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Dima020 [189]
3 years ago
7

What is the frequency of an xray with a wavelength of 2x10-10m?

Physics
1 answer:
Fudgin [204]3 years ago
7 0

Answer:

1.5 x 10¹⁸hz

Explanation:

Given parameters:

Wavelength  = 2 x 10⁻¹⁰m

Unknown:

Frequency   = ?

Solution:

To find the frequency, use the expression below;

  V  = f x wavelength

V is the speed of light = 3 x 10⁸m/s

f is the frequency

Now;

 Insert the parameters

     3 x 10⁸  = 2 x 10⁻¹⁰ x frequency

   Wavelength = \frac{3 x 10^{8} }{2 x 10^{-10} }    = 1.5 x 10¹⁸hz

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How much work is done when 0.0050 c is moved through a potential difference of 9.0 v? use w = qv?
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Answer:

0.045 J

Explanation:

The work done on a charge moving through a potential difference is given by

W=q\Delta V

where

W is the work done

q is the charge

\Delta V is the potential difference

In this problem, we have

q = 0.0050 C is the charge

\Delta V=9.0 V is the potential difference

Using the formula, we find the work done:

W=(0.0050 C)(9.0 V)=0.045 J

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Answer:

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Explanation:

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What wave on the electromagnetic spectrum has the highest frequency?
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Answer:

Explanation:

Answer: Gamma rays

Gamma rays have the highest frequency.

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An electromagnetic wave requires no medium for its propagation.

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Higher frequency rays have the highest energies.

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Gamma radiations are obtained from the decay of the atomic nucleus.

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It has the smallest wavelength and highest energy.

The frequency of gamma rays is more than 10^19 cycles per second and wavelength less than 100 picometers.

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An electron and a proton, starting from rest, are accelerated through an electric potential difference of the same magnitude. in
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A large crate with mass m rests on a horizontal floor. The static and kinetic coefficients of friction between the crate and the
Mariana [72]

a) F=\frac{\mu_k mg}{cos \theta - \mu_k sin \theta}

Here the crate is moving at constant velocity, so no acceleration:

a = 0

Let's analyze the forces acting along the horizontal and vertical direction.

- Vertical direction: the equation of the forces is

R-Fsin \theta - mg = 0 (1)

where

R is the normal reaction of the floor (upward)

F sin \theta is the component of the force F in the vertical direction (downward)

mg is the weight of the crate (downward)

- Horizontal direction: the equation of the forces is

F cos \theta - \mu_k R = 0 (2)

where

F cos \theta is the horizontal component of the force F (forward)

\mu_k R is the force of friction (backward)

From (1) we get

R=Fsin \theta +mg

And substituting into (2)

F cos \theta - \mu_k (Fsin \theta +mg) = 0\\F cos \theta -\mu _k F sin \theta = \mu_k mg\\F(cos \theta - \mu_k sin \theta) = \mu_k mg\\F=\frac{\mu_k mg}{cos \theta - \mu_k sin \theta}

b) \mu_s=cot(\theta)

In this second case, the crate is still at rest, so we have to consider the static force of friction, not the kinetic one.

The equations of the forces will be:

R-Fsin \theta - mg = 0 (1)

F cos \theta - \mu_s R = 0 (2)

In this second case, we want to find the critical value of \mu_s such that the woman cannot start the crate: this means that the force of friction must be at least equal to the component of the force pushing on the horizontal direction, F cos \theta.

Therefore, using the same procedure as before,

R=Fsin \theta +mg

F cos \theta - \mu_s (Fsin \theta +mg) = 0

And solving for \mu_s,

F cos \theta = \mu_s (Fsin \theta +mg) \\\mu_s = \frac{F cos \theta}{F sin \theta + mg}

Now we analyze the expression that we found. We notice that if the force applied F is very large, F sin \theta >> mg, therefore we can rewrite the expression as

\mu_s \sim \frac{F cos \theta}{F sin \theta}\\\mu_s=cot(\theta)

So, this is the critical value of the coefficient of static friction.

8 0
3 years ago
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