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Softa [21]
3 years ago
7

How do you think speed affects car and driver safety?

Physics
1 answer:
blondinia [14]3 years ago
4 0
An increase in average speed of 1 km/h typi- cally results in a 3% higher risk of a crash involving injury, with a 4–5% increase for crashes that result in fatalities. — Speed also contributes to the severity of the impact when a collision does occur. Hope this helps!
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What is the formula for kinetic energy​
Svetradugi [14.3K]
The equation for Kinetic Energy is KE = 1/2 m v^2, where m is the mass, and v is the velocity. The velocity in this equation is squared, which means that it is exponential. That means that as the velocity increases, you will be multiplying by a bigger and bigger number! KE = 18 joules!
6 0
3 years ago
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What factors cause a rock to become metamorphic
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Pressure and heat. I hope this helps
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3 years ago
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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
1. In an urban area, what can you expect
pantera1 [17]

Answer: Most inhabitants of urban areas have nonagricultural jobs meaning not of, relating to, or used in farming and agriculture nonagricultural products/land. Urban areas are very developed, meaning there is a density of human structures such as houses, commercial buildings, roads, bridges, and railways.

Explanation:

If you appreciate my answer please consider marking it Brainliest. Thank you!

8 0
3 years ago
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A compound microscope has an objective lens of focal length 1.40 cm and an eyepiece with a focal length of 2.20 cm. The objectiv
olchik [2.2K]

Answer:

magnification is - 159

objective distance is 3.85 cm

Explanation:

Given data

focal length f1 = 1.40 cm

focal length f2 = 2.20 cm

separated d = 19.6 cm

to find out

angular magnification and How far from the objective

solution

we know magnification formula that is

magnification = ( - L / f1 ) (D/f2)

here D = 25 cm put all value

magnification = ( - 19.6 / 1.40 ) (25/2.20)

magnification = - 159

and

now we apply lens formula

i/f = 1/q + 1/p

p = f2 = 2.20

so

q = f2 p / p -f2

q = 1.4(2.20) / ( 2.2 - 1.4 )

q = 3.85 cm

so objective distance is 3.85 cm

3 0
3 years ago
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