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
slavikrds [6]
2 years ago
11

What is the gravitational force between two students, John and Mike, if John has a mass of 81.0 kg, Mike has a mass of 93.0 kg,

and their centers are separated by a distance of .620 m?
Physics
1 answer:
Marianna [84]2 years ago
8 0

Answer:

1.31×10¯⁶ N

Explanation:

From the question given above, the following data were obtained:

Mass of John (M₁) = 81 Kg

Mass of Mike (M₂) = 93 Kg

Distance apart (r) = 0.620 m

Gravitational constant (G) = 6.67×10¯¹¹ Nm²/Kg²

Force (F) =?

The gravitational force between the two students, John and Mike, can be obtained as follow:

F = GM₁M₂ / r²

F = 6.67×10¯¹¹ × 81 × 93 / 0.62²

F = 6.67×10¯¹¹ × 7533 / 0.3844

F = 1.31×10¯⁶ N

Therefore, the gravitational force between the two students, John and Mike, is 1.31×10¯⁶ N

You might be interested in
Behavior is a product of both the situation and the person. Which of the following is an example of a situation that might influ
Montano1993 [528]
Self productive and it depends on how whom is behaving.
7 0
2 years ago
A man is standing on a weighing machine on a ship which is bobbing up and down with simple harmonic motion of period T=15.0s.Ass
STALIN [3.7K]

Well, first of all, one who is sufficiently educated to deal with solving
this exercise is also sufficiently well informed to know that a weighing
machine, or "scale", should not be calibrated in units of "kg" ... a unit
of mass, not force.  We know that the man's mass doesn't change,
and the spectre of a readout in kg that is oscillating is totally bogus.

If the mass of the man standing on the weighing machine is 60kg, then
on level, dry land on Earth, or on the deck of a ship in calm seas on Earth,
the weighing machine will display his weight as  588 newtons  or as 
132.3 pounds.  That's also the reading as the deck of the ship executes
simple harmonic motion, at the points where the vertical acceleration is zero.

If the deck of the ship is bobbing vertically in simple harmonic motion with
amplitude of M and period of 15 sec, then its vertical position is 

                                     y(t) = y₀ + M sin(2π t/15) .

The vertical speed of the deck is     y'(t) = M (2π/15) cos(2π t/15)

and its vertical acceleration is          y''(t) = - (2πM/15) (2π/15) sin(2π t/15)

                                                                = - (4 π² M / 15²)  sin(2π t/15)

                                                                = - 0.1755 M sin(2π t/15) .

There's the important number ... the  0.1755 M.
That's the peak acceleration.
From here, the problem is a piece-o-cake.

The net vertical force on the intrepid sailor ... the guy standing on the
bathroom scale out on the deck of the ship that's "bobbing" on the
high seas ... is (the force of gravity) + (the force causing him to 'bob'
harmonically with peak acceleration of  0.1755 x amplitude).

At the instant of peak acceleration, the weighing machine thinks that
the load upon it is a mass of  65kg, when in reality it's only  60kg.
The weight of 60kg = 588 newtons.
The weight of 65kg = 637 newtons.
The scale has to push on him with an extra (637 - 588) = 49 newtons
in order to accelerate him faster than gravity.

Now I'm going to wave my hands in the air a bit:

Apparent weight = (apparent mass) x (real acceleration of gravity)

(Apparent mass) = (65/60) = 1.08333 x real mass.

Apparent 'gravity' = 1.08333 x real acceleration of gravity.

The increase ... the 0.08333 ... is the 'extra' acceleration that's due to
the bobbing of the deck.

                        0.08333 G  =  0.1755 M

The 'M' is what we need to find.

Divide each side by  0.1755 :          M = (0.08333 / 0.1755) G

'G' = 9.0 m/s²
                                       M = (0.08333 / 0.1755) (9.8) =  4.65 meters .

That result fills me with an overwhelming sense of no-confidence.
But I'm in my office, supposedly working, so I must leave it to others
to analyze my work and point out its many flaws.
In any case, my conscience is clear ... I do feel that I've put in a good
5-points-worth of work on this problem, even if the answer is wrong .

8 0
3 years ago
Examples of energy balance: physical education
mart [117]

Answer:children burn calories to being a student  

Explanation:That mean when a children getting ready to go to high school

7 0
3 years ago
Charles Darwin,
grandymaker [24]
The answer is B artificial selection
3 0
3 years ago
Read 2 more answers
What’s is the relationship between energy and motion ?
LekaFEV [45]
Motion energy is the sum of potential and kinetic energy in an object that is used to do work.
5 0
3 years ago
Other questions:
  • Which best describes the transition from gas to liquid?
    10·2 answers
  • What is the most precise method for recording data points during an experiment? A. plotting the data point on an X-Y graph B. cr
    14·2 answers
  • What is unit of pressure
    13·2 answers
  • A forklift raises a crate weighing 8.35 × 102 newtons to a height of 6.0 meters. What amount of work does the forklift do?
    7·2 answers
  • A 15.0 cm object is 12.0 cm from a convex mirror that has a focal length of -6.0 cm. What is the height of the image produced by
    13·2 answers
  • Which of the following is the best evidence that Earth's continents were once in vastly different positions than they are today?
    11·2 answers
  • (1) Expansion of concrete
    9·1 answer
  • Explain
    8·1 answer
  • A train traveling at 6.4 m/s accelerates at 0.10 m/s 2 over a distance of 100 m. How large is the train’s final velocity?
    15·1 answer
  • Explain the difference between Longitudinal and Transverse waves.
    11·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!