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AnnyKZ [126]
3 years ago
11

One car is sitting still the other is moving at a velocity of 4m/s. If two cars have a mass of 0.04 kg run into each other, what

is the momentum of the pared cars?
Physics
1 answer:
Leni [432]3 years ago
7 0

Answer:

p = 0.16 kgm/s

Explanation:

the initial momentum combined of the two cars and the final momentum of the paired cares are the same, so we just need to find the initial momentum

p = m1v1 + m2v2

p = 0.04*4 + 0.04*0

p = 0.04*4

p = 0.16 kgm/s

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Observe: Air pressure is equal to the weight of a column of air on a particular location. Airpressure is measured in millibars (
lesya [120]

Answer:

a) When moving towards a high pressure center the pressure values ​​increase in the equipment

b) This area is called high prison since the weight of the atmosphere on top is maximum

Explanation:

A) A high atmospheric pressure system is an area where the pressure is increasing the maximum value is close to 107 Kpa, the other side as low pressure can have small values ​​85.5 kPa.

When moving towards a high pressure center the pressure values ​​increase in the equipment

B) This area is called high prison since the weight of the atmosphere on top is maximum

in general they are areas of good weather

6 0
3 years ago
A rubber ball is dropped from a height of 8m. After strikingthe floor, the ball bounces to a height of 5m. a. If the ball had bo
kifflom [539]

Answer:

a) This means the collision between the ball and the floor is elastic.

b) This points to a perfectly inelastic collision between the ball and the floor as they stick together after collision

c) Check Explanation.

Explanation:

Collision of bodies are analysed according to whether both momentum and kinetic energy of the system is conserved, that is, if these two quantities before collision are equal to their values after collision.

In all types of collisions, momentum is usually conserved, but kinetic energy is conserved only in an elastic collision.

A ball dropped from a height of 8 m bounces up back to a height of 5 m.

a. If the ball had bounced to a height of 8m, how would you describe the collision between the ball and the floor?

The ball not bouncing back to a height of 8 m shows energy loss at some point in the total motion of the ball (most likely at the collision). If kinetic energy was conserved, the ball would bounce back up to the height at which it fell from (8 m) after the collision with the floor.

b. If the ball had not bounced at all, how would you describe the collision between the ball and the floor?

If the ball had not bounced at all, this means it lost all of its kinetic energy to the floor, and this points to a perfectly inelastic collision between the ball and the floor as they stick together after collision.

c. What happened to the energy lost by the ball during thecollision?

The energy lost during the collision is converted to another form, most likely responsible for some deformation on the ball & a minute deformation on the floor, converted to some form of heat as a result of the collision or into sound energy, usually, it's a combination of all This!

Hope this Helps!!!

5 0
3 years ago
A vector A has a magnitude of 5 units and points in the −y-direction, while a vector B has triple the magnitude of A and points
Harman [31]

Answer:

A+B; 5√5 units, 341.57°

A-B; 5√5 units, 198.43°

B-A; 5√5 units, 18.43°

Explanation:

Given A = 5 units

By vector notation and the axis of A, it is represented as -5j

B = 3 × 5 = 15 units

Using the vector notations and the axis, B is +15i. The following vectors ate taking as the coordinates of A and B

(a) A + B = -5j + 15i

A+B = 15i -5j

|A+B| = √(15)²+(5)²

= 5√5 units

∆ = arctan(5/15) = 18.43°

The angle ∆ is generally used in the diagrams

∆= 18.43°

The direction of A+B is 341.57° based in the condition given (see attachment for diagrams

(b) A - B = -5j -15i

A-B = -15i -5j

|A-B|= √(15)²+(-5)²

|A-B| = √125

|A-B| = 5√5 units

The direction is 180+18.43°= 198.43°

See attachment for diagrams

(c) B-A = 15i -( -5j) = 15i + 5j

|B-A| = 5√5 units

The direction is 18.43°

See attachment for diagram

5 0
3 years ago
A proton in a particle accelerator is traveling at a speed of 0.99c
Gwar [14]

A proton in a particle accelerator is traveling at a speed of 0.99c has a speed magnitude of 2.97 x 10⁸ m/s.

<h3>What is speed of proton?</h3>

The speed of a proton is the rate at which a proton is moving through a given space.

The given speed of the proton is 0.99c

where;

  • c is speed of light

<h3>What is speed of light?</h3>

The speed of light in vacuum, commonly denoted c, is a universal physical constant that is important in many areas of physics.

The value of speed of light in a vacuum is given as 3 x 10⁸ m/s.

The speed of the proton is calculated as follows;

v = 0.99 x 3 x 10⁸ m/s.

v = 2.97 x 10⁸ m/s.

Thus, a proton in a particle accelerator is traveling at a speed of 0.99c has a speed magnitude of 2.97 x 10⁸ m/s.

Learn more about speed of proton here: brainly.com/question/14663642

#SPJ1

8 0
1 year ago
g Initially, the motorcycle travels along a straight road with a speed of 35 m/s (this is almost 80 mph). The maximum decelerati
astraxan [27]

Given:

Initial speed of the motorcycle (u) = 35 m/s

Final speed of the motorcycle (v) = 0 m/s (Complete Stop)

Maximum deceleration of the motorcycle (a) = -1.2 m/s²

Required Equation:

\boxed{\bf{ v = u + at}}

Answer:

By substituting values in the equation, we get:

\rm \longrightarrow 0 = 35 + ( - 1.2)t \\  \\  \rm \longrightarrow  0 = 35 - 1.2t \\  \\  \rm  \longrightarrow 35 - 1.2t = 0 \\  \\  \rm  \longrightarrow 35- 35 - 1.2t = 0 - 35 \\  \\  \rm  \longrightarrow  - 1.2t =  - 35 \\  \\  \rm \longrightarrow  \dfrac{ - 1.2t}{ - 1.2}  =  \dfrac{ - 35}{ - 1.2}  \\  \\  \rm \longrightarrow  t = 29.167 \: s

\therefore Time taken by motorcycle to come to a complete stop (t) = 29.167 s

4 0
3 years ago
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