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Anton [14]
3 years ago
11

What does it mean if a paint sample "matches" a known sample from a vehicle? Does this indicate the same source? Why or why not?

Physics
1 answer:
Mandarinka [93]3 years ago
4 0
It wouldn't indicate the same source if the paint was from a different place. If the paint was the same brand and color it would be but if not then no
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What happens to the kinetic energy of a snowball as it rolls across the lawn and gains mass.
LiRa [457]

If the ground is level, then the snowball can never have
any more kinetic energy than it hand when it left your hand.

If more mass sticks to it as it makes its way across the lawn,
then it must slow down, so that its

                 KE = (1/2) (present mass) (present speed)²

never exceeds the KE you gave it when you tossed it.

And we're not even talking yet about all the energy it loses
by scraping through the snow and mashing down the blades
of grass in its path.

5 0
3 years ago
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Can someone right me a flooding psa ?
Elan Coil [88]
What’s that? Sorry??
7 0
3 years ago
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Compared to its weight on Earth, a 10-kg object on the moon will weigh Question 9 options: less. the same amount. more.
Nostrana [21]

it will weigh less on the moon.

3 0
3 years ago
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Monochromatic light of variable wavelength is incident normally on a thin sheet of plastic film in air. The reflected light is a
BartSMP [9]

Answer:

thickness t = 528.433 nm

Explanation:

given data

wavelength λ1 = 477.1 nm

wavelength λ2 = 668.0 nm

n = 1.58

solution

we know for constructive interference condition will be

2 × t × μ = (m1+0.5) × λ1     ....................1

2 × t × μ = (m2+0.5) × λ2     ....................2

so we can say from equation 1 and 2

(m1+0.5) × λ1 = (m2+0.5) × λ2

so

\frac{\lambda 2}{\lambda 1} = \frac{m1+0.5}{m2+0.5}     ..............3

put here value and we get  

\frac{668.0}{477.1} = \frac{m1+0.5}{m2+0.5}  

\frac{m1+0.5}{m2+0.5}   = 1.4

\frac{m1+0.5}{m2+0.5}  = \frac{7}{5}   ...................4

so we here from equation 4

m1+0.5  = 7

m1 = 3    .................5

m2+0.5 = 4

m2 = 2    .................6

so now put value in equation  1

2 × t × μ = (m1+0.5) × λ1

2 × t × 1.58 = (3+0.5) ×  477.1

solve it we get

thickness t = 528.433 nm

4 0
3 years ago
A planet of mass 7.00 1025 kg is in a circular orbit of radius 6.00 1011 m around a star. The star exerts a force on the planet
BARSIC [14]

Answer:

A) 1.88 * 10^17 m

B) 1.22 * 10^34 J

C) 1.95 * 10^34 J

Explanation:

Parameters given:

Mass of planet = 7.00 * 10^25 kg

Radius of orbit = 6.00 * 10^11 m

Force exerted on planet = 6.51 * 10^22 N

Velocity of planet = 2.36 * 10^4 m/s

A) The distance traveled by the planet is half of the circumference of the orbit (which is circular).

The circumference of the orbit is

C = 2 * pi * R

R = radius of orbit

C = 2 * 3.142 * 6.0 * 10¹¹

C = 3.77 * 10¹² m

Hence, distance traveled will be:

D = 0.5 * 3.77 * 10¹²

D = 1.88 * 10 ¹² m/s

B) Work done is given as:

W = F * D

W = 652 * 10²² * 1.88 * 10¹¹

W = 1.22 * 10³⁴ J

C) Change in Kinetic energy is given as:

K. E. = 0.5 * m * v²

K. E. = 0.5 * 7 * 10^25 * (2.36 * 10^4)²

K. E. = 1.95 * 10³⁴ J

7 0
3 years ago
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