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3 years ago

The velocity time graph of a body is given below. Find the distance travelled by the body from A to B. *

Physics

1 answer:

Dovator [93]3 years ago

6 0

Answer:

100m

Explanation:

10 x 10 = 100m

area of velocity time graphs gives displacement.

im not sure if the distance is the same.

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Because there is no record of all things. As we have partially unknow information, it can never be held as a fact.

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2 years ago

Which actions are examples of conserving resources? Check all that apply.

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Answer:

1st, 2nd, and 4th

Explanation:

1st conserves gasoline/petroleum

2nd conserves electricity

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2 years ago

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Whose geocentric model of the solar system was accepted for 1400 years

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Answer:

Plato, Aristotle developed it further and used for 1400 years till Copernicus.

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2 years ago

A 3.00-kg object undergoes an acceleration given by a = (2.00 i + 5.00 j) m/s^2. Find (a) the resultant force acting on the obje

kobusy [5.1K]

Answer:

(a): The resultant force acting on the object are F= (5.99 i + 14.98 j).

(b): The magnitude of the resultant force are F= 16.4 N < 68.19º .

Explanation:

m= 3kg

a= 2 i + 5 j = 5 .38 < 68.19 º

F= m * a

F= 3* ( 5.38 < 68.19º )

F= 16.4 N < 68.19º

Fx= F * cos(68.19º)

Fx= 5.99

Fy= F* sin(68.19º)

Fy= 14.98

3 0

3 years ago

You stand on a straight desert road at night and observe a vehicle approaching. This vehicle is equipped with two small headligh

spayn [35]

To solve this problem we will apply the concepts related to Reyleigh's criteria. Here the resolution of the eye is defined as 1.22 times the wavelength over the diameter of the eye. Mathematically this is,

$\theta = \frac{1.22 \lambda }{D}$

Here,

D is diameter of the eye

$D = \frac{1.22 (539nm)}{5.11 mm}$

$D= 1.287*10^{-4}m$

The angle that relates the distance between the lights and the distance to the lamp is given by,

$Sin\theta = \frac{d}{L}$

For small angle, $sin\theta = \theta$

$sin \theta = \frac{d}{L}$

Here,

d = Distance between lights

L = Distance from eye to lamp

For small angle $sin \theta = \theta$

Therefore,

$L = \frac{d}{sin\theta}$

$L = \frac{0.691m}{1.287*10^{-4}}$

$L = 5367m$

Therefore the distance is 5.367km.

4 0

3 years ago