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joja [24]
3 years ago
11

What is the height of a 2.00 kg object if its potential energy is 100 J?

Physics
1 answer:
lara31 [8.8K]3 years ago
7 0

<u>Given that:</u>

       height (h) = ?

       mass  (m) = 2 Kg ,

       Potential energy (P.E.) = 100 J ;

    We know that P.E = m × g × h

                            100 J = 2 × 9.81 × h

                                h = (100) ÷ (2 × 9.81)

                              <em>  h  = 5.09 m</em>

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carlis lives 100 m away from his friend home what is his average speed if he reaches his friends home in 50s
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100m ÷ 50s = 2m/s
Just some simple divison.
8 0
4 years ago
The structure of which substances tends to exhibit the greatest hardness?
bagirrra123 [75]

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3 0
3 years ago
What is the danger to spacecraft and astronauts from micrometeoroids?
Sloan [31]

Answer:

Suppose the micrometeoroid weighed 1 g = .001 kg

Suppose also the spacecraft were moving at 18,000 mph (1.5 hrs per rev)

Usually, the smaller particle would be moving but for simplicity suppose that it were stationary wrt the ground

v = 18000 miles / hr * 1500 m/mile / 3600 sec/hr = 7500 m/s

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7 0
3 years ago
A sled of mass m is being pulled horizontally by a constant horizontal force of magnitude F. The coefficient of kinetic friction
rusak2 [61]

I'll bite:

-- Since the sled's mass is 'm', its weight is 'mg'.

-- Since the coefficient of kinetic friction is μk, the force acting opposite to the direction it's sliding is    (μk) times (mg) .

-- If the pulling force is constant 'F', then the horizontal forces on the sled
are 'F' forward and (μk · mg) backwards.

-- The net force on the sled is  (F - μk·mg).
(I regret the visual appearance that's beginning to emerge,
but let's forge onward.)

-- The sled's horizontal acceleration is  (net force) / (mass) = (F - μk·mg) / m.
This could be simplified, but let's not just yet.

-- Starting from rest, the sled moves a distance 's' during time 't'.
We know that  s = 1/2 a t² , and we know what 'a' is.  So we can write

           s = (1/2 t²)  (F - μk·mg) / m    .

Now we have the distance, and the constant force.
The total work is (Force x distance), and the power is (Work / time).
Let's put it together and see how ugly it becomes.  Maybe THEN
it can be simplified.

Work = (Force x distance) =  F x  (1/2 t²)  (F - μk·mg) / m
 
Power = (Work / time) =    <em>F (t/2) (F - μk·mg) / m </em>

Unless I can come up with something a lot simpler, that's the answer.


To simplify and beautify, make the partial fractions out of the
2nd parentheses:
                                   <em> F (t/2) (F/m - μk·m)</em>

I think that's about as far as you can go.  I tried some other presentations,
and didn't find anything that's much simpler.

Five points,ehhh ?


4 0
3 years ago
Read 2 more answers
An unbalanced force of 20N acts on a 4.0kg mass what is it's acceleration​
tankabanditka [31]

Hi there!

According to Newton's second law:

∑F = m · a, where:

∑F = net force (N = kgm/s²)

m = mass (kg)

a = acceleration (m/s²)

Rearrange to solve for acceleration:

F/m = a

20N / 4.0kg = 5 m/s²

4 0
3 years ago
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