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disa [49]
2 years ago
4

A photon of wavelength 18.0 pm is scattered through an angle of 120° by a stationary electron. What is the wavelength of the sca

ttered photon? (A) 19.2 pm (B) 20.4 pm (C) 21.6 pm (D) 22.9 pm
Physics
1 answer:
horsena [70]2 years ago
5 0

Answer:

(C) 21.6 pm

Explanation:

\lambda_i= Initial wavelength = 18 pm

\lambda_f= Final wavelength

mc² = Eloectron rest mass = 0.511 MeV

h = Planck's constant

c = Speed of light

hc = 1240 eV nm

\Delta \lambda=\lambda_f-\lambda_i=\frac{hc}{mc^2}(1-cos\theta)\\\Rightarrow \lambda_f-18=\frac{1240}{0.511}(1-cos120)\\\Rightarrow \lambda_f-18=3.63\\\Rightarrow \lambda_f=3.63+18\\\Rightarrow \lambda_f=21.63

∴ Wavelength of the scattered photon is 21.63 pm

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Object A with a mass of 500 kilograms hits stationary object B with a mass of 920 kilograms. If the collision is elastic, what h
Vika [28.1K]
In elastic collision, both the kinetic energy and momentum are conserved. Conservation means that both the kinetic energy and momentum will have the same values before and after elastic collision.

<span>As the object A has low mass than object B. Hence upon collision, object B moves forward, while object A will move backward. So option "C" is correct. </span>

5 0
2 years ago
If the briefcase hits the water 6.0 s later, what was the speed at which the helicopter was ascending?
vovikov84 [41]

Complete Question

In an action movie, the villain is rescued from the ocean by grabbing onto the ladder hanging from a helicopter. He is so intent on gripping the ladder that he lets go of his briefcase of counterfeit money when he is 130 m above the water. If the briefcase hits the water 6.0 s later, what was the speed at which the helicopter was ascending?

Answer:

The speed of the helicopter is u  =  7.73 \  m/s

Explanation:

From the question we are told that

   The height at which he let go of the brief case is  h =  130 m  

    The  time taken before the the brief case hits the water is  t =  6 s

Generally the initial speed of the  briefcase (Which also the speed of the helicopter )before the man let go of it is  mathematically evaluated using kinematic equation as

      s = h+  u t +  0.5 gt^2

Here s  is the distance covered by the bag at sea level which is zero

      0 = 130+  u * (6) +  0.5  *  (-9.8) * (6)^2

=>    0 = 130+  u * (6) +  0.5  *  (-9.8) * (6)^2

=>   u  =  \frac{-130 +  (0.5 * 9.8 *  6^2) }{6}

=>   u  =  7.73 \  m/s

     

7 0
2 years ago
This is a physical property of all visible light determined by the light's frequency and visible to the human eye.
motikmotik
Color <span>is a physical property of all visible light determined by the light's frequency and visible to the human eye.</span>
6 0
2 years ago
Un tubo de acero de 40000 kilómetros forma un anillo que se ajusta bien a la circunferencia de la tierra. Imagine que las person
Darina [25.2K]

Answer:

82.76m

Explanation:

In order to find the distance of the steel ring to the ground, when its temperature has raised by 1°C, you first calculate the radius of the steel tube before its temperature increases.

You use the formula for the circumference of the steel ring:

C=2\pi r    (1)

C: circumference of the ring = 40000 km = 4*10^7m (you assume the circumference is the length of the steel tube)

you solve for r in the equation (1):

r=\frac{C}{2\pi}=\frac{4*10^7m}{2\pi}=6,366,197.724m

Next, you use the following formula to calculate the change in the length of the tube, when its temperature increases by 1°C:

L=Lo[1+\alpha \Delta T]         (2)

L: final length of the tube = ?

Lo: initial length of the tube = 4*10^7m

ΔT = change in the temperature of the steel tube = 1°C

α: thermal coefficient expansion of steel = 13*10^-6 /°C

You replace the values of the parameters in the equation (2):

L=(4*10^7m)(1+(13*10^{-6}/ \°C)(1\°C))=40,000,520m

With the new length of the tube, you can calculate the radius of a ring formed with the tube. You again solve the equation (1) for r:

r'=\frac{C}{2\pi}=\frac{40,000,520m}{2\pi}=6,366,280.484m

Finally, you compare both r and r' radius:

r' - r = 6,366,280.484m - 6,366,197.724m = 82.76m

Hence, the distance to the ring from the ground is 82.76m

4 0
2 years ago
flat block is pulled along a horizontal flat surface by a horizontal rope perpendicular to one of the sides. The block measures
Lelu [443]

Answer:

Answered

Explanation:

v= 1 m/s

A= 1 m^2

m= 100 kg

y= 1 mm

μ = ?

ζ= viscosity of  SAE 20 crankcase oil of 15° C= 0.3075 N sec/m^2

forces acting on the block are  

                                 F_s   ←    ↓     →F_f

                                                mg

N= mg

F_s=  shear force = ζAv/y        F_f= friction force = μN

now in x- direction F_s= F_f

ζAv/y  =  μN

0.3075×1×1×1/1×10^{-3} = μ×100

⇒μ=0.313 (coefficient of sliding friction for the block)

Now, as the velocity is increased shear force also increases and due to this frictional force also increases.

Now, to compensate this frictional force friction coefficient must increase

as v∝μ

7 0
2 years ago
Read 2 more answers
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