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3 years ago

Speed=100m/sec Frequency=10 Hz Wavelength=?

Physics

1 answer:

elena-14-01-66 [18.8K]3 years ago

4 0

Answer:

Wavelength = 10 m

Explanation:

Given:

Speed = 100 m $s^{-1}$

Frequency = 10 Hz = 10 $s^{-1}$

To find : Wavelength = ?

We know that the relationship between wavelength λ, frequency f and speed v is given by the equation

v = fλ

Therefore wavelength λ = v/f

= 100 m $s^{-1}$ / 10 m $s^{-1}$

= 10 m

Hence wavelength = 10 m

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A fruit bat falls from the roof of a cave. We know that her potential energy was

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Answer:

v = 15.65 m/s

Explanation:

We use conservation of mechanical energy between initial (i) and final (f) states:

Pi + KEi = Pf + KEf

At the top of the cave at the instant the bat starts to fall, there is only potential energy since the bat's velocity is zero.

Pi = m g h = 600 J

and the KEi = 0 J (no velocity)

Knowing the height of the cave's roof (12.8 m) , we can find the mass of the bat:

m = 600 J / (g 12.5) = 4.9 kg

Using conservation of mechanical energy, the final state is:

Pf + KEf = 600 J

with Pf = 0 (just touching the ground)

KEf= 1/2 4.9 (v^2)

and we solve for the velocity:

600 J = 0 + 1/2 4.9 (v^2)

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3 years ago

Two children are pulling on opposite sides of a blanket. The brother is pulling with a force of 3 N. The sister is pulling with

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Let F1=Force exerted by the brother (+F1)
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3 years ago

Compare the wavelengths of an electron (mass = 9.11 × 10−31 kg) and a proton (mass = 1.67 × 10−27 kg), each having (a) a speed o

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Answer:

Part A:

The proton has a smaller wavelength than the electron.

$\lambda_{proton} = 6.05x10^{-14}m$ < $\lambda_{electron} = 1.10x10^{-10}m$

Part B:

The proton has a smaller wavelength than the electron.

$\lambda_{proton} = 1.29x10^{-13}m$ < $\lambda_{electron} = 5.525x10^{-12}m$

Explanation:

The wavelength of each particle can be determined by means of the De Broglie equation.

$\lambda = \frac{h}{p}$ (1)

Where h is the Planck's constant and p is the momentum.

$\lambda = \frac{h}{mv}$ (2)

Part A

Case for the electron:

$\lambda = \frac{6.624x10^{-34} J.s}{(9.11x10^{-31}Kg)(6.55x10^{6}m/s)}$

But $J = Kg.m^{2}/s^{2}$

$\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(9.11x10^{-31}Kg)(6.55x10^{6}m/s)}$

$\lambda = 1.10x10^{-10}m$

Case for the proton:

$\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(1.67x10^{-27}Kg)(6.55x10^{6}m/s)}$

$\lambda = 6.05x10^{-14}m$

Hence, the proton has a smaller wavelength than the electron.

<em>Part B </em>

For part b, the wavelength of the electron and proton for that energy will be determined.

First, it is necessary to find the velocity associated to that kinetic energy:

$KE = \frac{1}{2}mv^{2}$

$2KE = mv^{2}$

$v^{2} = \frac{2KE}{m}$

$v = \sqrt{\frac{2KE}{m}}$ (3)

Case for the electron:

$v = \sqrt{\frac{2(7.89x10^{-15}J)}{9.11x10^{-31}Kg}}$

but $1J = kg \cdot m^{2}/s^{2}$

$v = \sqrt{\frac{2(7.89x10^{-15}kg \cdot m^{2}/s^{2})}{9.11x10^{-31}Kg}}$

$v = 1.316x10^{8}m/s$

Then, equation 2 can be used:

$\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(9.11x10^{-31}Kg)(1.316x10^{8}m/s)}$

$\lambda = 5.525x10^{-12}m$

Case for the proton :

$v = \sqrt{\frac{2(7.89x10^{-15}J)}{1.67x10^{-27}Kg}}$

But $1J = kg \cdot m^{2}/s^{2}$

$v = \sqrt{\frac{2(7.89x10^{-15}kg \cdot m^{2}/s^{2})}{1.67x10^{-27}Kg}}$

$v = 3.07x10^{6}m/s$

Then, equation 2 can be used:

$\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(1.67x10^{-27}Kg)(3.07x10^{6}m/s)}$

$\lambda = 1.29x10^{-13}m$

Hence, the proton has a smaller wavelength than the electron.

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3 years ago

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Answer: Inertia!!

Explanation: I just completed the edg quiz and got that answer correct! Hope its not too late for you!

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