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3 years ago

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An electric generator has an 18-cmcm-diameter, 120-turn coil that rotates at 60 HzHz in a uniform magnetic field that is perpend

icular to the rotation axis. Part A What magnetic field strength is needed to generate a peak voltage of 330 VV

1 answer:

kow [346]3 years ago

7 0

Explanation:

<h2>OK ok nosepo </h2><h2> $2825.55$ </h2>

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What did the protoplanets become?

alexira [117]

What did the protoplanets become?

a. nebulae
b. planets
c. solar nebulae
d. planetesimals

The protoplanets become nebulae. The answer is letter A. The rest of the choices do not answer the question above.

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3 years ago

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What was the greatest discovery by galileo during his inclined-plane experiments?

Sergeeva-Olga [200]

Galileo discovered during his inclined-plane experiments that a ball rolling down an incline and onto a horizontal surface would roll indefinitely.

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3 years ago

A jet airliner moving initially at 406 mph (with respect to the ground) to the east moves into a region where the wind is blowin

astraxan [27]

Answer:

966 mph

Explanation:

Using as convention:

- East --> positive x-direction

- North --> Positive y-direction

The x- and y- components of the initial velocity of the jet can be written as

$v_{1x} = 406 mph\\v_{1y} = 0$

While the components of the velocity of the wind are

$v_{2x} = (568)(cos 15^{\circ})=548.6 mph\\v_{2y} = (568)(sin 15^{\circ})=147.0 mph$

So the components of the resultant velocity of the jet are

$v_x = v_{1x}+v_{2x}=406+548.6=954.6 mph\\v_y = v_{1y}+v_{2y}=0+147.0=147.0 mph$

And the new speed is the magnitude of the resultant velocity:

$v=\sqrt{v_x^2+v_y^2}=\sqrt{(954.6)^2+(147.0)^2}=965.8 mph \sim 966 mph$

6 0

3 years ago

What is the most appropriate tool to use when collecting data from the investigation? List, table, graph or chart?

forsale [732]

A table would be the most appropriate because that way you can compare the data.

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2 years ago

An electron enters a region with a speed of 5×10^6m/s and is slowed down at the rate of 1.25×10^-4m/s². How far does the electro

Mashutka [201]

1) The distance travelled by the electron is $1\cdot 10^{17} m$

2) The time taken is $4.0\cdot 10^{10}s$

Explanation:

1)

The electron in this problem is moving by uniformly accelerated motion (constant acceleration), so we can use the following suvat equation

$v^2-u^2=2as$

where

v is the final velocity

u is the initial velocity

a is the acceleration

s is the distance travelled

For the electron in this problem,

$u=5\cdot 10^6 m/s$ is the initial velocity

v = 0 is the final velocity (it comes to a stop)

$a=-1.25\cdot 10^{-4} m/s^2$ is the acceleration

Solving for s, we find the distance travelled:

$s=\frac{v^2-u^2}{2a}=\frac{0-(5\cdot 10^6)^2}{2(-1.25\cdot 10^{-4})}=1\cdot 10^{17} m$

2)

The total time taken for the electron in its motion can also be found by using another suvat equation:

$v=u+at$

where

v is the final velocity

u is the initial velocity

a is the acceleration

t is the time taken

Here we have

$u=5\cdot 10^6 m/s$

v = 0

$a=-1.25\cdot 10^{-4} m/s^2$

And solving for t, we find the time taken:

$t=\frac{v-u}{a}=\frac{0-5\cdot 10^6}{-1.25\cdot 10^{-4}}=4.0\cdot 10^{10}s$

Learn more about accelerated motion:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

#LearnwithBrainly

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2 years ago