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alexgriva [62]
3 years ago
14

Alonzo sprints for 500 meters along a straight track during a race. After crossing the finish line, he to walks back along the t

rack for 50 meters to cool down. What is Alonzo's displacement? 450 meters, straight ahead 500 meters, straight ahead 550 meters, straight ahead
Physics
2 answers:
Nata [24]3 years ago
6 0
450 meters, straight ahead is the answer.
allsm [11]3 years ago
3 0

Answer:

450 meters straight ahead

Explanation:

As we know that displacement is the total vector displacement of the object from initial point to final point

As here the Alonzo moves in a straight line during the finish line which is at distance of 500 m

so initial displacement is given as

d_1 = 500 m

now for cool down he again moves back 50 meter along the same track in backward direction

so it is given as

d_2 = -50 m

so we can say that total displacement will be

d = 500 - 50 = 450 m straight ahead

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A 60-W, 120-V light bulb and a 200-W, 120-V light bulb are connected in series across a 240-V line. Assume that the resistance o
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A. 0.77 A

Using the relationship:

P=\frac{V^2}{R}

where P is the power, V is the voltage, and R the resistance, we can find the resistance of each bulb.

For the first light bulb, P = 60 W and V = 120 V, so the resistance is

R_1=\frac{V^2}{P}=\frac{(120 V)^2}{60 W}=240 \Omega

For the second light bulb, P = 200 W and V = 120 V, so the resistance is

R_1=\frac{V^2}{P}=\frac{(120 V)^2}{200 W}=72 \Omega

The two light bulbs are connected in series, so their equivalent resistance is

R=R_1 + R_2 = 240 \Omega + 72 \Omega =312 \Omega

The two light bulbs are connected to a voltage of

V  = 240 V

So we can find the current through the two bulbs by using Ohm's law:

I=\frac{V}{R}=\frac{240 V}{312 \Omega}=0.77 A

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The power dissipated in the first bulb is given by:

P_1=I^2 R_1

where

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Substituting numbers, we get

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C. 42.7 W

The power dissipated in the second bulb is given by:

P_2=I^2 R_2

where

I = 0.77 A is the current

R_2 = 72 \Omega is the resistance of the bulb

Substituting numbers, we get

P_2 = (0.77 A)^2 (72 \Omega)=42.7 W

D. The 60-W bulb burns out very quickly

The power dissipated by the resistance of each light bulb is equal to:

P=\frac{E}{t}

where

E is the amount of energy dissipated

t is the time interval

From part B and C we see that the 60 W bulb dissipates more power (142.3 W) than the 200-W bulb (42.7 W). This means that the first bulb dissipates energy faster than the second bulb, so it also burns out faster.

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3 years ago
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