2 years ago

I need help for the first 3 plz.

Physics

2 answers:

madam [21]2 years ago

7 0

1) hypothesis
2) data
3) method

I think these are correct.

Sonbull [250]2 years ago

3 0

It's hypothesis then its data and then is method

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20 POINTS!!!

hram777 [196]

I think its inductance. If its not then I think its none of the above

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2 years ago

To see why an MRI utilizes iron to increase the magnetic field created by a coil, calculate the current needed in a 400-loop-per

gtnhenbr [62]

Answer:

B = 4.059 x 10¹⁵ T

Explanation:

Given,

Number of loop, N = 400

radius of loop, r = 0.65 x 10⁻¹⁵ m

Current, I = 1.05 x 10⁴ A

Magnetic field at the center of the loop

$B = \dfrac{\mu_0NI}{2R}$

$B = \dfrac{4\pi\times 10^{-7}\times 400 \times 1.05 \times 10^4}{2\times 0.65\times 10^{-15}}$

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3 years ago

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dezoksy [38]

Answer:1.301 s

Explanation:

Given

Initial Velocity(u)=30 m/s

Height of cliff=8.3 m

Time taken to cover 8.3 m

$h=ut+\frac{at^2}{2}$

here Initial vertical velocity is 0

$8.3=\frac{gt^2}{2}$

$t^2=1.69$

$t=1.301 s$

Horizontal distance

$R=u\times t$

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2 years ago

What is the magnitude of an electric field that balances the weight of a plastic sphere of mass 2.1 g that has been charged to -

Liula [17]

Answer:

Electric field, $E=6.86\times 10^6\ N/C$

Explanation:

It is given that,

Mass of sphere, m = 2.1 g = 0.0021 kg

Charge, $q=-3\ nC=-3\times 10^{-9}\ C$

We need to find the magnitude of electric field that balances the weight of a plastic spheres. So,

$ma=qE$

a = g

$E=\dfrac{mg}{q}$

$E=\dfrac{0.0021\ kg\times 9.8\ m/s^2}{-3\times 10^{-9}\ C}$

$E=6860000\ N/C$

$E=6.86\times 10^6\ N/C$

Hence, the magnitude of electric field that balances its weight is $6.86\times 10^6\ N/C$ . Hence, this is the required solution.

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2 years ago

Que relación encontramos entre el modelo atomico de borh y la tabla periódica

mash [69]

Answer:

Explanation:

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2 years ago