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
sergejj [24]
2 years ago
13

State the law of conversation of momentum​

Physics
1 answer:
Sliva [168]2 years ago
6 0
The total momentum before a collision is equal to the total momentum after the collision if no external forces act on the system.
Hope this helps
You might be interested in
Calculate the KE of a car which has a mass of 1000 kg and is moving at the rate of 20 m/s
Elan Coil [88]

Answer:

The kinetic energy of the car is 2.0\cdot 10^5 J

Explanation:

3 0
2 years ago
Which of the following statements is TRUE for high-visibility clothing? A. High-visibility clothing helps to reduce insect probl
Mazyrski [523]
C. High-visibility clothing is important to wear in areas with moving vehicles
5 0
3 years ago
Read 2 more answers
If a 60-g object has a volume of 30 cm", what is its density?
alisha [4.7K]

Answer:

Density = Mass/volume. D= 60/30.Divide it and you'll get ur answer as 2

5 0
3 years ago
A force of 6600 N is exerted on a piston that has an area of 0.010 m2
sveticcg [70]

Answer:

Choice A: approximately 0.015\; \rm m^2, assuming that the two pistons are connected via some confined liquid to form a simple machine.

Explanation:

Assume that the two pistons are connected via some liquid that is confined. Pressure from the first piston:

\displaystyle P_1 = \frac{F_1}{A_1} = \frac{6.600\times 10^3\; \rm N}{1.0\times 10^{-2}\; \rm m^{2}} = 6.6\times 10^{5}\; \rm N \cdot m^{-2}.

By Pascal's Principle, because the first piston exerted a pressure of 6.6\times 10^{5}\; \rm N \cdot m^{-2} on the liquid, the liquid will now exert the same amount of pressure on the walls of the container.

Assume that the second piston is part of that wall. The pressure on the second piston will also be 6.6\times 10^{5}\; \rm N \cdot m^{-2}. In other words:

P_2 = P_1 = 6.6\times 10^{5}\; \rm N \cdot m^{-2}.

To achieve a force of 9.900 \times 10^3\; \rm N, the surface area of the second piston should be:

\displaystyle A_2 = \frac{F_2}{P_2} = \frac{9.900\times 10^{3}\; \rm N}{6.6\times 10^5\; \rm N \cdot m^{-2}} \approx 0.015\; \rm m^{2}.

4 0
3 years ago
The standard wave format for any wave is wave. When depicting wave in standard wave format, the direction of motion must be rota
german

Answer:

Transverse wave  and Longitudinal wave  and Electromagnetic wave

Explanation:

  • An inverted wave is a wave in which the vibrations of the particles are perpendicular to the direction of wave motion.
  • Longitudinal waves, on the other hand, are waves in which the vibrations of the particles are parallel to the direction of wave motion.
  • Electromagnetic waves are waves that do not require medium media for transmission, including radio waves, microwaves, UV lights, etc.
  • Most electromagnetic waves are transverse in nature.
4 0
2 years ago
Other questions:
  • What is a locus of points
    13·2 answers
  • Which layer contains solid rock and magma
    8·1 answer
  • What type of energy is the energy of motion?
    11·2 answers
  • An object is launched at a velocity of 20 m/s in a direction making an angle of 25° upward with the horizontal.
    8·1 answer
  • If a shot is put an angle of 41 degrees relative to the horizontal with a velocity of 36 ft/s in the direction of the put, what
    9·1 answer
  • Which is an example of velocity?
    14·2 answers
  • What is the mass of an object that requires a force of 30 N to accelerate at a rate of 5 m/sec2 ?
    8·1 answer
  • 1. A block of aluminum occupies
    15·1 answer
  • Un coche de unos 500 kg viaja a 90 km/h. Percibe un obstáculo y debe frenar a tope. Por las marcas del suelo se sabe que el espa
    5·1 answer
  • Show your workikkkkkkkk
    12·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!