A car starts from rest at a stoplight and reaches 20 M/s in 3.5 seconds determine the acceleration of the car

1 answer:

4 0

$a = \frac{v}{t}$
so just plug in the velocity for v, which is 20 m/s
and plug in the time for seconds which is 3.5s

so
$a = \frac{20 \frac{m}{s} }{3.5s}$
that will give you your answer

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In which situations is elastic potential energy present? Check all that apply. A bow is drawn back from its equilibrium position

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1,4,6
a bow is drawn back
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3 years ago

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You are driving at 50 miles per hour. If you decrease the time it takes you to travel 1 mile by 8 seconds, what is your new spee

Dmitriy789 [7]

Answer:

The new speed is 56.25 miles/hour.

Explanation:

Since speed = distance/time;

time = distance/speed.

While driving at 50 miles/hour, time taken for one to complete 1 mile is (1/50) hour

(1/50) hour = (1/50) × 3600s = 72 seconds.

So, if this time to complete 1 mile (72 seconds) is reduced by 8 seconds,

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4 years ago

An object weighs 60.0 kg on the surface of the earth. How much does it weigh 4R from the surface? (5R from the center)

Alecsey [184]

"60 kg" is not a weight. It's a mass, and it's always the same
no matter where the object goes.

The weight of the object is

   (mass) x (gravity in the place where the object is) .

On the surface of the Earth,

   Weight = (60 kg) x (9.8 m/s²)

      =    588 Newtons.

Now, the force of gravity varies as the inverse of the square of the distance from the center of the Earth.
On the surface, the distance from the center of the Earth is 1R.
So if you move out to 5R from the center, the gravity out there is

(1R/5R)² = (1/5)² = 1/25 = 0.04 of its value on the surface.

The object's weight would also be 0.04 of its weight on the surface.

   (0.04) x (588 Newtons) = 23.52 Newtons.

Again, the object's mass is still 60 kg out there.
___________________________________________

If you have a textbook, or handout material, or a lesson DVD,
or a teacher, or an on-line unit, that says the object "weighs"
60 kilograms, then you should be raising a holy stink.
You are being planted with sloppy, inaccurate, misleading
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They owe you better material.

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3 years ago

Use Eq. cosϕ=R/Z to show that the average power delivered by the source in an L−R−C series circuit is given by Pav = I^2rmsR .

Evgen [1.6K]

Answer:

Explanation:

In a L C R circuit, the average power is given by

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