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3 years ago

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Someone help me solve this please and explain

1 answer:

vfiekz [6]3 years ago

8 0

Answer:

i think you minus The weights I think

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A thin, circular coil of wire of radius r consists of 100 turns of wire carrying a conventional current of I in the counter-cloc

CaHeK987 [17]

Answer:

Magnitude = B = (μ₀*N*I /2R)

Direction= The direction of the magnetic field will be find out by the right hand rule. We will curve the finger of right hand in the direction of the current and our thumb will show the direction of the magnetic field. which will be out of the screen. That is + z direction.

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3 years ago

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two identical springs of spring constant 7580 N/m are attached to a block of mass 0.245 kg. What is the frequency of oscillation

ad-work [718]

The frequency of oscillation on the frictionless floor is 28 Hz.

<h3>Frequency of the simple harmonic motion</h3>

The frequency of the oscillation is calculated as follows;

f = (1/2π)(√k/m)

where;

k is the spring constant
m is mass of the block

f = (1/2π)(√7580/0.245)

f = 28 Hz

Thus, the frequency of oscillation on the frictionless floor is 28 Hz.

Learn more about frequency here: brainly.com/question/10728818

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1 year ago

What is one main difference between animal cells and plant cells?

vekshin1

Answer:

No cell wall

Explanation:

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2 years ago

Two charged concentric spherical shells have radii of 11.0 cm and 14.0 cm. The charge on the inner shell is 3.50 ✕ 10−8 C and th

Sergio039 [100]

Answer:

The magnitude of the electric field are $2.38\times10^{4}\ N/C$ and $1.09\times10^{4}\ N/C$

Explanation:

Given that,

Radius of inner shell = 11.0 cm

Radius of outer shell = 14.0 cm

Charge on inner shell $q_{inn}=3.50\times10^{-8}\ C$

Charge on outer shell $q_{out}=1.60\times10^{-8}\ C$

Suppose, at r = 11.5 cm and at r = 20.5 cm

We need to calculate the magnitude of the electric field at r = 11.5 cm

Using formula of electric field

$E=\dfrac{kq}{r^2}$

Where, q = charge

k = constant

r = distance

Put the value into the formula

$E=\dfrac{9\times10^{9}\times3.50\times10^{-8}}{(11.5\times10^{-2})^2}$

$E=2.38\times10^{4}\ N/C$

The total charge enclosed by a radial distance 20.5 cm

The total charge is

$q=q_{inn}+q_{out}$

Put the value into the formula

$q=3.50\times10^{-8}+1.60\times10^{-8}$

$q=5.1\times10^{-8}\ C$

We need to calculate the magnitude of the electric field at r = 20.5 cm

Using formula of electric field

$E=\dfrac{kq}{r^2}$

Put the value into the formula

$E=\dfrac{9\times10^{9}\times5.1\times10^{-8}}{(20.5\times10^{-2})^2}$

$E=1.09\times10^{4}\ N/C$

Hence, The magnitude of the electric field are $2.38\times10^{4}\ N/C$ and $1.09\times10^{4}\ N/C$

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2 years ago

The __________ pedal in a stick shift vehicle enables a driver to shift gears.

mamaluj [8]

The answer is C. Clutch

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3 years ago

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