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2 years ago

Estimate the wavelength of electrons that have been accelerated from rest through a potential difference of 60 kV.

Physics

1 answer:

ivolga24 [154]2 years ago

8 0

Answer: 2.068* $10^{-14}$ m

Explanation: According to work energy-theorem , the workdone in accelerating the electron equals the energy it would give off in terms of light.

workdone= qV

energy = hc/λ

q=magnitude of an electronic charge= 1.602* $10^{-16}$

h= planck constant = 6.626* $10^{-34}$

c= speed of light =2.998* $10^{8}$

v= potential difference= 6* $10^{4}$

λ= wavelength=unknown

by making λ subject of formulae we have that

λ= $\frac{hc}{qv}$

λ = 6.626* $10^{-34}$ * 2.998* $10^{8}$ / 1.602* $10^{-16}$ * 6* $10^{4}$

λ = $\frac{19.878*10^{-26} }{9.612*10^{-12} }$

by doing the necessary calculations, we have that

λ = 2.068* $10^{-14}$ m

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2 years ago

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salantis [7]

t = 0.527 s

It accelerates for 0.527 s.

Explanation:

We use the formula:

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Given:

v = 106 m/s

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So applying the formula,

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2 years ago

Robin would like to shoot an orange in a tree with his bow and arrow. The orange is hanging yf=5.00 myf=5.00 m above the ground.

tensa zangetsu [6.8K]

Answer:

h' = 55.3 m

Explanation:

First, we analyze the horizontal motion of the projectile, to find the time taken by the arrow to reach the orange. Since, air friction is negligible, therefore, the motion shall be uniform:

s = vt

where,

s = horizontal distance between arrow and orange = 60 m

v = initial horizontal speed of the arrow = v₀ Cos θ

θ = launch angle = 30°

v₀ = launch speed = 35 m/s

Therefore,

60 m = (35 m/s)Cos 30° t

t = 60 m/30.31 m/s

t = 1.98 s

Now, we analyze the vertical motion to find the height if arrow at this time. Using second equation of motion:

h = Vi t + (1/2)gt²

where,

Vi = Vertical Component of initial Velocity = v₀ Sin θ = (35 m/s)Sin 30°

Vi = 17.5 m/s

Therefore,

h = (17.5 m/s)(1.98 s) + (1/2)(9.81 m/s²)(1.98 s)²

h = 34.6 m + 19.2 m

h = 53.8 m

since, the arrow initially had a height of y = 1.5 m. Therefore, its final height will be:

h' = h + y

h' = 53.8 m + 1.5 m

h' = 55.3 m

4 0

2 years ago

The illustration in figure below shows a uniform metre rule weighing 30 N pivoted on a wedge placed under the 40 cm mark and car

Nitella [24]

Answer:

W = 30 N

Explanation:

Applying the summation of torques about the wedge for equilibrium, taking the clockwise direction as negative. Since the ruler is balanced horizontally about the wedge. Therefore, the summation of all torques acting about the wedge must be equal to zero.

$(70\ N)(40\ cm - 10\ cm)-(30\ N)(50\ cm-40\ cm)-(W)(100\ cm - 40\ cm) = 0\\W(60\ cm) = (70\ N)(30\ cm)-(30\ N)(10\ cm)\\\\W = \frac{1800\ N.cm}{60\ cm}$

W = 30 N

6 0

2 years ago

A small car with mass m and speed 2v and a large car with mass 2m and speed v both travel the same circular section of an unbank

Alenkasestr [34]

Answer:

(D) F/2

Explanation:

Since the circular section is unbanked, the centripedal acceleration acting on each of the cars is

$a_s = \frac{v_s^2}{r} = \frac{(2v)^2}{r} = \frac{4v^2}{r}$

$a_l = \frac{v_l^2}{r} = \frac{v^2}{r}$

Therefore the centripetal force on each car

$F_s = m_sa_s = \frac{4mv^2}{r}$

$F_l = m_la_l = \frac{2mv^2}{r}$

Since $F_s = 2F_l$ this means the friction force required to keep the large car on the road is only half of the friction force required to keep the small car on road

So (D) F/2 is the correct answer

5 0

3 years ago