Describe your experience with the largest earthquake you have left

n alpha particle (q = +2e, m = 4.00 u) travels in a circular path of radius 5.94 cm in a uniform magnetic field with B = 1.10 T.

9966 [12]

Answer:

a). V = 3.13*10⁶ m/s

b). T = 1.19*10^-7s

c). K.E = 2.04*10⁵

d). V = 1.02*10⁵V

Explanation:

q = +2e

M = 4.0u

r = 5.94cm = 0.0594m

B = 1.10T

1u = 1.67 * 10^-27kg

M = 4.0 * 1.67*10^-27 = 6.68*10^-27kg

a). Centripetal force = magnetic force

Mv / r = qB

V = qBr / m

V = [(2 * 1.60*10^-19) * 1.10 * 0.0594] / 6.68*10^-27

V = 2.09088 * 10^-20 / 6.68 * 10^-27

V = 3.13*10⁶ m/s

b). Period of revolution.

T = 2Πr / v

T = (2*π*0.0594) / 3.13*10⁶

T = 1.19*10⁻⁷s

c). kinetic energy = ½mv²

K.E = ½ * 6.68*10^-27 * (3.13*10⁶)²

K.E = 3.27*10^-14J

1ev = 1.60*10^-19J

xeV = 3.27*10^-14J

X = 2.04*10⁵eV

K.E = 2.04*10⁵eV

d). K.E = qV

V = K / q

V = 2.04*10⁵ / (2eV).....2e-

V = 1.02*10⁵V

7 0

2 years ago

A marble at the front of a truck bed traveling at 20 m/s relative to the highway rolls toward the back of the truck bed with a s

irga5000 [103]

14 m/s in the direction of the truck

5 0

2 years ago

What is 6.02 x 10^4 in standard notation

Vesnalui [34]

I believe you mean 6.02*10^7 but you want to shift the decimal 7 times to the right which would be 60200000 (:

4 0

2 years ago

you stand on a straight desert road at night and observe a vehicle approaching. this vehicle is equipped with two small headligh

inysia [295]

Answer:

The distance that you marginally able to discern that there are two headlights rather than a single light source is 6.084 km

Explanation:

Given:

d = distance = 0.679 m

λ = wavelength of the light = 537 nm = 537x10⁻⁹m

dp = pupil diameter = 4.81 mm = 0.00481 m

Question: What distance, in kilometers, are you marginally able to discern that there are two headlights rather than a single light source, dx = ?

For the separation of the peak from the central maximum it is:

$sin\theta =\frac{\lambda }{d_{p} } =\frac{537x10^{-9} }{0.00481} =1.116x10^{-4}$

In this case, the two small sources of the headlights have the same angle as the images that form inside the eye

$d_{x} =\frac{d}{sin\theta } =\frac{0.679}{1.116x10^{-4} } =6.084x10^{3} m=6.084km$

7 0

2 years ago

Light of wavelength 530.00 nm is incident normally on a diffraction grating, and the first‑order maximum is observed to be 33.0∘

GarryVolchara [31]

Answer:

1028 slits/mm

Explanation:

We are given that

Wavelength of light, $\lambda=530nm=530\times 10^{-9} m$

1nm= $10^{-9} m$

$\theta=33^{\circ}$

n=1

We have to find the number of slits per mm are marked on the grating.

We know that

$dsin\theta=n\lambda$

Using the formula

$dsin33^{\circ}=1\times 530\times 10^{-9}$

$d=\frac{530\times 10^{-9}}{sin33^{\circ}}$

$d=9.731\times 10^{-7} m$

1m= $10^{3}mm$

$d=9.731\times 10^{-7}\times 10^3$ mm

$d=0.0009731mm$

Number of slits= $\frac{1}{d}$

Number of slits= $\frac{1}{0.0009731}$ /mm

Number of slits=1028/mm

Hence, 1028 slits/mm are marked on the grating.

8 0

2 years ago

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