As time goes on, the ENTROPY in a closed system should increase. This is because of which Law?

A spaceship starting from a resting position accelerates at a constant rate of 9.8 m/s. How far will the spaceship travel if its

Dvinal [7]

300 000 0 squared = 2 x 9.8 distance

KINEMATICS

Uniform or constant motion in a straight line (rectilinear). Speed or velocity constant and/or acceleration constant. If motion is up and down and/or has an up and down component then acceleration omn earth will be g. g is about 10m/s/s.

speed = distance/time

velocity = displacement/time

s=distance ... u=initial speed ... v = final speed ... a = acceleration ... t = time

v=u+at

v^2=u^2+2as

s=ut+1/2at^2

3 0

3 years ago

How many significant digits are in the measurement 4,032,010 m/s?

choli [55]

Since there is no decimal point in the number given above, the counting for the number of the significant figures will start from the left. Then, the first zero from the left is insignificant. Therefore, in this number there are 6 significant figures.

5 0

3 years ago

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A cue stick hits a cue ball with an average force of 22 N for a duration of 0.029 s. If the mass of the ball is 0.15 kg, how fas

Likurg_2 [28]

Answer:

4.25 m/s

Explanation:

Force, F = 22 N

Time, t = 0.029 s

mass, m = 0.15 kg

initial velocity of the cue ball, u = 0

Let v be the final velocity of the cue ball.

Use newton's second law

Force = rate of change on momentum

F = m (v - u) / t

22 = 0.15 ( v - 0) / 0.029

v = 4.25 m/s

Thus, the velocity of cue ball after being struck is 4.25 m/s.

8 0

3 years ago

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A quarter is flipped from a height of 1.45 m above the ground. How much time will it take to reach the ground if the person flip

Artemon [7]

It will take the quarter 0.151 seconds to reach the ground.

<u>Given the following data:</u>

Height = 1.45 meters

Initial velocity = 10.32 m/s

We know that acceleration due to gravity (a) for an object is equal to 9.8 meter per seconds square.

To find how much time it will take the quarter to reach the ground, we would use the second equation of motion.

Mathematically, the second equation of motion is given by the formula;

$S = ut + \frac{1}{2} at^2$

Where:

S is the height or distance covered.

u is the initial velocity.

a is the acceleration.

t is the time measured in seconds.

Substituting the values into the formula, we have;

$1.45 = 10.32(t) + \frac{1}{2} (9.8)t^2\\\\1.45 = 10.32t + 4.9t^2\\\\4.9t^2 + 10.32t - 1.45 = 0$

The standard form of a quadratic equation is:

$ax^2 + bx + c = 0$

a = 4.9, b = 10.32 and c = 1.45

We would solve the above quadratic equation by using the quadratic equation formula;

$x = \frac{-b\; \pm \;\sqrt{b^2 - 4ac}}{2a}$

Substituting the values, we have;

$t = \frac{-10.32\; \pm \;\sqrt{10.32^2\; - \;4(4.9)(1.45)}}{2(4.9)}\\\\t = \frac{-10.32\; \pm \;\sqrt{106.5024\; - \;28.42}}{9.8}\\\\t = \frac{-10.32\; \pm \;\sqrt{78.0824}}{9.8}\\\\t = \frac{-10.32\; \pm \;8.84}{9.8}\\\\t = \frac{-10.32\; + \;8.84}{9.8}\\\\t = \frac{1.48}{9.8}$

<em>Time, t = 0.151 seconds.</em>

Therefore, it will take the quarter 0.151 seconds to reach the ground.