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3 years ago

Bumper cars A and B undergo a collision during which the momentum of the combined system is conserved.

Physics

1 answer:

dexar [7]3 years ago

5 0

Answer:

1. P⃗ A,i+P⃗ B,i=P⃗ A,f+P⃗ B,f

Explanation:

Collision occurs when two bodies moving at different velocities exert force on each other through colliding. Collision can either be elastic or inelastic.

For elastic collision, both energy and momentum of the bodies is conserved i.e they separates after collision.

For inelastic collision, only momentum is conserved but not energy. The body sticks together after colliding and moves with a common velocity.

According law of conservation of momentum which states that the sum or momentum of two bodies before collision is equal to the momentum of the bodies after collision.

Given two bumper cars A and B that undergoes collision during which momentum is conserved, the type of collision that exists between the bumpers is inelastic collision

If initial momentum of A is PiA and initial momentum of B is PiB, the sum of their momentum before collision is expressed as;

PiA + PiB ... (1)

If final momentum of A is PfA and initial momentum of B is PfB, the sum of their momentum after collision is expressed as;

PfA+PfB ... (2)

According to the law, equation 1 is equal to equation 2

The fore the formula that states the law is expressed as;

PiA + PiB = PfA+PfB

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How and where is old oceanic crust destroyed?

sladkih [1.3K]

Answer:

It is destroyed in subduction zones. A Geologic process in which a tectonic plate made of dense lithospheric material melts or falls below a plate made of less-dense lithosphere at a convergent plate boundary

Explanation:

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4 years ago

Read 2 more answers

A glass bottle of soda is sealed with a screw cap. The absolute pressure of the carbon dioxide inside the bottle is 1.90 x 105 P

worty [1.4K]

Answer:

The magnitude of the force is 34.59 N.

Explanation:

Given that,

Inside pressure $P_{in}= 1.90\times10^{5}\ Pa$

Area $A=3.90\times10^{-4}\ m^2$

Outside pressure = 1 atm

We need to calculate the magnitude of the force

Using formula of force

$F_{net}=F_{in}-F_{out}$

$F_{net}=(P_{in}-P_{out})A$

Where, $P_{in}$ =inside Pressure

$P_{out}$ =outside Pressure

A = area

Put the value into the formula

$F_{net}=(1.90\times10^{5}-1.013\times10^{5})3.90\times10^{-4}$

$F_{net}=34.59\ N$

Hence, The magnitude of the force is 34.59 N.

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3 years ago

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3 years ago

A man and his dog are “walking” on flat Street. He is pulling on his stubborn dog with a force of 70 N Directed at a 30° angle f

Delvig [45]

Answer:

X component of force is 60.62 N.

Y component of force is 35 N.

Force of gravity on the dog is 245 N.

Magnitude of normal force is 210 N.

Explanation:

Given:

Force of pull is, $F=70\ N$

Mass of the dog is, $m=25\ kg$

Angle of inclination is, $\theta =30$ °

Acceleration due to gravity is, $g=9.8\ m/s^2$

The free body diagram of the dog is shown below.

The X and Y components of force of pull is given as:

$F_X=F\cos \theta=(70)\cos (30)=60.62\ N\\F_Y=F\sin \theta=(70)\sin(30)=35\ N$

Therefore, the X and Y components of the force are 60.62 N and 35 N respectively.

Force of gravity on the dog is the product of its mass and acceleration due to gravity. Thus,

$F_g=mg=25\times 9.8=245\ N$

Therefore, the force of gravity on the dog is 245 N.

Now, consider the motion in the vertical direction of the dog. As there is no motion in the vertical direction, the net force along the Y direction is 0. In other words, the total Upward force is equal to the total downward force.

From the free body diagram,

$N+F_Y=mg\\N=mg-F_Y\\N=245-35=210\ N$

Therefore, the normal force acting on the dog is 210 N.

6 0

4 years ago

Two charged particles are located on the x axis. The first is a charge 1Q at x 5 2a. The second is an unknown charge located at

sergejj [24]

Answer:

Q_2 = +/- 295.75*Q

Explanation:

Given:

- The charge of the first particle Q_1 = +Q

- The second charge = Q_2

- The position of first charge x_1 = 2a

- The position of the second charge x_2 = 13a

- The net Electric Field produced at origin is E_net = 2kQ / a^2

Find:

Explain how many values are possible for the unknown charge and find the possible values.

Solution:

- The Electric Field due to a charge is given by:

E = k*Q / r^2

Where, k: Coulomb's Constant

Q: The charge of particle

r: The distance from source

- The Electric Field due to charge 1:

E_1 = k*Q_1 / r^2

E_1 = k*Q / (2*a)^2

E_1 = k*Q / 4*a^2

- The Electric Field due to charge 2:

E_2 = k*Q_2 / r^2

E_2 = k*Q_2 / (13*a)^2

E_2 = +/- k*Q_2 / 169*a^2

- The two possible values of charge Q_2 can either be + or -. The Net Electric Field can be given as:

E_net = E_1 + E_2

2kQ / a^2 = k*Q_1 / 4*a^2 +/- k*Q_2 / 169*a^2

- The two equations are as follows:

1: 2kQ / a^2 = k*Q / 4*a^2 + k*Q_2 / 169*a^2

2Q = Q / 4 + Q_2 / 169

Q_2 = 295.75*Q

2: 2kQ / a^2 = k*Q / 4*a^2 - k*Q_2 / 169*a^2

2Q = Q / 4 - Q_2 / 169

Q_2 = -295.75*Q

- The two possible values corresponds to positive and negative charge Q_2.

7 0

3 years ago