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givi [52]
3 years ago
14

This is the number of complete movements of a wave per second.

Physics
1 answer:
Alona [7]3 years ago
5 0

A unit for measuring frequency, equal to one cycle per second. If a sound wave has a frequency of 20,000 Hz, this means that 20,000 waves are passing through a given point during the interval of one second.

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A spring gun is made by compressing a spring in a tube and then latching the spring at
inn [45]

Answer:

a)v=13.2171\,m.s^{-1}

b)H=8.9605\,m

Explanation:

Given:

mass of bullet, m=4.97\times 10^{-3}\,kg

compression of the spring, \Delta x=0.0476\,m

force required for the given compression, F=9.12 \,N

(a)

We know

F=m.a

where:

a= acceleration

9.12=4.97\times 10^{-3}\times a

a\approx 1835\,m.s^{-2}\\

we have:

initial velocity,u=0\,m.s^{-1}

Using the eq. of motion:

v^2=u^2+2a.\Delta x

where:

v= final velocity after the separation of spring with the bullet.

v^2= 0^2+2\times 1835\times 0.0476

v=13.2171\,m.s^{-1}

(b)

Now, in vertical direction we take the above velocity as the initial velocity "u"

so,

u=13.2171\,m.s^{-1}

∵At maximum height the final velocity will be zero

v=0\,m.s^{-1}

Using the equation of motion:

v^2=u^2-2g.h

where:

h= height

g= acceleration due to gravity

0^2=13.2171^2-2\times 9.8\times h

h=8.9129\,m

is the height from the release position of the spring.

So, the height from the latched position be:

H=h+\Delta x

H=8.9129+0.0476

H=8.9605\,m

4 0
3 years ago
How to calculate sound of an echo ​
bonufazy [111]

by an echo meter

please flw me and thank my answers

#Genius kudi

5 0
2 years ago
In the 1920s what did Edmund hubble notice about the galaxies
zlopas [31]
Hubble noticed that the galaxies were moving away from us, which meant the universe was expanding.

This is why constellations change over time. In some years, the Big Dipper won't actually look like a dipper anymore.
5 0
3 years ago
A dielectric-filled parallel-plate capacitor has plate area A = 30.0 cm2 , plate separation d = 9.00 mm and dielectric constant
PIT_PIT [208]

Answer:

9.96\cdot 10^{-10}J

Explanation:

The capacitance of the parallel-plate capacitor is given by

C=\epsilon_0 k \frac{A}{d}

where

ϵ0 = 8.85x10-12 C2/N.m2 is the vacuum permittivity

k = 3.00 is the dielectric constant

A=30.0 cm^2 = 30.0\cdot 10^{-4}m^2 is the area of the plates

d = 9.00 mm = 0.009 m is the separation between the plates

Substituting,

C=(8.85\cdot 10^{-12}F/m)(3.00 ) \frac{30.0\cdot 10^{-4} m^2}{0.009 m}=8.85\cdot 10^{-12} F

Now we can calculate the energy of the capacitor, given by:

U=\frac{1}{2}CV^2

where

C is the capacitance

V = 15.0 V is the potential difference

Substituting,

U=\frac{1}{2}(8.85\cdot 10^{-12}F)(15.0 V)^2=9.96\cdot 10^{-10}J

4 0
3 years ago
What is the longest wavelength of light that will emit electrons from a metal whose work function is 3.70 eV
SSSSS [86.1K]

Answer:

h f = Wf + K

where the total energy available is h f, Wf is the work function or the work needed to remove the electron and K is the kinetic energy of the removed electron

If K = zero then hf = Wf

Wf = h f = h c / λ    or

λ = h c / Wf = 6.63E-34 * 3.0E8 / (3.7 * 1.6E-19)

λ = 6.63 * 3 / (3.7 * 1.6) E-7 = 3.36E-7

This would be 3360 angstroms or 336 millimicrons

Visible light = 400-700 millimicrons

6 0
2 years ago
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