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2 years ago

6. Find the magnetic field strength at the centre of a solenoid with

Physics

1 answer:

inysia [295]2 years ago

8 0

Answer:

0.03 T

Explanation:

The magnetic field B at the center of a solenoid is given by B = μ₀Ni/L where

μ₀ = permeability of free space = 4π × 10⁻⁷ H/m, N = number of turns of solenoid = 5000 turns, i = current in solenoid = 5 A and L= length of solenoid. Since we are not given length of solenoid, let us assume it is 1 meter. So, L = 1 m

So, B = μ₀Ni/L

= 4π × 10⁻⁷ H/m × 5000 turns × 5 A/1m

= 4π × 10⁻⁷ H/m × 25000 A-turns/m

= 314159.27 × 10⁻⁷ T

= 0.031415927 T

≅ 0.03 T

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When an average force F is exerted over a certain distance on a shopping cart of mass m, its kinetic energy increases by 12mv2.

VMariaS [17]

Answer:

A) $d=\dfrac{1}{2F}mv^2$

B) $\Delta KE'=2\times \dfrac{1}{2}mv^2$

Explanation:

Given that

Force = F

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$\Delta KE=\dfrac{1}{2}mv^2$

we know that

Work done by all the forces =change in the kinetic energy

Lets distance = d

We know work done by force F

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$d=\dfrac{1}{2F}mv^2$

If the force become twice

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F'.d=ΔKE'

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Therefore the final kinetic energy will become the twice if the force become twice.

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3 years ago

How are the planets sizes related to their surface gravity

dexar [7]

Answer:

The surface gravity is inversely proportional to the square of the radius of the planet

Explanation:

The gravity at the surface of a planet is given by:

$g=\frac{GM}{R^2}$

where

G is the gravitational constant

M is the mass of the planet

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At the Earth's surface, the value of the surface gravity is approximately 9.81 m/s^2.

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2 years ago

Continuous sinusoidal perturbation Assume that the string is at rest and perfectly horizontal again, and we will restart the clo

Elena-2011 [213]

a) $3.14 \cdot 10^{-4} s$

b) See plot attached

c) 10.0 m

d) 0.500 cm

Explanation:

The position of the tip of the lever at time t is described by the equation:

$y(t)=(0.500 cm) sin[(2.00\cdot 10^4 s^{-1})t]$ (1)

The generic equation that describes a wave is

$y(t)=A sin (\frac{2\pi}{T} t)$ (2)

where

A is the amplitude of the wave

T is the period of the wave

t is the time

By comparing (1) and (2), we see that for the wave in this problem we have

$\frac{2\pi}{T}=2.00\cdot 10^4 s^{-1}$

Therefore, the period is

$T=\frac{2\pi}{2.00\cdot 10^4}=3.14 \cdot 10^{-4} s$

The sketch of the profile of the wave until t = 4T is shown in attachment.

A wave is described by a sinusoidal function: in this problem, the wave is described by a sine, therefore at t = 0 the displacement is zero, y = 0.

The wave than periodically repeats itself every period. In this sketch, we draw the wave over 4 periods, so until t = 4T.

The maximum displacement of the wave is given by the value of y when $sin(...)=1$ , and from eq(1), we see that this is equal to

y = 0.500 cm

So, this is the maximum displacement represented in the sketch.

When standing waves are produced in a string, the ends of the string act as they are nodes (points with zero displacement): therefore, the wavelength of a wave in a string is equal to twice the length of the string itself:

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