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Effectus [21]
3 years ago
10

Two blocks of masses 3.0 kg and 5.0 kg are connected by a spring and rest on a frictionless surface. They are given velocities t

oward each other such that the 3.0 kg block travels initially at 1.2 m/s toward the center of mass, which remains at rest. What is the initial speed of the other block?
Physics
1 answer:
miskamm [114]3 years ago
3 0

Answer:

-0.7 m/sec

Explanation:

Mass of first block = m1 =3.0 kg

Mass of second block = m2= 5.0 kg

Velocity of first block = V1= 1.2 m/s

Velocity of second block = V2 = ?

Momentum of Center of mass MVcom  is sum of both blocks momentum and is given by

MVcom= m1v1+m2v2

Where

M= mass of center of mass

Vcom= Velocity of center of mass=0 m/s (because center of mass is at rest , so Vcom = 0 m.sec)

Putting values, we get;

0= 3×1.2+5v2

==> v2=  3.6/5= - 0.7 m/s

-ve sign indicates that block 2 is moving in opposite direction of block 1

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Answer:

An example of kinetic energy is a <u><em>car coming to a stop</em></u>

Explanation:

Kinetic energy is the energy that a body or system possesses due to its movement. In physics this energy is defined as the amount of work necessary to accelerate a body of a certain mass and in rest position, until reaching a certain speed. This energy obtained will remain unchanged as long as this body does not vary its speed. That is, kinetic energy measures how many changes an object that is moving can cause.

<u><em>An example of kinetic energy is a car coming to a stop</em></u>. If the car is moving and comes to a stop, there is a change in speed, therefore in movement, eventually producing a change in kinetic energy. This energy depends on the mass of the body, in this case the car, and the speed.  As the speed decreases, the kinetic energy will decrease.

4 0
3 years ago
if two point charges are separated by 1.5 cm and have charge values of 2.0 and -4.0, respectively, what is the value of the mutu
RUDIKE [14]

Complete question:

if two point charges are separated by 1.5 cm and have charge values of +2.0 and -4.0 μC, respectively, what is the value of the mutual force between them.

Answer:

The mutual force between the two point charges is 319.64 N

Explanation:

Given;

distance between the two point charges, r = 1.5 cm = 1.5 x 10⁻² m

value of the charges, q₁ and q₂ = 2 μC and - μ4 C

Apply Coulomb's law;

F = \frac{k|q_1||q_2|}{r^2}

where;

F is the force of attraction between the two charges

|q₁| and |q₂| are the magnitude of the two charges

r is the distance between the two charges

k is Coulomb's constant = 8.99 x 10⁹ Nm²/C²

F = \frac{k|q_1||q_2|}{r^2} \\\\F = \frac{8.99*10^9 *4*10^{-6}*2*10^{-6}}{(1.5*10^{-2})^2} \\\\F = 319.64 \ N

Therefore, the mutual force between the two point charges is 319.64 N

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