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Irina18 [472]
3 years ago
10

A car has a speed of 20 m/s. If the speed of the car increases to 30 m/s in 5 seconds, what is the car’s acceleration?

Physics
1 answer:
Tema [17]3 years ago
7 0

2

Explanation:

2 times 5 is 10/ 20+10=30

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an object traveling 200 feet per second slows to 50 feet per second in 5 seconds. Calculate the acceleration of the object
emmainna [20.7K]

Answer:

The acceleration of the object is -30\ m/s^2

Explanation:

Given:

Initial velocity of object v_i = 200 feet/second

Final velocity of object v_f = 50 feet/second

Time of travel = 5 seconds

To calculate acceleration of the object we will find the rate of change of velocity with respect to time.

So, acceleration a is given by:

a=\frac{v_f-v_i}{t}

where v_f represents final velocity, v_i represents initial velocity and t is time of travel.

Plugging in values to evaluate acceleration.

a=\frac{50-200}{5}

a=\frac{-150}{5}

a=-30\ m/s^2

The acceleration of the object is -30\ m/s^2 (Answer). The negative sign shows the object is slowing down.

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nika2105 [10]

Answer:

gasoline and natural gas

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heating homes and gas

Explanation:

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3 years ago
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Explanation:

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6 0
3 years ago
A 1.0-cm-tall object is 13 cm in front of a converging lens that has a 40 cm focal length.
kicyunya [14]

A) Image position: -19.3 cm

B) Image height: 1.5 cm, upright

Explanation:

A)

In order to calculate the image position, we can use the lens equation:

\frac{1}{p}+\frac{1}{q}=\frac{1}{f}

where

p is the distance of the object from the lens

q is the distance of the image from the lens

f is the focal length

In this problem, we have:

p = 13 cm (object distance)

f = 40 cm (focal length, positive for a converging lens)

So the image distance is

\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{40}-\frac{1}{13}=-0.0519\\q=\frac{1}{-0.0519}=-19.3 cm

The negative sign means that the image is virtual.

B)

In order to calculate the image height, we use the magnification equation:

\frac{y'}{y}=-\frac{q}{p}

where

y' is the image height

y is the object height

In this problem, we have:

y = 1.0 cm (object height)

p = 13 cm

q = -19.3 cm

Therefore, the image heigth is

y'=-\frac{qy}{p}=-\frac{(-19.3)(1.0)}{13}=1.5 cm

And the positive sign means the image is upright.

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3 years ago
How does 'g' vary from place to place?​
r-ruslan [8.4K]

Explanation:

The acceleration g varies by about 1/2 of 1 percent with position on Earth's surface, from about 9.78 metres per second per second at the Equator to approximately 9.83 metres per second per second at the poles.

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