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3 years ago

PLZZ HELP Explain how you can tell the difference between a physical and a chemical change.

Physics

1 answer:

Alecsey [184]3 years ago

7 0

Answer:

You can tell the difference between physical and chemical change by looking at the change that takes place like whether it is a permanent or temporary change.

Explanation:

physical change is temporary change and chemical change is permanent change
the original shape of the object can be restored in physical change whereas the original composition of the object will be lost in chemical change
in physical change the change is reversible whereas he chemical change is irreversible and the object takes new shape and position in this change.

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Which sample has the lowest temperature?

Fittoniya [83]

Answer:

I believe its A

Explanation:

3 0

3 years ago

A 1200 kg car passes traffic light at a velocity of 10.2 m/s to the north and accelerates at a rate of 2.45 m/s^2. Calculate the

kumpel [21]

The car’s momentum after 4.21s is 24617.4 kgm/s

<h3>Newton's Second Law of Motion.</h3>

Newton's second law state that, the rate of change of momentum, is directly proportional to the applied force.

Given that a 1200 kg car passes traffic light at a velocity of 10.2 m/s to the north and accelerates at a rate of 2.45 m/s^2. To calculate the car’s momentum after 4.21 s, Let us first list all the parameters involved.

Velocity u = 10.2 m/s

Acceleration a = 2.45 m/s²

Mass m = 1200Kg

Time t = 4.21 s

From Newton's second law,

F = (mv - mu) / t

ma = (mv - mu) / t

Substitute all the parameters into the formula above.

1200 × 2.45 = ( mv - 1200 × 10.2 ) / 4.21

2940 = ( mv - 12240 ) / 4.21

Cross multiply

12377.4 = mv - 12240

Make mv the subject of the formula

mv = 12377.4 + 12240

mv = 24617.4 kgm/s

Therefore, the car’s momentum after 4.21s is 24617.4 kgm/s

Learn more about Momentum here: brainly.com/question/25121535

#SPJ1

3 0

1 year ago

49W

Nikolay [14]

Answer:

Explanation:

The magnitude of the electrostatic force between two charged particles is given by

$F=\frac{kq_1 q_2}{r^2}$

where

k is the Coulomb's constant

q1, q2 are the charges of the two particles

r is the separation between the particles

In this problem, the initial force between the particles is F.

Later, the distance between the two particles is increased by four, so

r' = 4r

So, the new force between the particles will be

$F'=\frac{kq_1 q_2}{(4r)^2}=\frac{kq_1 q_2}{16r^2}=\frac{1}{16}F$

So, the new force decreases by a factor of 16.

4 0

3 years ago

In the graph, during which time period does the particle undergo the greatest displacement?

VMariaS [17]

I'm not very good with Physics but of what I do know I'd go with point B / 4seconds

I apologize in advance if that is incorrect.

5 0

3 years ago

HELPP!

Marizza181 [45]

5000.700 is the answer good luck

4 0

2 years ago