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solmaris [256]
3 years ago
11

A car drives to the east in a time of 4 hours. Then, immediately (not realistic, but just assume this is the case for this probl

em), travels 12 km to the west in 4 hours. The average speed for the entire trip is 5 km/hr. What is the average speed of the car for the first part of the trip, in km/hr, while the car was traveling east?
Physics
1 answer:
diamong [38]3 years ago
4 0

Answer:

Average speed in east direction speed=\frac{distance}{time}=\frac{28}{4}=7km/hr            

Explanation:

We have given average speed of the entire trip = 5 km/hr

First the car 4 hours then travels 12 km in 4 hours

So total time of the trip = 4+4 = 8 hours

So total distance traveled in the trip d=speed\times time = 5\times 8=40km

As the car travel 12 km in 4 hours to the west

So distance traveled in 4 hour in east = 40 -12 = 28 km

Time in east direction = 4 hour

So average speed in east direction speed=\frac{distance}{time}=\frac{28}{4}=7km/hr

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What is the change in electric potential energy in moving an electron from a location 3 × 10-10 m from a proton to a location 7
joja [24]

Answer:

4.39 x 10^-19 J

Explanation:

q1 = 1.6 x 10^-19 C

q2 = - 1.6 x 10^-19 C

r1 = 3 x 10^-10 m

r2 = 7 x 10^-10 m

The formula for the potential energy is given by

U1 = k q1 q2 / r1 = - (9 x 10^9 x 1.6 x 10^-19 x 1.6 x 10^-19) / (3 x 10^-10)

U1 = - 7.68 x 10^-19 J

U2 = k q1 q2 / r2 = - (9 x 10^9 x 1.6 x 10^-19 x 1.6 x 10^-19) / (7 x 10^-10)

U2 = - 3.29 x 10^-19 J

Change in potential energy is

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A 12 v battery produces a current of 25 amps. What is the resistance <br>Please help me
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2 years ago
By what factor is the intensity of sound at a rock concert louder than that of a whisper when the two intensity levels are 120 d
lyudmila [28]

Answer:

d) 7.94\times 10^{9}

Explanation:

β₁ = sound level of sound at rock concert = 120 dB

β₂ = sound level of sound due to whisper = 21 dB

I₁ = Intensity of sound at rock concert

I₂ = Intensity of sound due to whisper

sound level of sound at rock concert is given as

\beta _{1} = 10 log\left ( \frac{I_{1}}{10^{-12}} \right )

120 = 10 log\left ( \frac{I_{1}}{10^{-12}} \right )

12 = log\left ( \frac{I_{1}}{10^{-12}} \right )               Eq-1

sound level due to whisper is given as

\beta _{2} = 10 log\left ( \frac{I_{2}}{10^{-12}} \right )

21 = 10 log\left ( \frac{I_{2}}{10^{-12}} \right )

2.1 = log\left ( \frac{I_{2}}{10^{-12}} \right )                          Eq-2

subtracting Eq-2 from Eq-1

12 - 2.1 = log\left ( \frac{I_{1}}{10^{-12}} \right ) - log\left ( \frac{I_{2}}{10^{-12}} \right )

9.9 = log\left ( \frac{I_{1}}{I_{2}} \right )

\left ( \frac{I_{1}}{I_{2}} \right ) = 7.94\times 10^{9}

6 0
3 years ago
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