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kaheart [24]
3 years ago
15

Mary starts from her house, walks 80 meters south, and stops to chat with her aunt on the sidewalk. After chatting for a few min

utes, she walks 125 meters to the west to meet a friend and then walks 45 meters north with her friend to a cafe for lunch. Overall, she takes 10 minutes to do all this. What is Mary’s average speed?
8 meters/second
1.25 meters/second
25 meters/second
4.5 meters/second
0.42 meters/second
Physics
1 answer:
maksim [4K]3 years ago
6 0

Answer:

0.42 meters/second

Explanation:

The total distance traveled by Mary is: 80 m + 125 m + 45 m = 250 m  

We want the speed in meters/second, so we have to convert the 10 minutes elapsed during the trip into seconds. Then we need a conversion factor, 1 minute is equivalent to 60 seconds.

time = 10 minutes * (60 seconds/1 minute) = 600 seconds (notice that minute is cancelled because is multiplying and dividing)

Now, we can compute the average speed is as follows:

Average speed = total distance/time = 250 m/600 s =  0.42 m/s

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A thin spherical shell with radius R1 = 4.00 cm is concentric with a larger thin spherical shell with radius R2 = 6.00 cm . Both
kakasveta [241]

Answer:

The right response will be "450 volts".

Explanation:

The given values are:

R1 = 4.00 cm

R2 = 6.00 cm

q1 = +6.00 nC

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As we know,

The potential difference between the two shell's difference will be:

⇒  \Delta V=K[(\frac{q1}{R1}+\frac{q2}{R2})-(\frac{q1}{R1} +(\frac{q2}{R2}))]

           =K[\frac{q1}{R2}-\frac{q1}{R1} ]

On substituting the values, we get

           =(9\times 10^9)[\frac{6\times 10^{-9}}{0.04}-\frac{6\times 10^{-9}}{0.06}]  

       Δ =450 \ volts

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Explanation:

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A vector → A has a magnitude of 56.0 m and points in a direction 30.0° below the negative x axis. A second vector, → B , has a m
MissTica

Answer:

  • The magnitude of the vector \vec{C} is 107.76 m

Explanation:

To find the components of the vectors we can use:

\vec{A} = | \vec{A} | \ ( \ cos(\theta) \ , \ sin (\theta) \ )

where | \vec{A} | is the magnitude of the vector, and θ is the angle over the positive x axis.

The negative x axis is displaced 180 ° over the positive x axis, so, we can take:

\vec{A} = 56.0 \ m \ ( \ cos( 180 \° + 30 \°) \ , \ sin (180 \° + 30 \°) \ )

\vec{A} = 56.0 \ m \ ( \ cos( 210 \°) \ , \ sin (210 \°) \ )

\vec{A} = ( \ -48.497 \ m \ , \ - 28 \ m \ )

\vec{B} = 82.0 \ m \ ( \ cos( 180 \° - 49 \°) \ , \ sin (180 \° - 49 \°) \ )

\vec{B} = 82.0 \ m \ ( \ cos( 131 \°) \ , \ sin (131 \°) \ )

\vec{B} = ( \ -53.797 \ m \ , \ 61.886\ m \ )

Now, we can perform vector addition. Taking two vectors, the vector addition is performed:

(a_x,a_y) + (b_x,b_y) = (a_x+b_x,a_y+b_y)

So, for our vectors:

\vec{C} = ( \ -48.497 \ m \ , \ - 28 \ m \ ) + ( \ -53.797 \ m \ ,  ) = ( \ -48.497 \ m \ -53.797 \ m , \ - 28 \ m \ + \ 61.886\ m \ )

\vec{C} = ( \ - 102.294 \ m , \ 33.886 m \ )

To find the magnitude of this vector, we can use the Pythagorean Theorem

|\vec{C}| = \sqrt{C_x^2 + C_y^2}

|\vec{C}| = \sqrt{(- 102.294 \ m)^2 + (\ 33.886 m \)^2}

|\vec{C}| =107.76 m

And this is the magnitude we are looking for.

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