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elena55 [62]
3 years ago
6

The minimum frequency of light needed to eject electrons from a metal is called the threshold frequency, ???? 0 ν0 . Find the mi

nimum energy needed to eject electrons from a metal with a threshold frequency of 4.23× 10 14 s1 . 4.23×1014 s−1.
Physics
1 answer:
Allisa [31]3 years ago
7 0

Answer:

<h2>E = 2.8028*10⁻¹⁹ Joules</h2>

Explanation:

The minimum energy needed to eject electrons from a metal with a threshold frequency fo is expressed as E = hfo

h = planck's constant

fo = threshold frequency

Given the threshold frequency fo = 4.23×10¹⁴ s⁻¹

h = 6.626× 10⁻³⁴ m² kg / s

Substituting this value into the formula to get the energy E

E = 4.23×10¹⁴ *  6.626 × 10⁻³⁴

E = 28.028*10¹⁴⁻³⁴

E = 28.028*10⁻²⁰

E = 2.8028*10⁻¹⁹ Joules

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Suppose you give a 10 Newton push to Ryan on skis (he weighs 50 kg), how much will he accelerate?
Talja [164]

Well we can just use F=ma. The force is 10N, the mass is 50 kg, solve for a. Well since we kg and N, no conversion is necessary. So just plugging in the numbers, we get

10N = 50 kg · a

\frac{10N}{50kg}=a

A newton is just \frac{kg·m}{s^{2}}

a=\frac{\frac{10kg·m}{s^{2}}}{50kg}

The s^2 and 50 kg you multiply

a=\frac{10kg·m}{50kg·s^{2}}

The kg's cancel and 10/50 is 1/5

\frac{1}{5}·\frac{m}{s^{2}}

So the acceleration is 1/5 m/s^2


3 0
2 years ago
A ball is thrown upward from the top of a building at an angle of 30.0° to the horizontal and with
Ray Of Light [21]

Answer:

See below

Explanation:

Vertical position = 45 +  20 sin (30) t  - 4.9 t^2

 when it hits ground this = 0

               0 = -4.9t^2 + 20 sin (30 ) t + 45

                0 = -4.9t^2 + 10 t +45 = 0     solve for t =4.22 sec

  max height is at  t= - b/2a = 10/9.8 =1.02

     use this value of 't' in the equation to calculate max height = 50.1 m

      it has  4.22 - 1.02 to free fall = 3.2 seconds free fall

           v = at = 9.81 * 3.2 = 31.39 m/s VERTICAL

      it will <u>also</u> still have horizontal velocity =  20 cos 30 = 17.32 m/s

        total velocity will be sqrt ( 31.39^2 + 17.32^2) = 35.85 m/s

Horizontal range = 20 cos 30  * t  =  20 *  cos 30  * 4.22 = 73.1 m

8 0
1 year ago
Consider a block on a spring oscillating on a frictionless surface. The amplitude of the oscillation is 11 cm, and the speed of
IRISSAK [1]

Answer:

The angular frequency of the block is ω = 5.64 rad/s

Explanation:

The speed of the block v = rω where r = amplitude of the oscillation and ω = angular frequency of the oscillation.

Now ω = v/r since v = speed of the block = 62 cm/s and r = the amplitude of the oscillation = 11 cm.

The angular frequency of the oscillation ω is

ω = v/r

ω = 62 cm/s ÷ 11 cm

ω = 5.64 rad/s

So, the angular frequency of the block is ω = 5.64 rad/s

6 0
3 years ago
What time period did fish develop
marysya [2.9K]

Answer:

They developed during the Cambrian time period, which was around 530 million years ago.

Explanation:

Hope this Helps!

5 0
3 years ago
Push the plunger in the two syringes. Is your prediction correct? Why?​
4vir4ik [10]

Answer:

This is worded strangely.

4 0
2 years ago
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