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2 years ago

15

During a field experiment about speed, a scientist created the chart above. The chart shows distance and time measurements for a

racing car on the straight section of a race track. What is the racing car's speed?

1 answer:

BaLLatris [955]2 years ago

6 0

Answer:

yes

Explanation:

its not a good thing for the rest of your life but you have a PS4

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a machine with a mechanical advantage of 2.5 requires an input force of 120 newtons. What output force is applied by this machin

creativ13 [48]

If your machine has a mechanical advantage of 2.5, then WHATEVER force you apply to the input, the force at the output will be 2.5 times as great.

If you apply 1 newton to the machine's input, the output force is

(2.5 x 1 newton) = 2.5 newtons.

If you apply 120 newtons to the machine's input, the output force is

(2.5 x 120 newtons) = 300 newtons.

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3 years ago

How much force is needed to accelerate a 1000Kg car at a rate of 3m/s2?

ikadub [295]

According to Newton (2nd law), Force = (mass) x (acceleration)

Substitute what we know : Force = (1,000 kg) x (3 m/s²)

Do the arithmetic: Force = 3,000 kg-m/s² = 3,000 newtons

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3 years ago

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The picture below shows the setup for an experiment involving a 1000 mL beaker, sitting on a burner, filled with 500 mL of water

Vlad1618 [11]

Answer:

bye

Explanation:

bye

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3 years ago

Examples of uniform speed

Stella [2.4K]

Answer:

A body is said to be moving with uniform speed, if it covers equal distances in equal intervals of time. ...

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3 years ago

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A 1.2 g pebble is stuck in a tread of a 0.76 m diameter automobile tire, held in place by static friction that can be at most 3.

Maksim231197 [3]

Answer:

$v=33.764m/s$

Explanation:

Given data

Mass m=1.2 g=0.0012 kg

diameter d=0.76 m

Friction Force F=3.6 N

To find

Velocity v

Solution

From the Centripetal force we know that

$F_{c}=\frac{mv^{2} }{r}$

Where m is mass

v is velocity

r is radius

Substitute the given values to find velocity v

So

$F_{c}=\frac{mv^{2} }{r}\\v^{2}=\frac{F_{c}(r)}{m}\\ v=\sqrt{\frac{F_{c}(r)}{m}}\\ v=\sqrt{\frac{F_{c}(diameter/2)}{m}}\\v=\sqrt{\frac{(3.6N)(0.76/2)m}{(0.0012kg)}}\\v=33.764m/s$

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3 years ago