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natka813 [3]
3 years ago
10

Firecrackers 1 and 2 are 600 m apart. You are standing exactly halfway between them. Bob is standing 300 m on the other side of

firecracker 1. You see the flashes of light from the two firecrackers exploding at the exact same instant.
a) Does Bob see both firecrackers explode at exactly the same time, firecracker 1 explode first, or firecracker 2 explode first? Explain.
Physics
1 answer:
Afina-wow [57]3 years ago
6 0

Answer:

the same time

Explanation:

since the speed of light is very fast (3×10^8 ms^-1) the difference in time arrived from firecrackers A and B to Bob is negligible, and our brain can distinguish such small time difference,l.

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How do I solve this step by step? I’m really confused
LekaFEV [45]

Step-#1:

Ignore the wire on the right.

Find the strength and direction of the magnetic field at P,

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Write it down.


Step #2:

Now, ignore the wire on the left.

Find the strength and direction of the magnetic field at P,

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Write it down.


Step #3:

Take the two sets of magnitude and direction that you wrote down

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4 0
3 years ago
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
The kinetic energy of a rocket is increased by a factor of eight after its engines are fired, whereas its total mass is reduced
Rudik [331]

The momentum increases by a factor of 2

Explanation:

We can solve this problem by rewriting the momentum of the rocket in terms of the kinetic energy and the mass.

The kinetic energy of the rocket is:

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From eq.(1) we get

v=\sqrt{\frac{2K}{m}}

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K' = 8K

- The mass is reduced by half:

m'=\frac{m}{2}

Substituting, we find the new momentum:

p'=\sqrt{2(\frac{m}{2}(8K)}=\sqrt{4(2mK)}=2\sqrt{2mK}=2p

So, the momentum increases by a factor of 2.

Learn more about momentum and kinetic energy:

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#LearnwithBrainly

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