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MissTica
3 years ago
15

Which option would it be

Physics
1 answer:
guapka [62]3 years ago
7 0
It would be 4 atm, because the way to figure out the final pressure is that (P1)(V1)=(P2)(V2)

meaning that the original pressure x original volume is equal to the final pressure x final volume. This gas law is called Boyle's law if you'd like to learn more about it.

But (1 atm)(40 mL)=(4 atm)(10 mL)

So it would be the second choice.
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Mary starts at the edge of a circular platform that is slowly rotating on a frictionless axle. She then walks towards the opposi
bulgar [2K]

Answer:

Explanation:

The motion of Mary along the circular path is a centripetal.

As Mary moves from one edge of the circular platform to the other edge, she is covering a distance which is the radius of the circular path at a velocity.

According to the relationship

w = v/r where

w is the angular velocity

r is the radius

v is the linear velocity

Initially, before Mary starts, her linear speed is zero and her angular velocity is also zero. As she move towards the opposite edge, she is covering a distance of radius r. According to the formula, increase in radius will leads to decrease in her angular velocity and vice versa. As Mary starts moving towards the centre of the circular path, her angular velocity increases, at the centre of the platform, her angular velocity is at maximum at this point. As she moves further from the center to the other edge, her angular velocity decreases due to increase in distance covered across the circular path.

6 0
3 years ago
A basketball has a mass of 1 kg and is traveling 12 m/s
cricket20 [7]

A) It would be doubled.

Why?

To answer the question, we just need to calculate the momentum of the basketball using the following formula:

Momentum=mass*velocity\\\\Momenum=1kg*12\frac{m}{s}=12\frac{kg.m}{s}

Now, we have calculated the momentum and the result is 12 kg.m/s, what would happen to the velocity if we double the momentum? Let's calculate it!

Momentum=mass*velocity\\\\(2)*12\frac{kg.m}{s}=1kg*velocity\\\\velocity=\frac{24\frac{kg.m}{s} }{1kg} =24\frac{s}{s}

Hence, we can see that if the momentum is doubled, the velocity will be doubled too.

Have a nice day!

7 0
3 years ago
A football is kicked into the air with a velocity of 32m/s at an angle of 25º. At the very top of the ball’s path, its vertical
Travka [436]
At the "very top" of the ball's path, there's a tiny instant when the ball
is changing from "going up" to "going down".  At that exact tiny instant,
its vertical speed is zero.

You can't go from "rising" to "falling" without passing through "zero vertical
speed", at least for an instant.  It makes sense, and it feels right, but that's
not good enough in real Math.  There's a big, serious, important formal law
in Calculus that says it.  I think Newton may have been the one to prove it,
and it's named for him.

By the way ... it doesn't matter what the football's launch angle was,
or how hard it was kicked, or what its speed was off the punter's toe,
or how high it went, or what color it is, or who it belongs to, or even
whether it's full to the correct regulation air pressure.  Its vertical speed
is still zero at the very top of its path, as it's turning around and starting
to fall.
7 0
3 years ago
a 2100-kg car drives with a speed of 18 m/s onb a flat road around a curve that has a radius of curvature of 83m. The coefficien
Law Incorporation [45]

Answer:

<u>The magnitude of the friction force is 8197.60 N</u>

Explanation:

Using the definition of the centripetal force we have:

\Sigma F=ma_{c}=m\frac{v^{2}}{R}

Where:

  • m is the mass of the car
  • v is the speed
  • R is the radius of the curvature

Now, the force acting in the motion is just the friction force, so we have:

F_{f}=m\frac{v^{2}}{R}

F_{f}=2100\frac{18^{2}}{83}

F_{f}=8197.60 \: N

<u>Therefore the magnitude of the friction force is 8197.60 N</u>

I hope it helps you!

7 0
3 years ago
How do we know how much<br> gravitational force objects have?
Rina8888 [55]

Answer:

With a force of 1 Newton, an object weighing 100 grams is dragged towards the planet's center. On Earth, an item with a mass of 100 grams has a gravitational pull of only 1%, or about kg.

Explanation:

i hope this helps

3 0
3 years ago
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