Ask question

Ask question

All categories

2 years ago

8

Martha (50 kg) is attracted to Stewart (70 Kg) who sits 4 m away. What is the gravitational attraction between them? G=6.67 x 10

^-11 (-11 is an exponent)

1 answer:

kodGreya [7K]2 years ago

8 0

Answer:

1.46× 10^-8 N

Explanation:

F=Gm₁m₂/r²

F=(6.67×10⁻¹¹ × 50× 70)/(4)²

F=1.46× 10^-8 N

Hope, this helps you.

You might be interested in

The cable of the 1800kg elevator cab in Fig. 8−56 snaps when the cab is at rest at the first floor, where the cab bottom is a di

drek231 [11]

a ) The speed of the cab just before it hits the spring = 7.4 m / s

b ) The maximum distance x that the spring is compressed = 0.9 m

c ) The distance that the cab will bounce back up the shaft = 2.8 m

a ) The speed of the cab just before it hits the spring,

Ki + Pi = K final + P final + W

1 / 2 m vo² + m g hi = 1 / 2 m v² + m g h + f d

0 + ( 1800 * 9.8 * 3.7 m ) = ( 1 / 2 * 1800 * v² ) + 0 + ( 4400 * 3.7 )

v = 7.4 m / s

b ) The maximum distance x that the spring is compressed,

Ki + Pi = K final + P final + W + Fs

1 / 2 m vo² + m g x = 1 / 2 m v² + m g h + f d + 1 /2 k x²

( 1/2 * 1800 * 7.4² ) + ( 1800 * 9.8 * x ) =0 + 0+ ( 4400 * x ) + ( 1/2 * 1800 * x² )

75000 x² + 13420 x - 50625 = 0

x = 0.9 m

c ) The distance that the cab will bounce back up the shaft,

Ki + Pi + Fs = K final + P final + W

1 / 2 m vo² + m g x + 1 /2 k x² = 1 / 2 m v² + m g h + f d

0 + 0 + ( 1 / 2 * 0.15 * 0.9² ) = 0 + ( 1800 * 9.8 * h ) + ( 4400 * h )

h = 2.8 m

Therefore,

a ) The speed of the cab just before it hits the spring = 7.4 m / s

b ) The maximum distance x that the spring is compressed = 0.9 m

c ) The distance that the cab will bounce back up the shaft = 2.8 m

To know more about law of conservation of energy

brainly.com/question/12050604

#SPJ4

3 0

1 year ago

Sergeu [11.5K]

The vertical component of the initial velocity is $v_0_y = \frac{y}{t} + \frac{1}{2} gt$

The horizontal component of the initial velocity is $v_0_x = \frac{x}{t}$

The horizontal displacement when the object reaches maximum height is $X = \frac{xy}{gt^2} + \frac{x}{2}$

The given parameters;

the horizontal displacement of the object, = x

the vertical displacement of the object, = y

acceleration due to gravity, = g

time of motion, = t

The vertical component of the initial velocity is given as;

$y = v_0_yt - \frac{1}{2} gt^2\\\\v_0_yt = y + \frac{1}{2} gt^2\\\\v_0_y = \frac{y}{t} + \frac{1}{2} gt$

The horizontal component of the initial velocity is calculated as;

$x = v_0_xt\\\\v_0_x = \frac{x}{t}$

The time to reach to the maximum height is calculated as;

$T = \frac{v_f_y -v_0_y}{-g} \\\\T = \frac{-v_0_y}{-g} \\\\T = \frac{v_0_y}{g} \\\\T = \frac{1}{g} (v_0_y)\\\\T = \frac{1}{g} (\frac{y}{t} + \frac{1}{2} gt)\\\\T = \frac{y}{gt} + \frac{1}{2} t$

The horizontal displacement when the object reaches maximum height is calculated as;

$X= v_0_x \times T\\\\X= \frac{x}{t} \times (\frac{y}{gt} + \frac{1}{2} t)\\\\X = \frac{xy}{gt^2} + \frac{x}{2}$

Learn more here: brainly.com/question/20689870

8 0

3 years ago

a hawk flies in a horizontal arc of radius 10.3 m at a constant speed of 4.8 m/s. find its centripetal acceleration. answer in u

n200080 [17]

The hawk’s centripetal acceleration is 2.23 m/s²

The magnitude of the acceleration under new conditions is 2.316 m/s²

radius of the horizontal arc = 10.3 m

the initial constant speed = 4.8 m/s

we know that the centripetal acceleration is given by

$a_{c}$ = $\frac{v^{2} }{r}$

$a_{c}$ = 23.04/10.3

$a_{c}$ = 2.23 m/s²

It continues to fly but now with some tangential acceleration

$a_{t}$ = 0.63 m/s²

therefore the net value of acceleration is given by the resultant of the centripetal acceleration and the tangential acceleration

so

$a_{net}$ = $\sqrt{a_{c} ^{2} +a_{t} ^{2} }$

$a_{net}$ = $\sqrt{4.97 + 0.396}$

$a_{net}$ = 2.316 m/s²

So the magnitude of net acceleration will become 2.316 m/s².

learn more about acceleration here :

brainly.com/question/11560829

#SPJ4

8 0

1 year ago

Optical astronomers need a clear, dark sky to collect good data. Explain why radio astronomers have no problem observing in the

gregori [183]

Answer:

In the clarification portion elsewhere here, the definition of the concern is mentioned.

Explanation:

So like optical telescopes capture light waves, introduce it to concentrate, enhance it, as well as make it usable through different instruments via study, so radio telescopes accumulate weak signal light waves, introduce that one to focus, enhance it, as well as make this information available during research. To research naturally produced radio illumination from stars, galaxies, dark matter, as well as other natural phenomena, we utilize telescopes.

Optical telescopes detect space-borne visible light. There are some drawbacks of optical telescopes mostly on the surface:

Mostly at night would they have been seen.
Unless the weather gets cloudy, bad, or gloomy, they shouldn't be seen.

Although radio telescopes monitor space-coming radio waves. Those other telescopes, when they are already typically very massive as well as costly, have such an improvement surrounded by optical telescopes. They should be included in poor weather and, when they travel through the surrounding air, the radio waves aren't obscured by clouds. Throughout the afternoon and also some at night, radio telescopes are sometimes used.

3 0

3 years ago

What do you think would happen to the force of attraction of two interacting charges if their distance apart is halved?

sweet [91]

Answer:

The new force becomes 4 times the initial force.

Explanation:

The force of attraction or repulsion is given by the relation as follows :

$F=k\dfrac{q_1q_2}{d^2}$

Where

d is the distance between the interacting charges

F is inversely proportional to the distance between charges.

If the distance is halved, d'=(d/2), new force is given by :

$F'=k\dfrac{q_1q_2}{d'^2}\\\\=k\dfrac{q_1q_2}{(\dfrac{d}{2})^2}\\\\=k\dfrac{q_1q_2}{\dfrac{d^2}{4}}\\\\=4\times \dfrac{kq_1q_2}{d^2}\\\\F'=4F$

So, the new force becomes 4 times the initial force.

4 0

3 years ago