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3 years ago

An electromagnetic wave transmits

Physics

1 answer:

pishuonlain [190]3 years ago

4 0

Answer:

Explanation:

I think so

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The block A and attached rod have a combined mass of 50 kg and are confined to move along the guide under the action of the 796-

Luba_88 [7]

Answer:

The bending moment is 459.16 N.m

Explanation:

From the given information;

Let's assume that the angle is 66°

Then, the free body diagram is draw and attached in the file below.

Now, the calculation of the acceleration from the first part of the free body diagram is:

$\sum F_x = ma_x \\ \\ 796 - 50(9.81) sin 66=50a \\ \\ 796 - 448.094 = 50 a \\ \\ a = \dfrac{347.906}{50} \\ \\ a = 6.96 \ m/s^2$

Bending moment M:

From the second part of the diagram:

$\sum M_B = mad \\ \\ M - (15 \times 9.81) (1.5) = (25 \times 6.96)(1.5 sin 66) \\ \\ M - 220.725 = 238.435 \\ \\ M = 238.435 + 220.725 \\ \\ \mathbf{M = 459.16 \ N.m}$

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3 years ago

An object 4 cm high is placed 20 cm in front of a convex lens of focal length 12 cm. What is the position and height of the imag

aleksandrvk [35]

It would be 12cm

hope this helps

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3 years ago

What is the energy of the photon that could cause (a) an electronic transition from then= 4 state to the n 5 state of hydrogen a

Fiesta28 [93]

Answer:

a) $E_photon =0.306 eV$

b) $E_photon =0.166 eV$

Explanation:

The energy of the photon (E) for $n^th$ orbit of the hydrogen atom is given as:

$E_photon = E_o(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}})$

where,

$E_o$ = 13.6 eV

n = orbit

a) Now for the transition from n = 4 to n = 5

$E_photon =13.6(\frac{1}{4^{2}}-\frac{1}{5^{2}})$

$E_photon =0.306 eV$

b) Now for the transition from n = 5 to n = 6

$E_photon =13.6(\frac{1}{5^{2}}-\frac{1}{6^{2}})$

$E_photon =0.166 eV$

7 0

3 years ago

What is Aerodynamic purpose in vhicle systems?

BigorU [14]

Aerodynamic is relating to reduces the drag from air moving past

8 0

2 years ago

The MRI (Magnetic Resonance Imaging) body scanners used in hospitals operate at a frequency of 400 MHz (M = 106 ). Calculate the

horrorfan [7]

Answer:

Energy, $E=2.65\times 10^{-25}\ J$

Explanation:

It is given that,

The MRI (Magnetic Resonance Imaging) body scanners used in hospitals operate at a frequency of 400 MHz,

$\nu=400\ MHz=400\times 10^6\ Hz=4\times 10^8\ Hz$

We need to find the energy for a photon having this frequency. The energy of a photon is given by :

$E=h\nu$

$E=6.63\times 10^{-34}\ J-s\times 4\times 10^8\ Hz$

$E=2.65\times 10^{-25}\ J$

So, the energy of the photon is $2.65\times 10^{-25}\ J$ . Hence, this is the required solution.

6 0

3 years ago