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Bad White [126]

2 years ago

9

The gravitational field strength at a distance R from the center of moon is gR. The satellite is moved to a new circular orbit t

hat is 2R from the center of the moon. What is the gravitational field strength of the moon at this new distance?

1 answer:

3241004551 [841]2 years ago

6 0

Answer:

$g'=\frac{g__R}{4}$

Explanation:

Given:

gravitational field strength of moon at a distance R from its center, $g__R$

Distance of the satellite from the center of the moon, $h=2R$

<u>Now as we know that the value of gravity of any heavenly body is at height h is given as:</u>

$g'=g__{R}} \times \frac{R^2}{(2R)^2}$

$g'=\frac{g__R}{4}$

∴The gravitational field strength will become one-fourth of what it is at the surface of the moon.

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A cart starts from rest and accelerates at 4.0 m/s2 for 5.0 s, then maintains that velocity for 10 s, and then decelerates at th

zhannawk [14.2K]

Answer:

Final speed of car = 12 m/s

Explanation:

We have equation of motion v = u + at, where v is final velocity, u is initial velocity, a is acceleration and t is time.

a) A cart starts from rest and accelerates at 4.0 m/s² for 5.0 s

v = ?

u = 0 m/s

a = 4.0 m/s²

t = 5 s

v = u + at = 0 + 4 x 5 = 20 m/s

b) Then maintains that velocity for 10 s

v = ?

u = 20 m/s

a = 0 m/s²

t = 10 s

v = u + at = 20 + 0 x 10 = 20 m/s

c) Then decelerates at the rate of 2.0 m/s² for 4.0 s

v = ?

u = 20 m/s

a = -2.0 m/s²

t = 4 s

v = u + at = 20 + -2 x 4 = 12 m/s

Final speed of car = 12 m/s

3 0

1 year ago

A child is sliding a toy block (with mass = m) down a ramp. The coefficient of static friction between the block and the ramp is

tiny-mole [99]

Answer:

$F=mg(sin(\theta )-0.25 cos(\theta ))$

Explanation:

The free body diagram of the block on the slide is shown in the below figure

Since the block is in equilibrium we apply equations of statics to compute the necessary unknown forces

N is the reaction force between the block and the slide

For equilibrium along x-axis we have

$\sum F_{x}=0\\\\mgsin(\theta )-\mu N-F=0\\\therefore F=mgsin(\theta)-\mu N......(\alpha )\\Similarly\\\sum F_{y}=0\\\\N-mgcos(\theta )=0\\\therefore N=mgcos(\theta ).......(\beta )\\\\$

Using value of N from equation β in α we get value of force as

$F=mg(sin(\theta )-\mu cos(\theta ))$

Applying values we get

$F=mg(sin(\theta )-0.25 cos(\theta ))$

8 0

1 year ago

Read 2 more answers

En el País Vasco los deportistas rurales levantan enormes piedras hasta su hombro. En un concurso , Jose levanta una piedra de 2

Mama L [17]

Answer:

Txomin lifted the stone with greater mass. (Txomin levantó la roca con mayor fuerza).

Explanation:

The sportsman that lifts the stone with a greater mass needs a higher force (El deportista que levanta la piedra con mayor masa necesita una mayor fuerza):

José

$F = (200\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)$

$F = 1961.4\,N$

Txomin

$F = (220\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)$

$F = 2157.54\,N$

Txomin lifted the stone with greater mass. (Txomin levantó la roca con mayor fuerza).

5 0

1 year ago

A 2.0 kg block on a horizontal frictionless surface is attached to a spring whose force constant is 590 N/m. The block is pulled

SVETLANKA909090 [29]

Answer:

The value is $v = -0.04 \ m/s$

Explanation:

From the question we are told that

The mass of the block is $m = 2.0 \ kg$

The force constant of the spring is $k = 590 \ N/m$

The amplitude is $A = + 0.080$

The time consider is $t = 0.10 \ s$

Generally the angular velocity of this block is mathematically represented as

$w = \sqrt{\frac{k}{m} }$

=> $w = \sqrt{\frac{590}{2} }$

=> $w = 17.18 \ rad/s$

Given that the block undergoes simple harmonic motion the velocity is mathematically represented as

$v = -A w sin (w* t )$

=> $v = -0.080 * 17.18 sin (17.18* 0.10 )$

=> $v = -0.04 \ m/s$

7 0

2 years ago

In 2014, the Rosetta space probe reached the comet Churyumov Gerasimenko. Although the comet's core is actually far from spheric

Viktor [21]

To solve this problem we will apply the concepts related to gravity according to the Newtonian definitions. From finding this value we will use the linear motion kinematic equations to find the time. Our values are

Comet mass $M = 1.0*10^{13} kg$

Radius $r = 1.6km = 1600 m$

Rock was dropped from a height 'h' from surface = 1m

The relation for acceleration due to gravity of a body of mass 'm' with radius 'r' is

$g = \frac{GM}{R^2}$

Where G means gravitational universal constant and M the mass of the planet

$g = \frac{(6.67408*10^{-11})(1*10^{13})}{1600^2}$

$g = 2.607*10^{-4} m/s^2$

Now calculate the value of the time

$h = \frac{1}{2} gt^2$

$t = \sqrt{\frac{2h}{g}}$

$t = \sqrt{\frac{2(1)}{2.607*10^{-4}}}$

$t = 87.58s$

The time taken for the rock to reach the surface is t = 87.58s

8 0

1 year ago