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3 years ago

Number of waves that pass a given point in one second

Physics

1 answer:

Studentka2010 [4]3 years ago

5 0

number of waves that pass a given point in one second is called frequency..

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A wire has a cross sectional area of 4.00 mm2 and is stretched by 0.100 mm by a certain force. How far will a wire of the same m

Nina [5.8K]

Answer: $0.05\ mm$

Explanation:

Given

Cross-sectional area of wire $A_1=4\ mm^2$

Extension of wire $\delta l=0.1\ mm$

Extension in a wire is given by

$\Rightarrow \delta l=\dfrac{FL}{AE}$

where, $E=\text{Youngs modulus}$

$\Rightarrow \delta_1=\dfrac{FL}{A_1E}\quad \ldots(i)$

for same force, length and material

$\Rightarrow \delta_2=\dfrac{FL}{A_2E}\quad \ldots(ii)$

Divide (i) and (ii)

$\Rightarrow \dfrac{0.1}{\delta_2}=\dfrac{A_2}{A_1}\\\\\Rightarrow \delta_2=0.1\times \dfrac{4}{8}\\\\\Rightarrow \delta_2=0.05\ mm$

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2 years ago

What is the difference between a neap and a spring tide in terms of size?

Mamont248 [21]

Spring tides occur when the moon is either new or full, and the sun, the moon, and the Earth are aligned. ... neap tide- A tide in which the difference between high and low tide is the least. Neap tides occur twice a month when the sun and moon are at right angles to the Earth.

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3 years ago

Two coils close to each other have a mutual inductance of 32 mH. If the current in one coil decays according to I=I0e−αt, where

fiasKO [112]

The emf induced in the second coil is given by:

V = -M(di/dt)

V = emf, M = mutual indutance, di/dt = change of current in the first coil over time

The current in the first coil is given by:

i = i₀ $e^{-at}$

i₀ = 5.0A, a = 2.0×10³s⁻¹

i = 5.0e^(-2.0×10³t)

Calculate di/dt by differentiating i with respect to t.

di/dt = -1.0×10⁴e^(-2.0×10³t)

Calculate a general formula for V. Givens:

M = 32×10⁻³H, di/dt = -1.0×10⁴e^(-2.0×10³t)

Plug in and solve for V:

V = -32×10⁻³(-1.0×10⁴e^(-2.0×10³t))

V = 320e^(-2.0×10³t)

We want to find the induced emf right after the current starts to decay. Plug in t = 0s:

V = 320e^(-2.0×10³(0))

V = 320e^0

V = 320 volts

We want to find the induced emf at t = 1.0×10⁻³s:

V = 320e^(-2.0×10³(1.0×10⁻³))

V = 43 volts

3 0

2 years ago

Please help with science! Describe the magnetic field in the vicinity of a bar magnet.

Katen [24]

The magnetic field is described mathematically as a vector field. This vector field can be plotted directly as a set of many vectors drawn on a grid. Each vector points in the direction that a compass would point and has length dependent on the strength of the magnetic force.

7 0

2 years ago

Two point charges are separated by 10 cm, with an attractive force between them of 15 N. Find the force between them when they a

suter [353]

Answer:

(a) the force is 8.876 N

(b) the magnitude of each charge is 4.085 μC

Explanation:

Part (a)

Given;

coulomb's constant, K = 8.99 x 10⁹ N.m²/C²

distance between two charges, r = 10 cm = 0.1 m

force between the two charges, F = 15 N

when the distance between the charges changes to 13 cm (0.13 m)

force between the two charges, F = ?

Apply Coulomb's law;

$F = \frac{Kq_1q_2}{r^2} \\\\let \ Kq_1q_2 = C\\\\F =\frac{C}{r^2} \\\\C = Fr^2\\\\F_1r_1^2 = F_2r_2^2\\\\F_2 =\frac{F_1r_1^2}{r_2^2} \\\\F_2 = \frac{15*0.1^2}{0.13^2} \\\\F_2 = 8.876 \ N$

Part (b)

the magnitude of each charge, if they have equal magnitude

$F = \frac{KQ^2}{r^2}$

where;

F is the force between the charges

K is Coulomb's constant

Q is the charge

r is the distance between the charges

$F = \frac{KQ^2}{r^2} \\\\Q = \sqrt{\frac{Fr^2}{K} } \\\\Q = \sqrt{\frac{15*(0.1)^2}{8.99*10^9} } = 4.085 *10^{-6} \ C\\\\Q = 4.085 \ \mu C$

4 0

3 years ago