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3 years ago

In which direction does earth's magnetic field move?

Physics

1 answer:

Natasha2012 [34]3 years ago

4 0

Answer:

Explanation:

because the repelling force of gravity is vertical

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A velocity selector in a mass spectrometer uses a 0.150 T magnetic field. (a) What electric field strength (in volts per meter)

Alekssandra [29.7K]

Answer:

The electric field strength is $6.6\times10^{5}\ V/m$

Explanation:

Given that,

Magnetic field = 0.150 T

Speed $v= 4.40\times10^{6}\ m/s$

We need to calculate the electric field strength

Using formula of velocity

$v=\dfrac{E}{B}$

$E=v\times B$

Where, v = speed

B = magnetic field

Put the value into the formula

$E=4.40\times10^{6}\times0.150$

$E=660000\ V/m$

$E=6.6\times10^{5}\ V/m$

Hence, The electric field strength is $6.6\times10^{5}\ V/m$

4 0

3 years ago

The amplitude of a simple harmonic oscillator will be doubled by:a) doubling only the initial speedb) doubling the initial displ

OLEGan [10]

Answer:

When both initial speed and initial displacement is doubled then amplitude will be doubled.

Explanation:

Given that :- Amplitude of simple harmonic Oscillator is doubled.

So,

Formula of Simple harmonic oscillator is $X=A\sin\ (2\pi ft +\phi)$ ...........(1)

Where X = Position in (m,cm,km.....)

A = Amplitude in (m,cm,km.....)

F = Frequency in (Hz)

T = Time in (sec.)

Ф = Phase in (rad)

For initial displacement taking t=0 we get,

Initial displacement = $A\sin(\phi)$ .................(2)

Now taking equation (1) and differentiating it w.r.t to (t) we get

$\frac{dx}{dt} = 2\pi fA\cos\ (2\pi ft+\phi)$

$V= 2\pi fA\cos\ (2\pi ft+\phi)$

taking t=0 for initial speed then we get,

Initial speed = $2\pi fA\cos\phi$ ...............(3)

observing equation (2) & (3) that the initial displacement and initial speed depends on the Amplitude of the Oscillator.

Hence,

when both initial speed and displacement is doubled then amplitude will be doubled.

4 0

4 years ago

Please help this is physics!!!

larisa [96]

I’m pretty sure u have it right

8 0

3 years ago

A particle of mass 4.00 kg is attached to a spring with a force constant of 100 N/m. It is oscillating on a frictionless, horizo

zloy xaker [14]

Solution :

Given :

Mass attached to the spring = 4 kg

Mass dropped = 6 kg

Force constant = 100 N/m

Initial amplitude = 2 m

Therefore,

a). $v_{initial} = A w$

$= 2 \times \sqrt{\frac{100}{4}}$

= 10 m/s

Final velocity, v at equilibrium position, v = 5 m/s

Now, $\frac{1}{2}(4+4)5^2 = \frac{1}{2} kA'$

A' = amplitude = 1.4142 m

b). $T=2 \pi \sqrt{\frac{m}{k}}$

m' = 2m

Hence, $T'=\sqrt2 T$

c). $\frac{\frac{1}{2}(4+4)5^2 + \frac{1}{2}\times 4 \times 10^2}{\frac{1}{2} \times 4 \times 10^2}$

$=\frac{1}{2}$

Therefore, factor $=\frac{1}{2}$

Thus, the energy will change half times as the result of the collision.

7 0

3 years ago

Electrical energy can be transformed into other types of energy. We often experience this transformation of energy in our everyd

vitfil [10]

Well I’m not sure because you don’t have anything listed

3 0

4 years ago