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3 years ago

How long Tina, a ballerina, in the air when she leaps straight up with a speed of 1.8 m/s?

1 answer:

3 0

The acceleration of gravity on or near the surface of the Earth is 9.8 m/s².
Anything acted on only by gravity loses 9.8 m/s of upward speed, or gains
9.8 m/s of downward speed, every second.

Leaping straight upward at 1.8 m/s, Tina keeps rising until she runs out of
upward speed. That happens in (1.8/9.8) = 0.1837 second after the leap.

After that, Finkel's First Law of Motion takes over:
"What goes up must come down."

The dropping part of the leap is symmetrical with the first. Please don't
make me go through proving it. Tina hits the floor at the same speed of
1.8 m/s with which she left it, and it takes the same amount of time to drop
from the peak to the floor as it took to rise from the floor to the peak.

So her total time out of contact with the floor is

2 x (0.1837 sec) = 0.367 second (rounded)

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3 years ago

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A 1.5kg cart moves along a track at 0.20m/s until it hits a fixed bumper at the end of the track. Find the average force exerted

DochEvi [55]

Answer:

Net force will be 4.875 N

Explanation:

We have given mass of the cart m = 1.5 kg

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And final velocity of the car = - 0.125 m/sec ( Negative direction is due to opposite direction )

Instant of time $\Delta t=0.1sec$

Change in momentum is given by

$\Delta P=1.5\times (0.2-(-0.125))=1.5\times 0.325=0.4875kgm/sec$

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2 years ago

How can you change the phases of matter using thermal energy

Leokris [45]

Answer:

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2 years ago

A plane is flying due east with the velocity of 90m/s. The wind is blowing out of the north at 4m/s. What is the magnitude of th

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<span> as we know that the velocity vectors are at right angles
magnitude = ?
hypotenuse of a right triangle.
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2 years ago

A student gives a 5.0 kg box a brief push causing the box to move with an initial speed of 8.0 m/s along a rough surface. The bo

WITCHER [35]

Answer:

The time taken to stop the box equals 1.33 seconds.

Explanation:

Since frictional force always acts opposite to the motion of the box we can find the acceleration that the force produces using newton's second law of motion as shown below:

$F=mass\times acceleration\\\\\therefore acceleration=\frac{Force}{mass}$

Given mass of box = 5.0 kg

Frictional force = 30 N

thus

$acceleration=\frac{30}{5}=6m/s^{2}$

Now to find the time that the box requires to stop can be calculated by first equation of kinematics

The box will stop when it's final velocity becomes zero

$v=u+at\\\\0=8-6\times t\\\\\therefore t=\frac{8}{6}=4/3seconds$

Here acceleration is taken as negative since it opposes the motion of the box since frictional force always opposes motion.

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3 years ago