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2 years ago

What is the period of a wave that has a frequency of 300 hz?

Physics

1 answer:

Butoxors [25]2 years ago

3 0

Answer:

T = 0.003 s

(Period is written as T)

Explanation:

Period = time it takes for one wave to pass (measured in seconds)

frequency = number of cycles that occur in 1 second

(measured in Hz / hertz / 1 second)

Period : T

frequency : f

So, if we know that the frequency of a wave is 300 Hz, we can find the period of the wave from the relation between frequency and period

T = $\frac{1}{f}$ f = $\frac{1}{T}$

to find the period (T) of this wave, we need to plug in the frequency (f) of 300

T = $\frac{1}{300}$

T = 0.00333333333

So, the period of a wave that has a frequency of 300 Hz is 0.003 s

[the period/T of this wave is 0.003 s]

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Which of the following would be a valid method to increase the buoyant force acting on an object?

shepuryov [24]

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3 years ago

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How is the voltage V across the resistor related to the current I and the resistance R of the resistor? (Use I for current and R

VladimirAG [237]

Answer:

This relationship is explained by Ohm's law

Explanation:

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8 0

4 years ago

Not in book

umka2103 [35]

Answer:

$x=2.4365\ m$

and

$x=-1.4365\ m$

Explanation:

Given:

first charge, $q_1=5\times 10^{-3}\ C$

second charge, $q_2=3\times 10^{-3}\ C$

position of first charge, $x_1=-2\ m$

position of second charge, $x_2=-1\ m$

Now since there are only 2 charges and of the same sign so they repel each other. This repulsion will be zero at some point on the line joining the charges.

<u>Now, according to the condition, electric field will be zero where the effects of field due to both the charges is equal.</u>

$E_1=E_2$

since first charge is greater than the second charge so we may get a point to the right of the second charge and the distance between the two charges is 1 meter.

$\frac{1}{4\pi.\epsilon_0} \frac{q_1}{(r+1)^2} =\frac{1}{4\pi.\epsilon_0} \frac{q_2}{(r)^2}$

$\frac{5\times 10^{-3}}{(r+1)^2} = \frac{3\times 10^{-3}}{(r)^2}$

$3(r^2+1+2r)=5r^2$

$2r^2-6r-3=0$

$r=3.4365 \&\ r=-0.4365$

Since we have assumed that the we may get a point to the right of second charge so we calculate with respect to the origin.

$x=-1+3.4365=2.4365\ m$

and

$x=-1-0.4365=-1.4365\ m$

6 0

3 years ago

Before the experiment, the total momentum of the system is 2.5 kg m/s to the right and the kinetic energy is 5J. After the exper

finlep [7]

Answer:

Option (b) is correct.

Explanation:

Elastic collision is defined as a collision where the kinetic energy of the system remains same. Both linear momentum and kinetic energy are conserved in case of an elastic collision.

Inelastic collision is defined as a collision where kinetic energy of the system is not conserved whereas the linear momentum is conserved. This loss of kinetic energy may due to the conversion to thermal energy or sound energy or may be due to the deformation of the materials colliding with each other.

As given in the problem, before the collision, total momentum of the system is $2.5~Kg~m~s^{-1}$ and the kinetic energy is $5~J$ . After the collision, the total momentum of the system is $2.5~Kg~m~s^{-1}$ , but the kinetic energy is reduced to $4~J$ . So some amount of kinetic energy is lost during the collision.

Therefor the situation describes an inelastic collision (and it could NOT be elastic).

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3 years ago