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2 years ago

Which instrument changes kinetic energy into electrical energy?

Physics

1 answer:

Schach [20]2 years ago

7 0

Answer:

dynamo

Explanation:

As we all know, dynamos are used in bicycles. When we cycle, the tire rotates, which means kinetic energy. And when the wheel rotates, the dynamo wheel starts to rotate. With that, the motor inside the dynamo starts to rotate. Then it generates electricity, which means electrical energy.

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Car 1 drives 20 mph to the south, and car 2 drives 30 mph to the north. From the frame of reference of car 1, what is the veloci

EleoNora [17]

We subtract the velocity of car 1 from the velocity of car 2:
$v=(30\ mph\ North)-(20\ mph\ South)$
$=(30\ mph\ North)+(20\ mph\ North)$
$=50\ mph\ North$

4 0

2 years ago

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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

What can happen overnight to soil?

UkoKoshka [18]

(GABS) Overnight, all of the particles settled down to the bottom , and the larger particles were on the bottom and the smaller particles were on the top. Therefore, clay was on top, hummus was in the middle, and soil was on the bottom.
Particles dissolve is an unique way

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3 years ago

A train is pulling four train cars and each car has a mass of 40,000 kg. The train is accelerating at 1.1 m/s^2. What is the for

IgorLugansk [536]

Answer:

176,000 N

Explanation:

Newton's second law:

∑F = ma

F = (4 × 40,000 kg) (1.1 m/s²)

F = 176,000 N

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3 years ago

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What situation would give you a mechanical advantage? *

const2013 [10]

D. Using a fixed-pulley system

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2 years ago