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vitfil [10]
3 years ago
10

What is the average speed of a car that travels 15 km in 4000. seconds?

Physics
1 answer:
Katyanochek1 [597]3 years ago
8 0

Answer:

266.66

Explanation:

and you divide 4000 by 15 to get your answer

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Es el conjunto de longitudes de onda de todas las radiaciones electromagnéticas
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Energy created by the flow of electrons through a conductor.
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The first one is electrical energy
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What is the correct answer?
damaskus [11]

Answer:

96 million of dollars

Explanation:

p=-6x^3+72x

p=-6(2)^3+72(2)

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p=-48+144

p=96

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2 years ago
A substance did not change its chemical nature in a reaction. Which most likely describes the reaction?
Andre45 [30]

B. It was hit with a hammer.

4 0
3 years ago
Read 2 more answers
A particle with charge − 2.74 × 10 − 6 C −2.74×10−6 C is released at rest in a region of constant, uniform electric field. Assum
s2008m [1.1K]

Answer:

241.7 s

Explanation:

We are given that

Charge of particle=q=-2.74\times 10^{-6} C

Kinetic energy of particle=K_E=6.65\times 10^{-10} J

Initial time=t_1=6.36 s

Final potential difference=V_2=0.351 V

We have to find the time t after that the particle is released and traveled through a potential difference 0.351 V.

We know that

qV=K.E

Using the formula

2.74\times 10^{-6}V_1=6.65\times 10^{-10} J

V_1=\frac{6.65\times 10^{-10}}{2.74\times 10^{-6}}=2.43\times 10^{-4} V

Initial voltage=V_1=2.43\times 10^{-4} V

\frac{\initial\;voltage}{final\;voltage}=(\frac{initial\;time}{final\;time})^2

Using the formula

\frac{V_1}{V_2}=(\frac{6.36}{t})^2

\frac{2.43\times 10^{-4}}{0.351}=\frac{(6.36)^2}{t^2}

t^2=\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}

t=\sqrt{\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}}

t=241.7 s

Hence, after 241.7 s the particle is released has it traveled through a potential difference of 0.351 V.

6 0
3 years ago
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