Ask question

Ask question

All categories

2 years ago

13

A student solving a physics problem to find the unknown has applied physics principles and obtained the expression: μkmgcosθ=mgs

inθ−ma, where g=9.80meter/second2, a=3.60meter/second2, θ=27.0∘, and m is not given. Which of the following represents a simplified expression for μk?
tanθ− ag
To avoid making mistakes, the expression should not be simplified until the numerical values are substituted.
gsinθ−agcosθ
The single equation has two unknowns and cannot be solved with the information given.

1 answer:

BaLLatris [955]2 years ago

4 0

Answer:

$\mu_k=\dfrac{gsin\theta -a}{gcos\theta}$

Explanation:

g = Acceleration due to gravity = 9.80 m/s²

a = Acceleration= 3.6 m/s²

$\theta$ = Angle = 27°

The equation is

$\mu_kmgcos\theta=mgsin\theta -ma$

Mass gets cancelled

$\\\Rightarrow \mu_kgcos\theta=gsin\theta -a$

Rearranging for $\mu_k$

$\\\Rightarrow \mu_k=\dfrac{gsin\theta -a}{gcos\theta}$

The simplified expression is

$\mathbf{\mu_k=\dfrac{gsin\theta -a}{gcos\theta}}$

*the options are incomplete. The above answer is the required solution

You might be interested in

2.What is the weight of an object that has a mass of 5 kg?<br>

Artemon [7]

Answer:

5 kg

or 11 lbs

Explanation:

4 0

2 years ago

I got this information for a lab but I don't know how to do the hypothesis and the conclusion please can you guys help me with i

pochemuha

Answer:

A hypothesis is what you think will happen.

A conclusion is the results of an experiment summarized.

Hope this helps.

8 0

3 years ago

X-rays with an energy of 300 keV undergo Compton scattering from a target. If the scattered rays are detected at 30 relative to

lys-0071 [83]

Answer:

a) $\Delta \lambda = \lambda' -\lambda_o = \frac{h}{m_e c} (1-cos \theta)$

For this case we know the following values:

$h = 6.63 x10^{-34} Js$

$m_e = 9.109 x10^{-31} kg$

$c = 3x10^8 m/s$

$\theta = 37$

So then if we replace we got:

$\Delta \lambda = \frac{6.63x10^{-34} Js}{9.109 x10^{-31} kg *3x10^8 m/s} (1-cos 37) = 4.885x10^{-13} m * \frac{1m}{1x10^{-15} m}= 488.54 fm$

b) $\lambda_0 = \frac{hc}{E_0}$

With $E_0 = 300 k eV= 300000 eV$

And replacing we have:

$\lambda_0 = \frac{1240 x10^{-9} eV m}{300000eV}=4.13 x10^{-12}m = 4.12 pm$

And then the scattered wavelength is given by:

$\lambda ' = \lambda_0 + \Delta \lambda = 4.13 + 0.489 pm = 4.619 pm$

And the energy of the scattered photon is given by:

$E' = \frac{hc}{\lambda'}= \frac{1240x10^{-9} eVm}{4.619x10^{-12} m}=268456.37 eV - 268.46 keV$

c) $E_f = E_0 -E' = 300 -268.456 kev = 31.544 keV$

Explanation

Part a

For this case we can use the Compton shift equation given by:

$\Delta \lambda = \lambda' -\lambda_0 = \frac{h}{m_e c} (1-cos \theta)$

For this case we know the following values:

$h = 6.63 x10^{-34} Js$

$m_e = 9.109 x10^{-31} kg$

$c = 3x10^8 m/s$

$\theta = 37$

So then if we replace we got:

$\Delta \lambda = \frac{6.63x10^{-34} Js}{9.109 x10^{-31} kg *3x10^8 m/s} (1-cos 37) = 4.885x10^{-13} m * \frac{1m}{1x10^{-15} m}= 488.54 fm$

Part b

For this cas we can calculate the wavelength of the phton with this formula:

$\lambda_0 = \frac{hc}{E_0}$

With $E_0 = 300 k eV= 300000 eV$

And replacing we have:

$\lambda_0 = \frac{1240 x10^{-9} eV m}{300000eV}=4.13 x10^{-12}m = 4.12 pm$

And then the scattered wavelength is given by:

$\lambda ' = \lambda_0 + \Delta \lambda = 4.13 + 0.489 pm = 4.619 pm$

And the energy of the scattered photon is given by:

$E' = \frac{hc}{\lambda'}= \frac{1240x10^{-9} eVm}{4.619x10^{-12} m}=268456.37 eV - 268.46 keV$

Part c

For this case we know that all the neergy lost by the photon neds to go into the recoiling electron so then we have this:

$E_f = E_0 -E' = 300 -268.456 kev = 31.544 keV$

3 0

2 years ago

Read 2 more answers

A 7 ft tall person is walking away from a 20 ft tall lamppost at a rate of 5 ft/sec. Assume the scenario can be modeled with rig

aleksandr82 [10.1K]

Answer:

The rate of change of the shadow length of a person is 2.692 ft/s

Solution:

As per the question:

Height of a person, H = 20 ft

Height of a person, h = 7 ft

Rate = 5 ft/s

Now,

From Fig.1:

b = person's distance from the lamp post

a = shadow length

Also, from the similarity of the triangles, we can write:

$\frac{a + b}{20} = \frac{a}{7}$

$a = \frac{7}{13}b$

Differentiating the above eqn w.r.t t:

$\frac{da}{dt} = \frac{7}{13}.\frac{db}{dt}$

Now, we know that:

Rate = $\frac{db}{dt} = 5\ ft/s$

Thus

$\frac{da}{dt} = \frac{7}{13}.\times 5 = 2.692\ ft/s$

5 0

2 years ago

Which type of mountain forms because of two oceanic plates colliding?

Snowcat [4.5K]

<span>''Aelutian'' is the type of mountain that forms.</span>

8 0

3 years ago

Read 2 more answers