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Inessa05 [86]
3 years ago
10

A bicycle traveled 150 meters west from point A to point B. Then It took the same route and came back to point A. It took a tota

l of 2 minutes for
the bicycle to return to point A. What is the average speed and average velocity of the bicycle?
A.
The average speed is 2.5 meters/second, and the average velocity is
2.5 meters/second east.
B.
The average speed Is 0 meters/second, and the average velocity is
2.5 meters/second east.
C. The average speed and average velocity are both 0 meters/second.
D.
The average speed is 2.5 meters/second, and the average velocity is
O meters/second.
Physics
1 answer:
Julli [10]3 years ago
6 0

Answer:

Option D.  The average speed is 2.5 meters/second, and the average velocity is  0 meters/second.

Explanation:

we know that

To find out the average speed divide the total distance by the total time

Let

d -----> the total distance in meters

t -----> the time in seconds

s ----> the speed in meters per second

s=\frac{d}{t}

Remember that

1\ min=60\ sec

we have

t=2\ min=2(60)=120\ sec

d=150(2)=300\ m

substitute

s=\frac{300}{120}

s=2.5\frac{m}{sec}

<u><em>Find out the average velocity</em></u>

To find out the average velocity divide the displacement) by the time

The displacement is the distance from the start point to the end point regardless of the route

In this problem

The start point is A and the end point is A

so

The displacement  is equal to zero

therefore

The average velocity is 0 m/sec

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Answer:

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Explanation:

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8 0
3 years ago
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A circular area with a radius of 6.50 cm lies in the xy-plane. What is the magnitude of the magnetic flux through this circle du
White raven [17]

Answer:

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Explanation:

#Consider a circular area of radius R=2.98cm in the xy-plane at z=0. This means all the are vector points toward the +ve z-axis.

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\phi _B=\int {\bar B} . d\bar A \\=\int BdAcos(\theta)=BAcos(0)=BA\\\\=\pi R^2B=\pi(6.50\times10^-^3m)^2(0.230T)\\=3.0528\times10^-^3 T\ m^2

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Magnetic flux is calculated as:

\phi _B=\int\bar B \bar dA\\=\int BdAcos (\theta)=BAcos(0)=BA\\=\pi R^2B=\pi(6.50\times 10^-^2m)^2(0.230T)\\=1.83299\times 10^-^3 T \ m^2

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c. We now need to find the magnetic flux if the field has a magnitue of B=0.230T and points in the direction of +y-direction. As with the previous parts, the magnetic flux will be calculated as:

\phi_B= \int\bar B \times d\bar A\\=\int BdAcos(\theta)\\=BAcos(90\textdegree)\\=0

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