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Fantom [35]
3 years ago
6

How far is a light year?? ​

Physics
2 answers:
zavuch27 [327]3 years ago
8 0

Answer

Au=63241.1

Explanation:

light-year = 63241.1

Gala2k [10]3 years ago
6 0

Answer:

1 light-year =

9.461 × 1015 metres

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There are both enviromental and food stimuli that can disrupt sleep true or false
konstantin123 [22]
The answer is true. I hope that this helps you out!!
4 0
3 years ago
What the unit of work?​
ra1l [238]

Answer:

yes

Explanation:

5 0
2 years ago
The speed of a car during a trip is represented in the graph. Compare distance and displacement at four hours into the trip.
mezya [45]

Answer:

The answer to your question is: D.

Explanation:

Distance refers to the amount of space between two points, it is a scalar quantity.

Displacement refers to the space between two points, measure from the minimum path linking them, it is a vector quantity.

I'm not agree with these answers, because the total distance  is approximately 500km.

A) The distance traveled is 300 km.  This answer is not correct.

B) Distance is 300 km and displacement is 0 km.  This answer is not correct because the displacement is also 500 km.

C) Distance is 300 km/hour and displacement is 300 km. 300 km/h is a measure of speed not of distance, this option is wrong.

D) Both distance traveled and displacement are 300 km. I think this is the correct answer because distance and displacement measure the same. but I think both measure 500 km.

7 0
3 years ago
Read 2 more answers
A pasta factory made 3.84 pounds of pasta in 4 minutes. How much pasta did the factory make in each minute? plzz help me
den301095 [7]

{\underline{\green{\textsf{\textbf{ Answer : }}}}}

➡ 0.96 pounds in a minute.

5 0
3 years ago
The velocity of a particle moving along the x-axis varies with time according to v(t) = A + Bt−1 , where A = 2 m/s, B = 0.25 m,
kondaur [170]

Answer:

a= -2\ m/s^2

a=-12.5\ m/s^2

x=2.17 m

x=8.4 m

Explanation:

Given that

v=A+Bt^{-1}

v=2+0.25t^{-1}

To find acceleration :

we know that

a=\dfrac{dv}{dt}

\dfrac{dv}{dt}=0-0.5t^{-2}

a=-0.5t^{-2}

Acceleration at t= 2 s

a=-0.5\times 2^{-2}

a= -2\ m/s^2

Acceleration at t= 5 s

a=-0.5\times 5^{-2}

a=-12.5\ m/s^2

We know that

v=\dfrac{dx}{dt}

dx=\left(2+\dfrac{1}{4t}\right)dt

Position at t= 2 s:

\int_{0}^{x}dx=\int_{1}^{2} \left(2+0.25\dfrac{1}{t}\right)dt

x=\left [2t+0.25\ lnt \right ]_{1}^{2}

x=2+0.25 ln2

x=2.17 m

Position at t= 5 s:

\int_{0}^{x}dx=\int_{1}^{5} \left(2+0.25\dfrac{1}{t}\right)dt

x=\left [2t+0.25\ lnt \right ]_{1}^{5}

x=8+0.25 ln5

x=8.4 m

4 0
3 years ago
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