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Aloiza [94]
4 years ago
15

What is the relationship between the strength of an

Physics
1 answer:
Anastaziya [24]4 years ago
3 0

Answer:

As the number of turns in the coil increases, the strength  of the electromagnet increases.

Explanation:

When current flows through a coil the coil behaves as an electromagnet. The strength of electromagnet depend the amount of current, no of turns of coil and the core of coil.

B=μ₀ N I

μ₀ = permeability of the core

N = Number of turns of the coil

I = Current flowing through the coil

Increasing the current and number of coils increase the strength of electromagnet.

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Atmospheric pressure decreases with increment in height.<br> give reason​
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pressure=height × density×acc due to gravity

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pressire is directly proportional to height hence it decreses with decrease in height

here air column height is measured upside down so decreases witn increment

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3 years ago
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HELP!!!! 20 pointsss
PSYCHO15rus [73]

Answer:

The answer will be the first one.

Explanation:

Divide

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5. A sled with no initial velocity accelerates at a rate of 5.0 m/s2 down a hill. How
ad-work [718]

Answer:

60 m/s

Explanation:

a =  \frac{vf - vi}{t}

5.0 =  \frac{vf - 0}{12}

5.0 \times 12 = vf

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3 years ago
The direction of an electric field is always in the direction _______________ would naturally move.
Masja [62]

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Im not really sure but Id say weather .

Explanation:

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3 years ago
In lab, your instructor generates a standing wave using a thin string of length L = 1.65 m fixed at both ends. You are told that
erik [133]

Answer:

On the standing waves on a string, the first antinode is one-fourth of a wavelength away from the end. This means

\frac{\lambda}{4} = 0.275~m\\\lambda = 1.1~m

This means that the relation between the wavelength and the length of the string is

3\lambda/2 = L

By definition, this standing wave is at the third harmonic, n = 3.

Furthermore, the standing wave equation is as follows:

y(x,t) = (A\sin(kx))\sin(\omega t) = A\sin(\frac{\omega}{v}x)\sin(\omega t) = A\sin(\frac{2\pi f}{v}x)\sin(2\pi ft) = A\sin(\frac{2\pi}{\lambda}x)\sin(\frac{2\pi v}{\lambda}t) = (2.45\times 10^{-3})\sin(5.7x)\sin(59.94t)

The bead is placed on x = 0.138 m. The maximum velocity is where the derivative of the velocity function equals to zero.

v_y(x,t) = \frac{dy(x,t)}{dt} = \omega A\sin(kx)\cos(\omega t)\\a_y(x,t) = \frac{dv(x,t)}{dt} = -\omega^2A\sin(kx)\sin(\omega t)

a_y(x,t) = -(59.94)^2(2.45\times 10^{-3})\sin((5.7)(0.138))\sin(59.94t) = 0

For this equation to be equal to zero, sin(59.94t) = 0. So,

59.94t = \pi\\t = \pi/59.94 = 0.0524~s

This is the time when the velocity is maximum. So, the maximum velocity can be found by plugging this time into the velocity function:

v_y(x=0.138,t=0.0524) = (59.94)(2.45\times 10^{-3})\sin((5.7)(0.138))\cos((59.94)(0.0524)) = 0.002~m/s

4 0
3 years ago
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