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3 years ago

_________ forces exist between objects that are not in contact with each other

Physics

1 answer:

valina [46]3 years ago

8 0

Answer:

Gravity as well as electrostatic and magnetic attraction and repulsion provide real life examples of forces being exerted by one object on another without them being in contact with each other. Many children are aware of magnetism and have played with fridge magnets.

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Write down the boiling point and freezing point of mercury an alcohol

nekit [7.7K]

Mercury has a high boiling point of 357 degrees C.
Mercury has a freezing point of −39 degrees C.

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3 years ago

A car with a velocity of 22 m/s is accelerated at a rate of 1.6m/s2 for 6.8s. determine the final velocity

Ivahew [28]

A car with a velocity of 22 m/s is accelerated at a rate of 1.6 $m/s^2$ for 6.8s has the final velocity t be 32.88 m/s.

The acceleration means the amount of velocity changing per unit time.

The given data:

initial velocity, u = 22 m/s

time, t = 6.8 s

acceleration, a = 1.6 $m/s^2$

We will be using the equation of motion:

v = u + at

$\therefore v=22+1.(6.8)$

$\Rightarrow v=22+10.88$

$\Rightarrow v=32.88 \ m/s$

The final velocity become 32.88 m/s.

To learn more about Attention here:

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2 years ago

What type of weathering is best expressed in this weathered limestone?

DENIUS [597]

It seems that you have missed the necessary options for us to answer this question so I had to look for it. Anyway, here is the answer. The type of weathering that is best expressed in this weathered limestone is DISSOLUTION. Hope this answers your question. Have a great day!

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3 years ago

A(n) 1700 kg car is moving along a level road at 21 m/s. The driver accelerates, and in the next 10 s the engine provides 22000

allochka39001 [22]

The final speed of the car at the given conditions is 30.1 m/s.

The given parameters:

Mass of the car, m = 1700 kg
Velocity of the car, v = 21 m/s
Time of motion, t = 10 s
Additional energy provided by the engine, E₁ = 22,000 J
Energy used in overcoming friction, E₂ = 3,666.67 J

The change in the energy applied to the car is calculated as;

$\Delta E = E_1 - E_2\\\\\Delta E = 22,000 \ J \ - \ 3,666.67 \ J\\\\\Delta E = 18,333.33 \ J$

The final speed of the car is calculated as follows;

$\Delta E = \frac{1}{2} m(v_f^2 - v_0^2)\\\\v_f^2 - v_0^2 = \frac{2\Delta E}{m} \\\\v_f^2 = \frac{2\Delta E}{m} + v_0^2\\\\v_f = \sqrt{ \frac{2\Delta E}{m} + v_0^2} \\\\v_f = \sqrt{ \frac{2\times 18,333.4}{1700} + (21)^2} \\\\v_f = 30.1 \ m/s$

Thus, the final speed of the car at the given conditions is 30.1 m/s.

Learn more about change in kinetic energy here: brainly.com/question/6480366

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3 years ago

PLEASE HELP ME!!!!! I'LL MARK YOU BRAINLIEST IF YOU ANSWER THIS CORRECTLY

Pachacha [2.7K]

Answer: a

Explanation:

3 0

4 years ago

Read 2 more answers