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weqwewe [10]
3 years ago
9

A 357 kg box is pulled 8.00 m up a 30° frictionless, inclined plane by an external force of 4925 N that acts parallel to the pla

ne. Calculate the work done by the external force. work done by the external force: J Calculate the work done by gravity. work done by gravity: J Calculate the work done by the normal force.
Physics
1 answer:
Sidana [21]3 years ago
4 0
I'm so tremble right now
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A positive kaon (K+) has a rest mass of 494 MeV/c² , whereas a proton has a rest mass of 938 MeV/c². If a kaon has a total energ
vitfil [10]

Answer:

<em>0.85c </em>

Explanation:

Rest mass of Kaon M_{0K} = 494 MeV/c²

Rest mass of proton M_{0P}  = 938 MeV/c²

The rest energy is gotten by multiplying the rest mass by the square of the speed of light c²

for the kaon, rest energy E_{0K} = 494c² MeV

for the proton, rest energy E_{0P} = 938c² MeV

Recall that the rest energy, and the total energy are related by..

E = γE_{0}

which can be written in this case as

E_{K} = γE_{0K} ...... equ 1

where E = total energy of the kaon, and

E_{0} = rest energy of the kaon

γ = relativistic factor = \frac{1}{\sqrt{1 - \beta ^{2} } }

where \beta = \frac{v}{c}

But, it is stated that the total energy of the kaon is equal to the rest mass of the proton or its equivalent rest energy, therefore...

E_{K} = E_{0P} ......equ 2

where E_{K} is the total energy of the kaon, and

E_{0P} is the rest energy of the proton.

From E_{K} = E_{0P} = 938c²    

equ 1 becomes

938c² = γ494c²

γ = 938c²/494c² = 1.89

γ = \frac{1}{\sqrt{1 - \beta ^{2} } } = 1.89

1.89\sqrt{1 - \beta ^{2} } = 1

squaring both sides, we get

3.57( 1 - \beta^{2}) = 1

3.57 - 3.57\beta^{2} = 1

2.57 = 3.57\beta^{2}

\beta^{2} = 2.57/3.57 = 0.72

\beta = \sqrt{0.72} = 0.85

but, \beta = \frac{v}{c}

v/c = 0.85

v = <em>0.85c </em>

7 0
3 years ago
You grab a hot cookie sheet out of the oven this is an example of ______ energy transfer.
Maru [420]
I would say Conduction because you are touching the cookie sheet, even though it is hot (so heat) you are physically touching it so it would not be radiation
6 0
3 years ago
Highlight two factors which shows that heat from the sun does reach the earth's surface by convection.
MAXImum [283]

radiation

when the suns radiation fall on the earth and its objects they receive heat energy and hence get heated. Thus the suns heat reaches the earth by. the process of radiation

3 0
3 years ago
Which waves have wavelengths longer than those of visible light? Give an example of how each kind of wave is used.
Kazeer [188]
1. Radio Waves
ex. Wi-Fi
2. Microwaves
ex. Mobile Phones
3. Infrared Radiation
ex. Heat Lamps
6 0
3 years ago
Read 2 more answers
A 7.5 cm object is 14 cm from a concave lens, which has a focal length of –7 cm. Its image is 5 cm in front of the lens. What is
Rama09 [41]
Given conditions:
height of object = 7.5cmdistance of object from mirror  = 14 cmfocus length = -7 cmimage distance =  ?
Using mirror formula: 
1/(focus length) = 1/(object distance) + 1/(image distance)
or, -1/7 = 1/14 + 1/(image distance)
or, image distance = -4.66cm (the image formed is a virtual image)


Also, magnification of image is:
image height /height of object = - image distance /object distance
or, image height = - image distance / object distance * height of object
or, image height = -(-4.66) / 14 * 7.5 = 2.49 = 3(nearest whole number)
3 0
3 years ago
Read 2 more answers
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