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shepuryov [24]
3 years ago
7

What is the equation for frequency, wavelength, and speed of a wave?

Physics
1 answer:
Nastasia [14]3 years ago
7 0

Answer:

       λ = v/f

Explanation:

frequency=f

wavelength = λ

speed of a wave=v

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Sergeeva-Olga [200]

Answer:

(a) 32.5 kgm/s

(b) 32.5 Ns

(c) 10.8 N

Explanation:

The change in momentum can be calculated from the definition of linear momentum:

\Delta p=\Delta (mv)= m \Delta v\\\\\Delta p= (5.00kg)(9.75m/s-3.25m/s)\\\\\Delta p=32.5kgm/s

Then, the change in momentum of the body is of 32.5 kgm/s (a).

Now, from the impulse-momentum theorem, we know that the change in momentum of a body \Delta p is equal to the impulse I exerted to it. So, the impulse produced by the force equals 32.5 kgm/s (or 32.5 Ns) (b).

Finally, since we know the value of the impulse and the interval of time, we can easily solve for the magnitude of the force:

I=F\Delta t\\\\\implies F=\frac{I}{\Delta t}\\\\F=\frac{32.5Ns}{3.0s}\\\\F=10.8N

It means that the magnitude of the force is of 10.8N (c).

8 0
3 years ago
Read 2 more answers
In a simplified model of a hydrogen atom, the electron moves around the proton nucleus in a circular orbit of radius 0.53×10−10m
Ksenya-84 [330]

Answer

given,

radius of the circular orbit, r = 0.53 x 10⁻¹⁰ m

mass of electron, M = 9.11 x 10⁻³¹ Kg

charge of electron, q₁ = 1.6 x 10⁻¹⁹ C

                                q₂ = 1.6 x 10⁻¹⁹ C

we know, force between two charges

F = \dfrac{kq_1q_2}{r^2}

F = \dfrac{9\times 10^9\times (1.6\times 10^{-19})^2}{(0.53\times 10^{-10})^2}

  F = 8.20 x 10⁻⁸ N

b) using newton's second law

F = m a

m a =  8.20 x 10⁻⁸

a =\dfrac{8.20\times 10^{-8}}{9.11\times 10^{-31}}

    a = 9 x 10²² m/s²

c) speed of the electron

 a =\dfrac{v^2}{r}

 9\times 10^{22} =\dfrac{v^2}{0.53\times 10^{-10}}

   v² = 4.77 x 10¹²

  v = 2.18 x 10⁶ m/s

d) the period of the circular motion.

    T=\dfrac{2\pi}{\omega}

    T=\dfrac{2\pi r}{v}

    T=\dfrac{2\pi\times 0.53\times 10^{-10}}{2.18\times 10^6}

          T = 1.53 x 10⁻¹⁶ s

8 0
3 years ago
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can someone please please please help me find the spring force constant thank you thank you thank you thank you
Levart [38]

Answer:

spring force frictonal

3 0
3 years ago
If the work required to stretch a spring 1 ft beyond its natural length is 12 ft-lb, how much work is needed to stretch the same
Angelina_Jolie [31]

Answer:W=\frac{3}{4} ft-lb

Explanation:

Given

Work required to stretch 1 ft is 12 ft-lb

and we have to find work required to stretch 3 in.

i.e. \frac{1}{4} ft

12=\frac{1}{2}K\left ( 1\right )^2 ------(1)

W=\frac{1}{2}k\left ( \frac{1}{4}\right )^2-----(2)

divide (1)&(2)

\frac{12}{W}=\left ( \frac{4}{1}\right )^2

W=\frac{12}{4\times 4}

W=\frac{3}{4} ft-lb

6 0
3 years ago
Which one of these is a chemical change?
satela [25.4K]
Burning a log because you are turning the log into ash from wood.
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3 years ago
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