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Verizon [17]
3 years ago
6

Firecrackers A and B are 600 m apart. You are standing exactly halfway between them. Your lab partner is 300 m on the other side

of firecracker A. You see two flashes of light, from the two explosions, at exactly the same instant of time.
Define event 1 to be "firecracker 1 explodes" and event 2 to be "firecracker 2 explodes." According to your lab partner, based on measurements he or she makes, does event 1 occur before, after, or at the same time as event 2? Explain.
Physics
1 answer:
pishuonlain [190]3 years ago
8 0

Answer:

See the explanation

Explanation:

Given:

Distance of Firecrackers A and B = 600 m

Event 1 = firecracker 1 explodes

Event 2 = firecracker 2 explodes

Distance of lab partner from cracker A = 300 m

You observe the explosions at the same time

to find:

does event 1 occur before, after, or at the same time as event 2?

Solution:

Since the lab partner is at 300 m distance from the firecracker A and Firecrackers A and B are 600 m apart

So the distance of fire cracker B from the lab partner is:

600 m  + 300 m = 900 m

It takes longer for the light from the more distant firecracker to reach so

Let T1 represents the time taken for light from firecracker A to reach lab partner

T1 = 300/c

It is 300 because lab partner is 300 m on other side of firecracker A

Let T2 represents the time taken for light from firecracker B to reach lab partner

T2 = 900/c

It is 900 because lab partner is 900 m on other side of firecracker B

T2 = T1

900 = 300

900 = 3(300)

T2 = 3(T1)

Hence lab partner observes the explosion of the firecracker A before the explosion of firecracker B.

Since event 1 = firecracker 1 explodes and event 2 = firecracker 2 explodes

So this concludes that lab partner sees event 1 occur first and lab partner is smart enough to correct for the travel time of light and conclude that the events occur at the same time.

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dsp73

Answer:

(i) See attached image for the drawing

(ii) net force given in component form: (20, 20)N with magnitude: \sqrt{800} \,\,\,N

Explanation:

First try to write all forces in  vector component form:

The force F1 acting at 45 degrees would have multiplication factors of \frac{\sqrt{2} }{2} on both axes, to take care of the sine and cosine projections. Therefore, the:

x-component of F1 is    F1_x=12\,\sqrt{2} \frac{\sqrt{2} }{2} =12\,\,N

y-component of F1 is    F1_y=12\,\sqrt{2} \frac{\sqrt{2} }{2} =12\,\,N

As far as force F2, it is given already in x and y components, then:

x-component of F2 = 8 N

y-component of F2 = -6 N (negative meaning pointing down the y-axis)

Force F3 has only component (upwards) in the y-direction

x-component of F3 = 0 N

y-component of F3 =14 N

The additions of all these component by component, gives the resultant force (R) acting on the 5 kg mass:

x-component of R = 12 + 8 = 20 N

y-component of R = 12 + 14 - 6 = 20 N

Therefore, the acceleration that the mass receives due to this force is given in component form as:

x-component of acceleration: 20 N / 5 kg = 4\,\,\,m/s^2

y-component of acceleration: 20 N / 5 kg = 4\,\,\,m/s^2

Now we can calculate the components of the velocity of this mass after 2 seconds of being accelerated by this force, using the formula of acceleration times time:

x-component of the velocity is:     v_x=4\,*\,2=8\,m/s

y-component of the velocity is:     v_y=4\,*\,2=8\,m/s

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3 years ago
A parallel-plate capacitor is connected to a battery and becomes fully charged. The capacitor is then disconnected, and the sepa
AnnZ [28]

Answer:

Increases

Explanation:

The expression for the capacitance is as follows as;

C=\frac{\epsilon _{0}A}{d}

Here, C is the capacitance, \epsilon _{0} is the permittivity of free space, A is the area and d is the distance between the parallel plate capacitor.

It can be concluded from the above expression, the capacitance is inversely proportional to the distance. According to the given problem, the capacitor is disconnected from the battery and the distance between the plates is increased. Then, the capacitance of the given capacitor will decrease in this case.

The expression for the energy stored in the parallel plate capacitor is as follows;

E=\frac{Q^{2}}{2C}

Here, E is the energy stored in the capacitor, C is the capacitance and Q is the charge.

Energy stored in the given capacitor is inversely proportional to the capacitor. The charge on the capacitor is constant. In the given problem, as the distance between the parallel plates is being separated, the energy stored in this capacitor increases.

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8 0
4 years ago
Energy from the Sun arrives at the top of the Earth's atmosphere with an intensity of 1.36 kW/m 2 . How long does it take for 3.
alina1380 [7]

Answer:

The time taken by 3.55\times 10^9\ J to arrive on an area of 4.25\ m^2 is 6.14\times 10^5\ seconds.

Explanation:

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Also, the area in which energy is being arriving is 4.25\ m^2

Now, we will use relation between energy (E), intensity of energy (p), area (A) and time (T).

Where energy is in Joule, intensity is in kW/m^2, area is in m^2 and time is in seconds.

The equation is

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So, the time taken by 3.55\times 10^9\ J to arrive on an area of 4.25\ m^2 is 6.14\times 10^5\ seconds.

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3 years ago
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