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PolarNik [594]
3 years ago
14

1. An airplane flies 800 kilometers east from Denver to Wichita in one hour. Then the plane flies west back to Denver in the sam

e amount of time. Which of the following best relates the speed and velocity of the two flights?
A.
The speed and velocity of the flights are not related for either flight.
B.
The velocity of the second flight is negative compared to the speed.
C.
The speed of the second flight is negative compared to the velocity.
D.
The velocity and speed of both flights are equal.
Physics
1 answer:
Bumek [7]3 years ago
3 0

B. The velocity of the second flight is negative compared to the speed.

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A 920-kg compact car moving at 92 m/s has approximately 3,893,440 Joules of kinetic energy. What is the change in kinetic energy
tensa zangetsu [6.8K]

Answer:

Change in kinetic energy = 3297280 J

Explanation:

Given that,

Mass, m = 920 kg

Speed of a car, v = 92 m/s

Kinetic energy, K = 3,893,440 J

If the speed of a car, V = 36 m/s

Net kinetic energy is given by :

K_n=\dfrac{1}{2}mV^2\\\\=\dfrac{1}{2}\times 920\times (36)^2\\\\K_n=596160\ J

The change in kinetic energy = 3,893,440 - 596160

= 3297280 J

So, the change in kinetic energy of the car is 3297280 J.

3 0
3 years ago
A car's fuel economy is 30 mpg. a warning light is displayed if the remaining distance that can be driven before the car runs ou
Alchen [17]
So 10 gallons of gas would let you travel 300 Miles. 

x gallons = 50 Miles

10 : 300 :: x : 50 

x = 500/300

x = 1.66667 gallons. 

So, the car would run 10 - 1.6666 gallons = 8.33 gallons.

After that, the warning light turns ON! 

Hope this helps!!
7 0
3 years ago
Read 2 more answers
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
gulaghasi [49]

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

B. 142.3 W

The power dissipated in the first bulb is given by:

P_1=I^2 R_1

where

I = 0.77 A is the current

R_1 = 240 \Omega is the resistance of the bulb

Substituting numbers, we get

P_1 = (0.77 A)^2 (240 \Omega)=142.3 W

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.

7 0
3 years ago
The temperature of a body of water influences _____
Maslowich
The temperature of the air above it
5 0
3 years ago
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i wanna learn a little bit of physics so if anyone could teach me a thing or two it would be much appreciated
pentagon [3]

Answer:

... message me. what do u want to know

4 0
3 years ago
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