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4 years ago

Urgent please help. it’s kinematic equations

Physics

2 answers:

weqwewe [10]4 years ago

5 0

Answer:

Explanation:

Key Points

The four kinematic equations involve five kinematic variables:

Each equation contains only four of the five variables and has a different one missing.

It is important to choose the equation that contains the three known variables and one unknown variable for each specific situation.

vovangra [49]4 years ago

5 0

Answer:

The displacement is 0 meters because it just moved back to its original position.

The distance traveled(in meters) is just the speed times the time, which is 18 m/s * 3600 s = 64800 meters.

In this case, the displacement and the distance traveled are the same because the object didn't change direction.

The distance traveled(in meters) and the displacement is just the speed times the time, which is 16 m/s * 15 s = 240 meters.

In this case, the displacement and the distance traveled are the same because the object didn't change direction.

We can use the equation (Final Velocity)^2 = (Initial Velocity)^2 + 2*Acceleration*Distance.

Now we can plug the known values: (5)^2 = (20)^2 + 2*(-4)*d, and now we solve for d: 25 = 400 + -8d -> -375 = -8d -> d = 46.875 meters.

Again, in this case, the displacement and the distance traveled are the same because the object didn't change direction.

We can use the equation Distance = (Initial Velocity) * (Time) + 1/2 * (Acceleration) * (Time)^2.

Plugging in the values we know, we get d = 0*20 + 1/2*3.7*20^2 -> d = 740 meters.

Again, in this case, the displacement and the distance traveled are the same because the object didn't change direction.

We can use the equation Distance = (Initial Velocity) * (Time) + 1/2 * (Acceleration) * (Time)^2.

Plugging in the values we know, we get d = 12*26 + 1/2*1.5*26^2 -> d = 819 meters.

If you have further questions please don't hesitate to ask.

I really hoped this helped you, and if it did, could you please consider giving me a brainliest? :)

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4 years ago

A U-shaped tube open to the air at both ends contains some mercury. A quantity of water is carefully poured into the left arm of

Gekata [30.6K]

Answer:

a) $P=2450\ Pa$

b) $\delta h=23.162\ cm$

Explanation:

Given:

height of water in one arm of the u-tube, $h_w=25\ cm=0.25\ m$

Gauge pressure at the water-mercury interface,:

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$P=1000\times 9.8\times 0.25$

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3 years ago

A 2 kg object with 100 J potential energy (with respect to the ground) is released from rest from the top of a building, the spe

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velocity of object just before hitting the ground = 10 m/s

10 m/sExplanation:

here's the solution : -

=》mechanical energy is always constant, when the object is at the top of the building , the object has potential energy but no kinetic energy,

so, at the top of the building the mechanical energy = potential energy = 100 J

but at when it is dropped from the building, the whole potential energy get converted into kinetic energy, just before reaching the bottom .

so, mechanical energy in the object just before hitting the ground = kinetic energy

( potential energy = 0 )

So, kinetic energy = 100 J

now, we know

$kinetic \: energy = \frac{1}{2} \times m {v}^{2}$

=》

$100 = \frac{1}{2} \times 2 \times {v}^{2}$

( since, mass = 2 kg )

=》

$100 = {v}^{2}$

=》

$v = \sqrt{100}$

=》

$v = 10$

so, velocity of object just before hitting the ground = 10 m/s

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