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soldier1979 [14.2K]

3 years ago

5

A motorcyclist changes the velocity of his bike from 20.0 meters/second to 35.0 meters/second under a constant acceleration of 4

.00 meters/second2. How long does it take the bike to reach the final velocity?

1 answer:

shutvik [7]3 years ago

8 0

You can use the equation V=Vo+at since the acceleration is constant. Plugging in the values you know, you will get an answer of 3.75 seconds

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A classroom has 24 fluorescent bulbs, each of which is 32 W. how much energy does it take to light the room for a minute?(unit=J

cestrela7 [59]

Answer:

Energy= 46.08KJ

Explanation:

Given that the power needed to light each bulb is 32W

We know that Power = $\frac{energy}{time}$

The energy needed to light one bulb= $power*time$

Given time = 1minute = 60 seconds

Energy = $32W*60sec$ =1920J

Therefore energy needed to light one bulb is 1920J

The energy needed to light 24 bulbs = $1920*24$ =46080J=46.08KJ

3 0

3 years ago

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What is the acceleration of a Ford Mustang GT that can go from 0.00 to 27.8<br> m/s in 5.15 seconds?

bixtya [17]

Finding acceleration= final speed-initial speed/time taken (or A=V-U\T)

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So..

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2 years ago

What is the acceleration of this object? The object's mass is 60 kg.

andre [41]

The acceleration is 3.3 m/s2

8 0

3 years ago

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Two point charges are arranged in a line with point P, which is on the right. The

MAVERICK [17]

Answer:

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3 years ago

You find an unmarked blue laser on your way to physics class. When you get to class you realize you can determine the wavelength

Ray Of Light [21]

Answer:

wavelength $\lambda$ = 437.27 nm

Explanation:

given data

first bright fringe = 2.96 mm

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distance D = 2.20 m

solution

we know that this is double slit experiment

so we apply here Fringe width formula that is

β = $\frac{D\lambda}{d}$ ....................1

$\lambda$ is Wavelength of light and D is Distance between screen and slit and d is slit width

so put here value and we get

$\lambda$ = $\frac{2.96*0.325*10^{-6}}{2.20}$

$\lambda$ = 437.27 × $10^{-9}$ m

wavelength $\lambda$ = 437.27 nm

4 0

3 years ago