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umka21 [38]
1 year ago
10

ts, x and y, move toward one another and eventually collide. object x has a mass of 2m and is moving at a speed of 2v0 to the ri

ght before the collision. object y has a mass of m and is moving at a speed of v0 to the left before the collision. which of the following describes the magnitude of the forces f the objects exert on each other when they collide?
Physics
1 answer:
Brrunno [24]1 year ago
3 0

They will accelerate the same amount but in opposite directions. In physics, a collision is any event wherein or greater our bodies exert forces on each other in a pretty brief time.

The mean loose time for a molecule in a fluid is the common time between collisions. The imply unfastened path of the molecule is made of the average velocity and the mean unfastened time. These principles are used inside the kinetic concept of gases to compute shipping coefficients including viscosity.

Collision, also known as an effect, in physics, is the surprising, forceful coming collectively in direct touch of two bodies, along with, as an example,  billiard balls, a golf membership, and a ball, a hammer, and a nail head, two railroad vehicles while being coupled collectively, or a falling item and a ground.

The verb collide has roots in the Latin word collider, which comes from col- or collectively and leader, to strike or harm, like planes that collide in mid-air.

To know more about Collision:

brainly.com/question/29611796

#SPJ4

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Compare the collision between two baseballs and a catcher's mitt.
Nookie1986 [14]

The applied force is different for the two cases

The case A with a greater force involves the greatest momentum change

The case A involves the greatest force.

<h3>What is collision?</h3>
  • This is the head-on impact between two object moving in opposite or same direction.

The initial momentum of the two ball is the same.

P = mv

where;

  • m is the mass of each
  • v is the initial velocity of each ball

Since the force applied by the arm is different, the final velocity of the balls before stopping will be different.

Thus, the final momentum of each ball will be different

The impulse experienced by each ball is different since impulse is the change in momentum of the balls.

J = ΔP

The force applied by the rigid arm is greater than the force applied by the relaxed arm because the force applied by the rigid arm will cause the ball to be brought to rest faster.

Thus, we can conclude the following;

  • The applied force is different for the two cases
  • The case A with a greater force involves the greatest momentum change
  • The case A involves the greatest force.

Learn more about impulse here: brainly.com/question/25700778

3 0
3 years ago
You leave a 3kW heater on in your room. You put it on at 8am and leave it on until 4pm. If a unit (1kWh) costs 12.5p, how much w
natali 33 [55]

Answer:

bpc

Explanation:

8 0
3 years ago
A rocket is fired upwards with an acceleration of 32 m/s2 . If the 200kg rocket experiences an air resistance force of 12000N, w
shutvik [7]

Answer: 20360 N

<em>If it served you, give me 5 stars please, thank you!</em>

________________________________________________________

6 0
3 years ago
A fixed system of charges exerts a force of magnitude 18 n on a 9.0 c charge. the 9.0 c charge is replaced with a 4.0 c charge.
laila [671]
According to the law of gravitational force:
the force between any two objects is directly proportional to the masses of the objects and inversely proportional to the square of the distance between these two objects.

Based on this:
if a 9 c charge is replaced with a 4 c charge while the distance between the charges is kept constant:
18/9 = F/4
2 = F/4
F = 2 x 4 = 8
7 0
3 years ago
The free-electron density in a copper wire is 8.5×1028 electrons/m3. The electric field in the wire is 0.0520 N/C and the temper
meriva

Answer:

(a) 1.87 x 10⁻⁴ m/s

(b) 0.013V

Explanation:

(a) Drift speed, v_{d} , is the average velocity that a charged particle can have due to an electric field. For a given current, I, the drift velocity is given by;

v_{d} = \frac{I}{qnA}             ----------------(i)

Where;

q = amount of charge

n = free charge density

A = cross-sectional area of the wire

But current density, J, is the electric current per unit cross-section area. This  is also equal to the ratio of the electric field, E, to the resistivity, p, of the material of the wire. i.e

J = \frac{I}{A} = \frac{E}{p}

Equation (i) can then be written as follows;

v_{d} = \frac{J}{qn} = \frac{E}{qnp}

v_{d} = \frac{E}{qnp}      ---------------------(ii)

From the question;

E = 0.0520N/C

p = 1.72 x 10⁻⁸ Ωm

n = 8.5 x 10²⁸ electrons/m³

c = charge on electron = 1.9 x 10⁻¹⁹C

Substitute these values into equation (ii) as follows;

v_{d} = \frac{0.0520}{1.9*10^{-19} * 8.5*10^{28} * 1.72*10^{-8}}

v_{d} = 1.87 x 10⁻⁴ m/s

(b) The potential difference, V, is given by the product of the electric field and the distance, d, between the two points in the wire. i.e

V = E x d        [where d = 25.0cm = 0.25m]

V = 0.0520 x 0.25

V = 0.013V

4 0
3 years ago
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