1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Vinvika [58]
3 years ago
12

What is the gravitational force between two 1 kg objects that are 1 m apart?

Physics
1 answer:
lapo4ka [179]3 years ago
3 0

Answer:

F=6.67\times 10^{-11}\ N

Explanation:

Given that,

The masses of two objects, m₁ = m₂ = 1 kg

The distance between masses, d = 1 m

We need to find the gravitational force between two masses. The force is given by the relation as follows :

F=G\dfrac{m_1m_2}{d^2}\\\\F=6.67\times 10^{-11}\times \dfrac{1\times 1}{(1)^2}\\F=6.67\times 10^{-11}\ N

So, the force between two masses of 1 kg is 6.67\times 10^{-11}\ N.

You might be interested in
When elements react they can form molecules. What else might they form?
MrMuchimi

Answer:

they react together and form compounds

Explanation:

3 0
3 years ago
Light takes 8 minutes to reach the Earth, and the speed of light is 3.0×10^8 m/s. a) What is the orbital speed of the Earth arou
spin [16.1K]

Answer:

(a) 28690 m/s (b) 2.46x10^{33}J

Explanation:

The orbital speed is define as:

v = \frac{2 \pi r}{T}   (1)

Where r is the radius of the trajectory and T is the orbital period.

To determine the orbital speed of the Earth it is necessary to know the orbital period and the radius of the trajectory. That can be done by means of the Kepler's third law and average velocity equation.

The average velocity in a Uniform Rectilinear Motion is defined as:

v = \frac{d}{t}   (2)

Where v is the velocity, d is the covered distance and t is the time.

Equation 2 can be rewritten for d to get:

d = vt   (3)

In this case, v will be the speed of light and t, the 8 minutes that takes to reach the Earth.

The time will be converted to seconds so the units in equation 3 can match:

8min . \frac{60s}{1min} ⇒  480s

t = 480s

Replacing all those values in equation 3 it is gotten:

d = (3.0x10^{8}m/s)(480s)

d = 1.44x10^{11}m

Kepler’s third law is defined as:

T^{2} = r^{3}

Where T is orbital period and r is the radius of the trajectory.

T = \sqrt{r^{3}}

T = \sqrt{(1.44x10^{11}m)^{3}}

It is necessary to pass from meters to astronomical unit (AU), 1 AU is defined as the distance between the Earth and the Sun.

T = \sqrt{1AU}

T = 1AU

That can be expressed in units of years.

T = 1AU . \frac{1year}{1AU}

T = 1year

But there are 31536000 seconds in one year:

T = 1year . \frac{31536000s}{1year}

T = 31536000s

Finally, equation  1 can be used:

v = \frac{2 \pi (1.44x10^{11}m)}{(31536000s)}

v = 28690 m/s

<u>So Earth orbital speed around the Sun is 28690 m/s.</u>

<em>b) What is its kinetic energy?</em>

The kinetic energy is defined as:

E = \frac{1}{2}mv^{2}  (4)

Notice that it is necessary to found the mass of the Earth, that can be done combining the Universal law of gravity and Newton's second law:

F = \frac{GMm}{r^{2}}

ma = \frac{GMm}{r^{2}}  (5)

M will be isolated in equation 5:

M = \frac{r^{2}a}{G}

Where r is the radius of the Earth (6.38x10^{6}m)

M = \frac{(6.38x10^{6}m)^{2}(9.8m/s^{2})}{6.67x10^{-11}kg.m/s^{2}.m^{2}/Kg^{2}})

M = 5.98x10^{24} Kg

E = \frac{1}{2}( 5.98x10^{24} Kg))(28.690m/s)^{2}

E = 2.46x10^{33}Kg.m^{2}/s^{2}

E = 2.46x10^{33}J

<u>Hence, the kinetic energy of Earth is 2.46x10^{33}J.</u>

8 0
3 years ago
There is a tree standing in the desert; the sun is rising to the east.
aksik [14]

a. As sun peeks, it creates a straight line- sun to tree to shadow, angle=180°

b. During morning, the sun-tree-shadow changes from 180° to 90°, during noon, angle =90°.During sunset the angle changes from 90 to 0°.

c. Acute angles created in afternoon.

d. A right angle would be created at noon.

e. Obtuse angle would be created in morning.

f. Yes, a straight angle would be created at sunrise.

g. When sun is at mid-morning, the angle=45°+90°= 135°

5 0
3 years ago
During a tennis match, a player serves the ball at a speed s, with the center of the ball leaving the racquet at angle θ below t
Pachacha [2.7K]

Answer: Maximum distance

= {s²/g} * sine(2*theta)unit

Explanation: This is a projectile motion problem. The horizontal distance between the tennis player and where the tennis reaches over the net is given by the horizontal Range.

Range = {s² * sine2*theta}/g

(s)is the initial speed of projection

Theta is the angle of projection

g is acceleration due to gravity 10m/s².

5 0
3 years ago
a super man I sitting on a tree 98m high with a he has rescued from a claws of a tiger. unfortunately the child shift and falls
nadezda [96]

Superman was a man who fell of the tree because he was up high at 98m and the. Tried to rescured Annie
5 0
2 years ago
Other questions:
  • What are the two factors that determine the period of a pendulum? acceleration of gravity and the mass of the bob mass of the bo
    5·2 answers
  • Why does icing sugar dissolve more rapidly than granular sugar?
    14·1 answer
  • Brendan drew a diagram to illustrate the centripetal force acting on a satellite.
    11·2 answers
  • 5) [Honors]A seagull, ascending straight upward at 5.2 m/s, drops a shell when it is 12.5m above the ground. (A)
    10·1 answer
  • Where should an object be placed in front of a concave mirror’s principal axis to form an image that is real, inverted, larger t
    11·2 answers
  • Which of the following statements are true of solar energy?
    6·1 answer
  • How information is transfered via electromagnetic waves
    9·1 answer
  • A man with mass m1 = 53.0 kg stands at the left end of a uniform boat with mass m2 = 172.0 kg and a length L = 2.5 m. Let the or
    13·1 answer
  • Why we have leap year?
    5·1 answer
  • The parasympathetic nervous system is part of the
    10·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!