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iogann1982 [59]
1 year ago
7

In each part calculate the kinetic energy of the given objects in joules a) an automobile whose mass is 1260kg moving at a speed

of 66.3km/h b) a runner whose mass is 64.7 kg, moving at a speed of 8.52m/s c) an electron (mass 9.11x10^31kg) moving at a speed of 2.74x10^7m/s
Physics
1 answer:
Contact [7]1 year ago
5 0

a)

Kinetic energy (KE) = 1/2 m v^2

Where:

m = mass = 1260 kg

v = speed = 66.3 km/h = 18.42 m/s

Replacing:

KE = 1/2 (1260 kg ) (18.42 m/s )^2 = 213,756.7 J

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An object is held 24.8 cm from a lens of focal length 16.0 cm. What is the magnification of the image?
Gwar [14]

Answer: 1.8

Explanation:

You are given

the object distance U = 24.8 cm

Focal length F = 16.0 cm

First find the image distance by using the formula:

1/f = 1/u + 1/v

Where V = image distance

Substitute u and f into the formula

1/16 = 1/24.8 + 1/v

1/ v = 1/16 - 1/24.8

1/v = 0.0625 - 0.04032258

1/v = 0.022177

Reciprocate both sides by dividing both sides by one

V = 45.09 cm

Magnification M is the ratio of image distance to the object distance. That is,

M = V/U

Substitute V and U into the formula

M = 45.09/24.8

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3 years ago
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zavuch27 [327]

Answer:

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Explanation:

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I believe the answer is Nuclear energy
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Joe is hiking through the woods when he decides to stop and take in the view. He is particularly interested in three objects: a
Novosadov [1.4K]

Answer:

A) correct answer is C,   B)   correct answer is b  and C) The correct answer is b

Explanation:

In the exercises of geometric optics, the equation of the constructor tells us the location of the image.

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where f is the focal length of the cornea-crystalline system, p and q are the distances to the object and the image.

In this case, the distance to the image on the retina is constant, about 3 cm. Therefore depending on the distance to the object) p = the focal length must change

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let's apply this expression to our case

A) indicates that the tree is at a medium distance

so that the image is formed on the retina THE SAME AS

correct answer is C

B) The squirrel is at a smaller distance (p ') than the tree (p), therefore if we substitute in the equation above we find that q must decrease. Consequently the image is in front of the retina

The mountain is very far, suppose in infinity, so the image is BEHIND THE RETINA

therefore the correct answer is b

C) The squirrel is very close so the curvature of the lens INCREASES, resulting in a DECREASE in the focal length

The correct answer is b

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3 years ago
Can a particle with constant speed be accelarating?<br>what if it has constant velocity?​
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Explanation:

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