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3 years ago

Calculate the potential energy at the top of the giant drop if the car weights 1000 kg

1 answer:

6 0

The gravitional potential energy, relative to the bottom of the giant drop, in joules, is (9800) times (the height of the drop in meters). That's the PE of the empty car only, not counting any hapless screaming souls who may be trapped in it at that moment.

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According to Newton's 3rd law of motion, if you hit a baseball with a bat it will _____.

Anna [14]

Fly in a straight line unless an outside force changes its course because i tried it once in a baseball game that my mommy rekt me in.

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3 years ago

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Which of the following is an expression of Newton's second law? O A. The acceleration of an object is determined by its mass and

dangina [55]

Answer: is A

The acceleration of an object is determined by its mass and the net force acting on it.

Newton's second law says that "F=ma". what this means is that the force you apply to an object is equal to its mass times how fast it is accelerating.

Hope it helped.

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2 years ago

A metal disk of radius 6.0 cm is mounted on a frictionless axle. Current can flow through the axle out along the disk, to a slid

Galina-37 [17]

Answer:

0.09 N

Explanation:

We are given that

Radius of disk,r=6 cm= $\frac{6}{100}=0.06 m$

1 m=100 cm

B=1 T

Current,I=3 A

We have to find the frictional force at the rim between the stationary electrical contact and the rotating rim.

$dF=IBdr$

$\tau=rdF=IBrdr$

$\tau=\int_{0}^{R}IBr dr$

$\tau=IB(\frac{R^2}{2}$

Torque due to friction

$\tau=R\times F$

Where friction force=F

$R\times F=\frac{IBR^2}{2}$

$F=\frac{IBR}{2}$

Substitute the values

$F=\frac{3\times 1\times 0.06}{2}$

$F=0.09 N$

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3 years ago

A cat dozes on a stationary merry-go-round, at a radius of 4.4 m from the center of the ride. The operator turns on the ride and

monitta

Answer:

The coefficient of static friction is 0.29

Explanation:

Given that,

Radius of the merry-go-round, r = 4.4 m

The operator turns on the ride and brings it up to its proper turning rate of one complete rotation every 7.7 s.

We need to find the least coefficient of static friction between the cat and the merry-go-round that will allow the cat to stay in place, without sliding. For this the centripetal force is balanced by the frictional force.

$\mu mg=\dfrac{mv^2}{r}$

v is the speed of cat, $v=\dfrac{2\pi r}{t}$

$\mu=\dfrac{4\pi^2r}{gt^2}\\\\\mu=\dfrac{4\pi^2\times 4.4}{9.8\times (7.7)^2}\\\\\mu=0.29$

So, the least coefficient of static friction between the cat and the merry-go-round is 0.29.

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