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3 years ago

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What current risks are there to Earth from NEOs?

1 answer:

velikii [3]3 years ago

4 0

current risks are there to Earth from NEO

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If the potential across two parallel plates, separated by 3 cm, is 12 volts, what is the electric field strength in volts per me

Mrrafil [7]

There are 100cm in 1m, divide 100 by 3 and you get 33, multiply that by 12 and you get 396 volts/m.

Hope this is correct and helps.

6 0

4 years ago

Read 2 more answers

A 40-turn coil has a diameter of 11 cm. The coil is placed in a spatially uniform magnetic field of magnitude 0.40 T so that the

seropon [69]

Answer:

(a) emf = 0.507 V

(b) emf = 0.0507 V

(c) emf = 0.00234 V

Explanation:

Given;

number of turns of the coil, N = 40 turns

diameter of the coil, d = 11 cm

radius of the coil, r = 5.5 cm = 0.055 m

magnitude of the magnetic field, B = 0.4 T

The magnitude of the induced emf is calculated as;

$emf = - N\frac{d\phi}{dt} \\\\where;\\\\\phi \ is \ magnetic \ flux= BA \\\\A \ is the \ area \ of \ the \ coil = \pi r^2 = \pi (0.055)^2 = 0.0095 \ m^2\\\\emf = - N \frac{dB.A}{dt} = -NA\frac{dB}{dt} \\\\emf = -NA\frac{(B_2 - B_1)}{t} \\\\emf = NA \frac{(B_1 - B_2)}{t} \\\\the \ final \ magnetic \ field \ is \ reduced \ to \ zero;\ B_2 = 0\\\\emf = \frac{NAB_1}{t}$

(a) when the time, t = 0.3 s

$emf = \frac{NAB_1}{t} = \frac{40\times 0.0095\times 0.4}{0.3} = 0.507 \ V$

(b) when the time, t = 3.0 s

$emf = \frac{NAB_1}{t} = \frac{40\times 0.0095\times 0.4}{3} = 0.0507 \ V$

(c) when the time, t = 65 s

$emf = \frac{NAB_1}{t} = \frac{40\times 0.0095\times 0.4}{65} = 0.00234 \ V$

7 0

3 years ago

It is weigh-in time for the local under 85 kg rugby team. The bathroom scale used to assess eligibility can be described by Hook

Grace [21]

15 m/n is the answer

4 0

3 years ago

Rosalinda’s take-home pay is 1700 a month she spends 11% of her take-home pay on her cable bill how much is Rosalinda’s monthly

Marizza181 [45]

Answer:

Explanation:

To solve this simply put 11% in its decimal form (0.11) and multiply it by her take-home pay and boom that's yer answer 187

6 0

3 years ago

shutvik [7]

Answer:

a) P = 149140[w]; b) 1491400[J]; c) v = 63.06[m/s]

Explanation:

As the solution to the problem indicates, we must convert the power unit from horsepower to kilowatts.

P = 200 [hp]

$200[hp] * 745.7 [\frac{watt}{1 hp}]\\149140[watt]$

Now the power definition is known as the amount of work done in a given time

P = w / t

where:

w = work [J]

t = time [s]

We have the time, and the power therefore we can calculate the work done.

w = P * t

w = 149140 * 10 = 1491400 [J]

And finally, we can calculate the velocity using, the expression for kinetic energy

$E_{k}=w=0.5*m*v^{2}\\ where:\\v = velocity[m/s]\\m=mass=750[kg]\\w=work=1491400[J]\\$

The key to solving this problem is to recognize that work equals kinetic energy

$v=\sqrt{\frac{w}{0.5*m}} \\v=\sqrt{\frac{1491400}{0.5*750}} \\v=63.06[m/s]$

3 0

3 years ago