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kompoz [17]
3 years ago
13

Newton's first law of motion says (in part) that an object in motion will remain in motion at a constant speed and in a straight

line if no _____ forces are acting on it. A. applied B. balanced C. unbalanced D. gravitational
Physics
2 answers:
kolbaska11 [484]3 years ago
6 0

Your answer is C. Unbalanced

unbalanced forces change the direction and speed of objects, therefore it would change the constant speed of an object in motion.

Hope this helps u!

vesna_86 [32]3 years ago
6 0

Answer:

C. unbalanced

Explanation:

I just took the quiz! :)

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If you rub an inflated balloon against your hair and place it against a door, by what mechanism does it stick? explain.
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When you rub inflated balloon with your hair or your kitten's fur, charge is induced on all over the balloon's surface. This is called "charging by friction" because you developed charges by rubbing to bodies with each other. It will also stick on your wall you can check it out. This is because of "unlike charges attract each other". Rubbed balloon and wall possessed unlike charges which made them stick together. 
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Which gas has molecules with the greatest average molecular speed at 25°c?
8_murik_8 [283]

Answer: the options to the questions are

a. 1.0 moles of N2

b.0.5 moles of New

c.0.2 moles of CO2

d.2 moles of He

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The average molecular speed v of gas is given by =√(8RT,/πM)

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3 years ago
Parallel rays of monochromatic light with wavelength 571nm illuminate two identical slits and produce an interference pattern on
Natasha2012 [34]

Answer:

8.8\times 10^{-6} W/m^2

Explanation:

We are given that

Wavelength,\lambda=571 nm=571\times 10^{-9} m

1 nm=10^{-9} m

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1 m=100 cm

d=0.640 mm=0.64\times 10^{-3} m

1 mm=10^{-3} m

a=0.434 mm=0.434\times 10^{-3} m

y=0.830 mm=0.830\times 10^{-3} m

I_0=5.00\times 10^{-4}W/m^2

tan\theta=\frac{y}{R}

\theta=\frac{0.830\times 10^{-3}}{0.75\times 10^{-2}}

\theta=1.1\times 10^{-3}rad

tan\theta\approx \theta

Because \theta is small.

\phi=\frac{2\pi dsin\theta}{\lambda}

\sin\theta\approx \theta,

Therefore

\phi=\frac{2\times\pi\times 0.64\times 10^{-3}\times 1.1\times 10^{-3}}{571\times 10^{-9}}

\phi=7.74 rad

\beta=\frac{2\pi a\theta}{\lambda}

\beta=5.3 rad

I=I_0cos^2(\frac{\phi}{2})(\frac{sin\frac{\beta}{2}}{\frac{\beta}{2}})^2

I=5\times 10^{-4}cos^2(\frac{7.74}{2})(\frac{sin\frac{5.3}{2}}{\frac{5.3}{2}})^2

I=8.8\times 10^{-6} W/m^2

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IV) A boy sits to the side of a campfire. He is 10 feet away, but still feels warm.

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The answer is A) 1 and 2.

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3 years ago
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