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Umnica [9.8K]
3 years ago
5

While John is traveling along a straight interstate highway , he notices that the mile marker reads 248 km. John travels until h

e reaches the 149 km marker and then retraces his path to the 167 km marker . What is John's resultant displacement from the 248 km marker

Physics
1 answer:
sergiy2304 [10]3 years ago
3 0

Answer:

I'm just going to tell you the information you need but not the answer so you can learn from the problem.

Explanation:

So he was at 248 km mark and traveled 99 km to get to the 149km mark. Then he turns around to go back 18 km to the 167 km mark. That is all the information you need to complete the question I recommend drawing it out in your notes.

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Incident rays parallel to the axis of a concave mirror reflect parallel to the axis.
coldgirl [10]
No they don't.  Incident rays parallel to the axis of a concave mirror
reflect from the mirror's surface and converge at its focal point.
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3 years ago
Pls help!! i don’t understand what i am supposed to do.
liq [111]

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7 0
3 years ago
The air pressure inside a car tire is
Valentin [98]

Answer:

0.137m²

Explanation:

Pressure = Force/Area

Given

Force = 41,500N

Pressure = 3.00atm

since 1atm = 101325.00 N/m²

3atm = 3(101325.00)

3atm = 303,975N/m²

Pressure = 303,975N/m²

Get the area

Area = Force/Pressure

Area = 41500/303,975

Area = 0.137m²'

Hence the surface area of the  inside of the tire is 0.137m²

7 0
3 years ago
Which list places the layers of the sun in the correct order from outermost to innermost?
Valentin [98]
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4 0
3 years ago
Assume the average value of the vertical component of Earth's magnetic field is 42 μT (downward) in some region that has an area
Oliga [24]

Answer:

The net magnetic flux through the rest of Earth's surface is  5.782\times10^{-2}\ T

Explanation:

Given that,

Magnetic field = 42 μT

Area A=3.71\times10^{5}\ km^2

A=3.71\times10^{11}\ m^2

We need to calculate the flux per unit area

flux\ per\ unit\ area=\dfrac{42\times10^{-6}}{3.71\times10^{11}}

flux\ per\ unit\ area=1.132\times10^{-16}\ T/m^2

We need to calculate the total earth's surface area

A'=4\pi r^2

A'=4\times\pi\times(6.3781\times10^{6})^2

A'=5.1120\times10^{14}\ m^2

We need to calculate the rest of earth's area

A''=A-A'

Put the value into the formula

A''=5.1120\times10^{14}-3.71\times10^{11}

A''=5.10829\times10^{14}\ m^2

We need to calculate the net magnetic flux through the rest of Earth's surface

B'=5.10829\times10^{14}\times1.132\times10^{-16}

B'=5.782\times10^{-2}\ T

Hence, The net magnetic flux through the rest of Earth's surface is  5.782\times10^{-2}\ T

3 0
4 years ago
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