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poizon [28]
3 years ago
6

Firecracker A is 300 m from you. Firecracker B is 600 m from you in the same direction. You see both explode at the same time. D

efine event 1 to be "firecracker A explodes" and event 2 to be "firecracker B explodes." Does.event 1 occur before, after, or at the same time as event 2? Explain.
Physics
1 answer:
ss7ja [257]3 years ago
3 0

Answer:

e see that the distances are different, the only way that the two beams of light approach simultaneously is that event 2 (farthest) occurs first than event 1

Explanation:

This is an ejercise in special relativity, where the speed of light is constant.

Let's carefully analyze the approach, we see the two events at the same time.

The closest event time is

       c = (x₁-300) / t

       t = (x₁-300) / c

The time for the other event is

       t = (x₂- 600) / c

since they tell us that we see the events simultaneously, we can equalize

        (x₁ -300) / c = (x₂ -600) / c

         x₁ = x₂ - 300

We see that the distances are different, the only way that the two beams of light approach simultaneously is that event 2 (farthest) occurs first than event 1

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Tanzania [10]

Answer:

12°F

Explanation:

Calculation for how much subcooling is there in the condenser

Since the CONDENSING TEMPERATURE for 417.4 psig discharge pressure is 120 degrees (120°) which means that the amount of subcooling that is there in the condenser will be calculated using this formula

Amount of Condenser subcooling= Condensing Temperature discharge pressure -Condenser outlet temperature

Let plug in the formula

Amount of Condenser subcooling=120°-108f

Amount of Condenser subcooling=12°F

Therefore the amount of subcooling that is there in the condenser will be 12°F

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3 years ago
Spider-Man and Ned were testing the distance he could shoot his web depending on the angle at which he points his web shooter. H
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Answer:

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3 years ago
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A body is at aheight of 81m and is ascending upwards with a velocity of 12m/s .a body of 2 kg weight is dropped from it.if g=10m
butalik [34]
First the velocity drops to zero in 1.2 secs. In those seconds it went upwards for 7.2 m, then it went from 87.2 to 0m. x-x0=v0*t+1/2*g*t^2 ergo t=sqrt(2x/g) that is 4.1761 s. Finally the total time required is 5.3761 s
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Can some answer 7 a and b please
pav-90 [236]

Answer:

a.After 15 second Mr Comer's speed =  1.7 m/s

b.Distance travelled by Mr.Comer in 15 seconds  =   18.75\ m

Explanation:

a. Lets recall our first equation of motion V_{f} =V_{i} + at

Now we know that V_{i} = 0.8m/s , a = 0.06\ m/s^2 and

t = 15 \ sec

Plugging the values we have.

V_{f} =V_{i} + at

V_{f} =0.8 + 0.06 \times 15

V_{f} =0.8+ 0.9

V_{f} =1.7 m/s

Then Mr.Comer's speed after 15 sec = 1.7ms^-1

b.

Lets find the distance and recall our third equation of motion.

V_{f} ^2-V{i}^2 = 2as

So s = distance covered.

Dividing both sides with 2a we have.

\frac{V_{f} ^2-V{i}^2}{2a} = 2as

Plugging the values.

\frac{(1.7)^2-(0.8)^2}{2\times0.06} = 2as

s= 18.75\ m

So Mr.Comer will travel a distance of s= 18.75\ m.

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