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3 years ago

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A diathermy machine, used in physiotherapy, generates electromagnetic radiation that gives the effect of "deep heat" when absorb

ed in tissue. If the radiation's frequency is 27.17 MHz, what is its wavelength?

1 answer:

zloy xaker [14]3 years ago

8 0

Answer:

The wavelength of the radiation is 11.04 m.

Explanation:

Given that,

The radiation's frequency is 27.17 MHz.

$1\ MHz=10^6\ Hz\\\\27.17\ Mhz=27.17\times 10^6\ Hz$

We need to find the wavelength of the radiation. It can be calculated using the formula of speed of electromagnetic wave as :

$c=f\lambda\\\\\lambda=\dfrac{c}{f}\\\\\lambda=\dfrac{3\times 10^8\ m/s}{27.17\times 10^6\ Hz}\\\\\lambda=11.04\ m$

So, the wavelength of the radiation is 11.04 m.

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A + 8.42 nC point charge and a - 3.75 nC point charge are 2.73 cm apart. What is the electric field strength at the midpoint bet

aev [14]

Given:

The charge q1 = 8.42 nC

The charge q2 = -3.75 nC

The distance between the charges is d = 2.73 cm

To find the electric field strength at the midpoint between the two charges.

Explanation:

The distance between the charges and the midpoint is d' =d/2

The electric field strength can be calculated by the formula

$E\text{ = }\frac{k|q|}{d^{\prime2}}$

The electric field strength at the midpoint due to the charge q1 will be

$\begin{gathered} E_1=\frac{9\times10^9\times8.42\times10^{-9}\text{ C}}{(\frac{2.73}{2}\times10^{-2})^2} \\ =\text{ 4.07}\times10^5\text{ V/m} \end{gathered}$

The electric field strength at the midpoint due to the charge q2 will be

$\begin{gathered} E_2=\frac{9\times10^9\times3.75\times10^{-9}}{(\frac{2.73}{2}\times10^{-2})^2} \\ =1.8\times10^5\text{ V/m} \end{gathered}$

The electric field strength at the midpoint between the charges will be

$\begin{gathered} E=E_1+E_2 \\ =4.07\times10^5+1.8\times10^5 \\ =\text{ 5.87}\times10^5\text{ V/m} \end{gathered}$

The electric field strength at the midpoint between the two charges is 5.87 x 10^(5) m.

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1 year ago

A sled is pulled up to the top of a hill. At thetop of the hill the sled is released from rest and allowed to coastdown the hill

Stolb23 [73]

Answer:

a.) the speed at the bottom is greater for the steeperhill

Explanation:

since the energy at the bottom of the steeper hilis greater

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

As we can see from above that v is higher when h ishigher.

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3 years ago

The electric field strength in the space between two closely spaced parallel disks is 1.0 105 N/C. This field is the result of t

balu736 [363]

Answer:

$D=2.996\times 10^{-2} m$

Explanation:

*Assume the parallel disks have equal diameters.

Given the electric strength as $1.0\times 10^5 N/C.$ transferring $3.9\times 10^9$ electrons, the disk's Area can be calculated using the formula:

$E=\frac{\eta}{\epsilon_o}=\frac{Q}{A\epsilon_o}\\\\A=\frac{Q}{E\epsilon_o}\\\\=\frac{(3.9\times 10^9)\times (1.6\times10^{-19})}{(1.0\times 10^5 )\times (8.85\times10^{-12})}\\\\A=7.0508\times 10^{-4} \ m^2$

#We now calculate the disks diameter:

$A=\pi(D/2)^2\\\\2\sqrt{\frac{A}{\pi}}=D\\\\=2\sqrt{7.0508\times 10^{-4}/\pi}\\\\D=2.996\times 10^{-2} \ m$

Hence, the diameter of the disks is $D=2.996\times 10^{-2} m$

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4 years ago

Choose the correct statement below that accurately describes the shear and normal stresses in a beam. A. Shear stresses are maxi

Scorpion4ik [409]

Answer:

A. Shear stresses are maximum at the neutral axis and normal stresses are maximum furthest from the neutral axis.

Explanation:

Normal stress :

Normal stress is defined as the stress or the restoring force that occurs on the plane when an external axial load is applied on it. For a beam the normal stress is maximum at the point furthest from the neutral axis and is zero at the neutral axis of the beam.

Shear stress :

Shear stress is a stress which occurs when the force acts on the surface of the member in a parallel direction. It changes the shape of the member. For a beam, the shear stress is maximum at the neutral axis.

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3 years ago

The following questions present a twist on the scenario above to test your understanding. Suppose another stone is thrown horizo

Ipatiy [6.2K]

The first part of the text is missing, you can find on google:

"A ball is thrown horizontally from the roof of a building 45 m. If it strikes the ground 56 m away, find the following values."

Let's now solve the different parts.

(a) 3.03 s

The time of flight can be found by analyzing the vertical motion only. The vertical displacement at time t is given by

$y(t) = h -\frac{1}{2}gt^2$

where

h = 45 m is the initial height

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

When y=0, the ball reaches the ground, so the time taken for this to happen can be found by substituting y=0 and solving for the time:

$0=h-\frac{1}{2}gt^2\\t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2(45)}{9.8}}=3.03 s$

(b) 18.5 m/s

For this part, we need to analyze the horizontal motion only, which is a uniform motion at constant speed.

The horizontal position is given by

$x=v_x t$

where

$v_x$ is the horizontal speed, which is constant

t is the time

At t = 3.03 s (time of flight), we know that the horizontal position is x = 56 m. By substituting these numbers and solving for vx, we find the horizontal speed:

$v_x = \frac{x}{t}=\frac{56}{3.03}=18.5 m/s$

The ball was thrown horizontally: this means that its initial vertical speed was zero, so 18.5 m/s was also its initial overall speed.

(c) 35.0 m/s at 58.1 degrees below the horizontal

At the impact, we know that the horizontal speed is still the same:

$v_x = 18.5 m/s$

we need to find the vertical velocity. This can be done by using the equation

$v_y = u_y -gt$

where

$u_y =0$ is the initial vertical velocity

g is the acceleration of gravity

t is the time

Substituting t = 3.03 s, we find the vertical velocity at the time of impact:

$v_y = -(9.8)(3.03)=-29.7 m/s$

So the magnitude of the velocity at the impact (so, the speed at the impact) is

$v=\sqrt{v_x^2+v_y^2}=\sqrt{18.5^2+(-29.7)^2}=35.0 m/s$

The angle instead can be found as:

$\theta=tan^{-1}(\frac{v_y}{v_x})=tan^{-1}(\frac{-29.7}{18.5})=-58.1^{\circ}$

so, 58.1 degrees below the horizontal.

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3 years ago