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3 years ago

8

A car travels 48 km in 1.2 hours. What is the average speed of the car in km/hr

1 answer:

Alborosie3 years ago

5 0

Answer:

average speed of car is 40 km/h

Explanation:

if car goes 48 km per 1.2 hours , the car goes 40 km per hour. so average speed of car is 40km/h

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Is the the last half of the month waxing or waning( talking about moon phases

rewona [7]

Its not really definable. Say, you can have a full moon on the last day of November, and have the waning phases throughout the first half. Or you can have a new moon on the last day of December, and have the waxing phases during the first half of the month.
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3 years ago

Joy uses 20n of force to shovel the snow 10 meters. how much work does she do?

GarryVolchara [31]

Work = Force * distance

Work = 20 N * 10 m = 200 Nm

Work = 200 Joules.

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3 years ago

Two 10-cm-diameter charged rings face each other, 25.0cm apart. Both rings are charged to +20.0nC. What is the electric field st

Black_prince [1.1K]

Answer:

Part A:

$E_{midpoint}=0$

Part B:

$E_{center}=2711.7558 N/C$

Explanation:

Part A:

Formula of Electric Field Strength:

$E=\frac{1}{4\pi\epsilon}\frac{xQ}{(x^2+R^2)^{3/2}}$

Where:

x is the distance from the ring

R is the radius of the ring

$\epsilon$ is constant permittivity of free space=8.854*10^-12 farads/meter

Q is the charge

For right Ring E at the midpoint can be calculated as:

x for right plate=25/2=12.5 cm=0.125 m

Radius=R=10/2=5 cm=0.05 m

$E_{right}=\frac{1}{4\pi8.854*10^{-12}}\frac{(0.125)*(20*10^{-19})}{((0.125)^2+(0.05)^2)^{3/2}}\\E_{right}=9208.1758 N/C$

For Left Ring E at the midpoint can be calculated as:

Since charge on both plates is +ve and same in magnitude, the electric field will be same for both plates.

$E_{left}=\frac{1}{4\pi8.854*10^{-12}}\frac{(0.125)*(20*10^{-19})}{((0.125)^2+(0.05)^2)^{3/2}}\\E_{left}=9208.1758 N/C$

Electric Field at midpoint:

Both rings have same magnitude but the direction of fields will be opposite as they have same charge on them.

$E_{midpoint}=E_{left}-E_{right}\\E_{midpoint}=9208.1758-9208.1758\\E_{midpoint}=0$

Part B:

At center of left ring:

Due to left ring Electric field at center is zero because x=0.

$E_{left}=\frac{1}{4\pi8.854*10^{-12}}\frac{(0)*(20*10^{-19})}{((0)^2+(0.05)^2)^{3/2}}\\E_{left}=0 N/C$

Due to right ring Electric field at center of left ring:

Now: x=25 cm= o.25 m (To the center of left ring)

$E_{right}=\frac{1}{4\pi8.854*10^{-12}}\frac{(0.25)*(20*10^{-19})}{((0.25)^2+(0.05)^2)^{3/2}}\\E_{right}=2711.7558 N/C$

Electric Field Strength at center of left ring is same as that of right ring.

$E_{center}=2711.7558 N/C$

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3 years ago

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valentinak56 [21]

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2 years ago

Help please <br> hhjshwjsjejjenrhrhfhhfheisiw

DanielleElmas [232]

Answer:

A

Explanation:

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3 years ago