is changing?

Sati [7]

Answer:

B, the car travels around a circular track at 30 m.

Explanation:

7 0

2 years ago

magine an astronaut on an extrasolar planet, standing on a sheer cliff 50.0 m high. She is so happy to be on a different planet,

Mama L [17]

Answer:

$\Delta t=(\frac{20}{g'}+\sqrt{\frac{400}{g'^2}+\frac{100}{g'} } )-(\frac{20}{g}+\sqrt{\frac{400}{g^2}+\frac{100}{g} } )$

Explanation:

Given:

height above which the rock is thrown up, $\Delta h=50\ m$

initial velocity of projection, $u=20\ m.s^{-1}$

let the gravity on the other planet be g'

The time taken by the rock to reach the top height on the exoplanet:

$v=u+g'.t'$

where:

$v=$ final velocity at the top height = 0 $m.s^{-1}$

$0=20-g'.t'$ (-ve sign to indicate that acceleration acts opposite to the velocity)

$t'=\frac{20}{g'}\ s$

The time taken by the rock to reach the top height on the earth:

$v=u+g.t$

$0=20-g.t$

$t=\frac{20}{g} \ s$

Height reached by the rock above the point of throwing on the exoplanet:

$v^2=u^2+2g'.h'$

where:

$v=$ final velocity at the top height = 0 $m.s^{-1}$

$0^2=20^2-2\times g'.h'$

$h'=\frac{200}{g'}\ m$

Height reached by the rock above the point of throwing on the earth:

$v^2=u^2+2g.h$

$0^2=20^2-2g.h$

$h=\frac{200}{g}\ m$

The time taken by the rock to fall from the highest point to the ground on the exoplanet:

$(50+h')=u.t_f'+\frac{1}{2} g'.t_f'^2$ (during falling it falls below the cliff)

here:

$u=$ initial velocity= 0 $m.s^{-1}$

$\frac{200}{g'}+50 =0+\frac{1}{2} g'.t_f'^2$

$t_f'^2=\frac{400}{g'^2}+\frac{100}{g'}$

$t_f'=\sqrt{\frac{400}{g'^2}+\frac{100}{g'} }$

Similarly on earth:

$t_f=\sqrt{\frac{400}{g^2}+\frac{100}{g} }$

Now the required time difference:

$\Delta t=(t'+t_f')-(t+t_f)$

$\Delta t=(\frac{20}{g'}+\sqrt{\frac{400}{g'^2}+\frac{100}{g'} } )-(\frac{20}{g}+\sqrt{\frac{400}{g^2}+\frac{100}{g} } )$

3 0

3 years ago

Choose the false statement concerning parallel circuits.

alisha [4.7K]

Current can vary in different branches of a circuit

8 0

4 years ago

You have a stopped pipe of adjustable length close to a taut 62.0-cm, 7.25-g wire under a tension of 3910 N . You want to adjust

Ber [7]

Answer:

L = 6 cm

Explanation:

Second overtone of the wire is same as third harmonic

so its frequency is given as

$f = \frac{3}{2L}\sqrt{\frac{T}{\mu}}$

here we know that

$\mu = \frac{M}{L}$

$\mu = \frac{7.25 \times 10^{-3}}{0.62}$

$\mu = 0.0117 kg/m$

now we have

$f = \frac{3}{2\times 0.62}\sqrt{\frac{3910}{0.0117}}$

$f = 1398.6 Hz$

Now fundamental frequency of sound in a pipe is given as

$f = \frac{v}{4L}$

$1398.6 = \frac{340}{4L}$

$L = 0.06 m$

$L = 6 cm$

7 0

3 years ago

A tiger paces back and forth along the front of its cage, which is 8 m wide. The tiger starts from the right side of the cage, p

Black_prince [1.1K]

Answer:

The total distance is 24 m.

The tiger's resultant displacement is 8 m.

Explanation:

Given that,

Width of tiger= 8 m

Tiger starts from the right side of the cage,

Paces to the left side

$\Delta x_{1}=+8\ m$

Then, back to the right side,

$\Delta x_{2}=-8\ m$

finally, back to the left

$\Delta x_{3}=+8\ m$

We need to calculate the total distance covered

Using formula of distance

$d= x_{1}+x_{2}+x_{3}$

Put the value into the formula

$d=8+8+8$

$d=24\ m$

We need to calculate the tiger's resultant displacement

Using formula of displacement

$D=\Delta x_{1}+\Delta x_{2}+\Delta x_{3}$

Put the value into the formula

$D=8+(-8)+8$

$D=8\ m$

Hence, The total distance is 24 m.

The tiger's resultant displacement is 8 m.

3 0

3 years ago