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Blizzard [7]
3 years ago
9

A students walks one hour. She walks 4 blocks east and 2 blocks west. Her average velocity is?

Physics
1 answer:
Rama09 [41]3 years ago
6 0

Let us consider the following figure

<em>figure</em><em> </em><em>is</em><em> </em><em>in</em><em> </em><em>pict</em><em>ure</em>

So, his/her total displacement = 2 blocks.

total time = 1 hour

Average Velocity

= Total displacement/ total time

= 2 blocks/ 1 hour

= 2 blocks/ h

So, her/his average velocity is 2 blocks per hour.

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An egg is thrown horizontally off the top row bleachers at the Brickyard,
Ludmilka [50]

Answer:

4

Explanation:

Divide 30 meters by 7.5 and you´re answer is 4. This is how I would think you solve the problem

6 0
2 years ago
to measure the static friction coefficient between a block and a vertical wall, a spring is attached to the block, is pushed on
Stolb23 [73]

Answer:

μ = mg/kx

Explanation:

Since the bock does not slip, the frictional force equals the weight of the block. So, F = mg. Now, the frictional force, F = μN where μ = coefficient of static friction and N = Normal force.

Now, the normal force equals the spring force F' = kx where k = spring constant and x = compression of spring.

N = F' = kx

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8 0
2 years ago
Graph the following data tables on different graphs.
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8 0
2 years ago
Charge Q is distributed uniformly throughout the volume of an insulating sphere of radius R = 4.00 cm. At a distance of r = 8.00
Elena L [17]

Answer:

2.62898\times 10^{-6}\ C/m^3

1979.99974\ N/C

Explanation:

k = Coulomb constant = 8.99\times 10^{9}\ Nm^2/C^2

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r = Distance = 8 cm

R = Radius = 4 cm

Electric field is given by

E=\dfrac{kQ}{r^2}\\\Rightarrow Q=\dfrac{Er^2}{k}\\\Rightarrow E=\dfrac{990\times 0.08^2}{8.99\times 10^{9}}\\\Rightarrow Q=7.04783\times 10^{-10}\ C

Volume charge density is given by

\sigma=\dfrac{Q}{\dfrac{4}{3}\pi R^3}\\\Rightarrow \sigma=\dfrac{7.04783\times 10^{-10}}{\dfrac{4}{3}\pi (0.04)^3}\\\Rightarrow \sigma=2.62898\times 10^{-6}\ C/m^3

The volume charge density for the sphere is 2.62898\times 10^{-6}\ C/m^3

E=\dfrac{kQr}{R^3}\\\Rightarrow E=\dfrac{8.99\times 10^9\times 7.04783\times 10^{-10}\times 0.02}{0.04^3}\\\Rightarrow E=1979.99974\ N/C

The magnitude of the electric field is 1979.99974\ N/C

8 0
3 years ago
Can someone help me with this science question
Anettt [7]
Mechanical layers of the earth:

-Lithosphere
-Asthenosphere
-Mesosphere
-Outer Core
-Inner Core


Chemical layers of the earth:

-Crust
-Mantle
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Hope this helps

Have a great weekend! :)
6 0
3 years ago
Read 2 more answers
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