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
Anarel [89]
3 years ago
12

A 25-kg wagon has a momentum of 300kg m/s. What is it’s acceleration?

Physics
1 answer:
arsen [322]3 years ago
8 0

The answer is 12 ....

You might be interested in
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
PLEASE HELP show the calculation of the number of neutrons of hydrogen​
xz_007 [3.2K]
The answer is 2.3 hope this helps texted me and tell me if it’s right
7 0
2 years ago
Kinetic friction acts on a baseball player sliding into first base. Will the player's velocity change?
Law Incorporation [45]
Yes, his velocity will decrease the further he slides.
7 0
3 years ago
The conversion of thermal energy into mechanical energy requires
emmainna [20.7K]
Temperature difference is required, so i’m guessing - a. thermometer - would be required to check that temperature.
8 0
3 years ago
What is the maximum acceleration the belt can have without the crate slipping? express your answer using two significant figures
Montano1993 [528]

To prevent the crate from slipping, the maximum force that the belt can exert on the crate must be equal to the static friction force.


Ff = 0.5 * 16 * 9.8 = 78.4 N

a = 4.9 m/s^2


If acceleration of the belt exceeds the value determined in the previous question, what is the acceleration of the crate?


In this situation, the kinetic friction force is causing the crate to decelerate. So the net force on the crate is 78.4 N minus the kinetic friction force.


Ff = 0.28 * 16 * 9.8 = 43.904 N

Net force = 78.4 – 43.904 = 34.496 N

To determine the acceleration, divide by the mass of the crate.

a = 34.496 ÷ 16 = 2.156 m/s^2



8 0
3 years ago
Other questions:
  • what is the potential energy of the ball when it gets to its maximum height just before falling back to the ground​
    14·1 answer
  • Based on Archimedes' principle, we know that if an object displaces a given weight of water, then the object is being buoyed up
    8·2 answers
  • 1) Rephrase the objective. (limit 2 sentences). Give a brief explanation of the concepts of linear motion, its characteristics a
    10·2 answers
  • Which of the following people believed that light behaved like a wave?
    8·1 answer
  • Most scientists believe the Big Bang Theory explains which of the following questions?
    12·2 answers
  • A 2kg hockey puck is sliding across the ice skating rink at 2 m/s. A player hits the puck so it's velocity increases to 10 m/s.
    12·1 answer
  • A 10 Kg ball is rolling at 2.5 m/s. It is then hit from behind with a bat that puts a 300 N force on the ball for a quick .3 sec
    13·1 answer
  • Seesaw unbalanced force explain
    7·1 answer
  • chris and molly wins $2400 in a competition they share the money in the ratio2:3 how much money do they each receive ?​
    12·1 answer
  • ****PLEASE HELP**** THERE ARE TWO QUESTIONS (ITS EASY)
    13·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!