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Alenkasestr [34]

3 years ago

15

(PHYSC) HELPPPP Individuals suffering from somatoform disorders often see several doctors about their symptoms because _________

_.
A.
doctors often refer their patients for a second opinion
B.
their symptoms often fall under several specialties
C.
doctors get frustrated with them and refuse to treat them
D.
they are looking to find a doctor to give them a diagnosis

2 answers:

Inessa05 [86]3 years ago

8 0

D ! ! ! ! ! ! ! ! !

Sonbull [250]3 years ago

4 0

Answer:

D.

they are looking to find a doctor to give them a diagnosis

Explanation:

cause I had this answer

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The brakes of an automobile are suddenly applied at the instant when its velocity is 20 m/s. If the automobile comes to a stop a

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Data:

Time (t) = 5 seconds
Final Velocity (v) = 0 [as the car comes to rest]
<span>Initial velocity (u) = 20 m/s
</span>Acceleration = ? <span>(Negative because it is braking)</span>

$Acceleration = \frac{v-u}{t}$
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$\boxed{Acceleration = - 4m/s^2}$

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3 years ago

(n = 1.33). When visible light (400nm – 700 nm) shines normally on the film an observed from above, the bright reflected light l

Yuki888 [10]

Complete Question

A thin film in air is made by putting a thin layer of acetone (n = 1.25) on a layer of water

(n = 1.33). When visible light (400 nm – 700 nm) shines normally on the film an observed from above, the bright reflected light looks kind of “purplish” with red light of wavelength 650 nm mixed some blue giving it that appearance and “greenish” light of wavelength 520 nm is completely destroyed. Determine the thickness of the acetone film. (you only need to worry about the red and green wavelengths, not the blue)

Answer:

The thickness of acetone is $d = 104 nm$

Explanation:

A diagram showing this process is shown on the first uploaded image

From the question we are told that

The refractive index of acetone is $n_a =1.25$

The refractive index of water is $n_w = 1.33$

The wavelength of the reflected light is $\lambda_r = 650nm = 650 *10^{-9}m$

The wavelength of the destroyed light is $\lambda_g = 520nm = 520 *10^{-9}m$

Looking at the given data we can see that the

$n_a < n_w$

This means that the light which the acetone-water layer would reflect will have a phase shift of $\pi$

Again this make us to understand that the light reflected at the acetone layer will also have a phase shift of $\pi$

Since they would be having the same phase shift the two light would interfere

For interference the condition for minima is mathematically represented as

$2 n_a d = (m + \frac{1}{2} ) \lambda_g$

Where d is the thickness of acetone

$d = \frac{\lambda_g}{4 n_a}$

Substituting values

$d = \frac{520 *10^{-9}}{4 * 1.25}$

$d = 104 *10^{-9}m$

$d = 104 nm$

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3 years ago

As shown in the diagram, two massless wires connect a rotating pole to a sphere of mass 7.2 kg. The sphere revolves in a horizon

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Use Fnet=ma. Apply it to the vertical components of all the forces acting on the ball and since tension is the same on both wires you should be able to factor it out and solve for it. Picture below

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4 years ago

There are two particles with charges +Q and +q. The electric force applied to one of the charges is 10 N. Now, +Q is replaced by

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Answer:

F = 10 N

Explanation:

Assuming that we can treat to both particles as point charges, the magnitude of the force that one charge exerts on the other, must obey Coulomb's Law.
The expression for this force, applied to the charges +Q and +q, separated by a distance d, is as follows:

$F_{0} = \frac{k*q*Q}{d^{2}}$

If we replace +Q by +4Q, and d by 2*d, the new expression for the magnitude of the force is as follows:

$F_{1} = \frac{k*q*4*Q}{(2*d)^{2} } = \frac{4}{4} *\frac{k*q*Q}{d^{2} } = F_{0} = 10 N$

As it can be seen, the force will be as same as before making changes, so F= 10 N.

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3 years ago

at certain times the demand for electric energy is low and electric energy is used to pump water to a reservoir 45 m above the g

Readme [11.4K]

The mass of water that must be raised is $5.25\cdot 10^7 kg$

Explanation:

Since the process is 70% efficiency, the power in output to the turbine can be written as

$P_{out} = 0.70 P_{in}$

where $P_{in}$ is the power in input.

The power in input can be written as

$P_{in} = \frac{W}{t}$

where

W is the work done in lifting the water

t = 3 h = 10,800 s is the time elapsed

The work done in lifting the water is given by

$W=mgh$

where

m is the mass of water

$g=9.8 m/s^2$ is the acceleration of gravity

h = 45 m is the height at which the water is lifted

Combining the three equations together, we get:

$P_{out} = 0.70 \frac{mgh}{t}$

Where

$P_{out} = 150 MW = 150\cdot 10^6 W$

And solving for m, we find:

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Learn more about power:

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4 years ago