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faust18 [17]
2 years ago
15

The work function for magnesium is 3.70 ev. what is its cutoff frequency?

Physics
1 answer:
alexandr402 [8]2 years ago
7 0

The cutoff frequency for magnesium is 8.93 x 10¹⁴ Hz.

<h3>What is cutoff frequency?</h3>

The work function is related to the frequency as

W0 = h x fo

where, fo = cutoff frequency and h is the Planck's constant

Given is the work function for magnesium is  3.70 eV.

fo = 3.7 x 1.6 x 10⁻¹⁹ / 6.626 x 10⁻³⁴

fo = 8.93 x 10¹⁴ Hz.

Thus, the cut off frequency is 8.93 x 10¹⁴ Hz.

Learn more about cutoff frequency.

brainly.com/question/14378802

#SPJ1

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A baseball is moving at a speed of 2.2m/s when it strikes the catchers glove. The paddding of the glove is compressed by 24mm be
Andre45 [30]

Average acceleration of the baseball: -101 m/s^2

Explanation:

Since the motion of the baseball is a uniformly accelerated motion, we can use the following suvat equation:

v^2-u^2=2as

where

u is the initial velocity

v is the final velocity

a is the acceleration

s is the displacement of the object

For the baseball in this problem, we have:

u = 2.2 m/s is the initial velocity

v = 0 is the final velocity (it comes to a stop)

s = 24 mm = 0.024 m is the displacement of the ball while decelerating

Therefore, we can solve for a to find the acceleration:

a=\frac{v^2-u^2}{2s}=\frac{0-2.2^2}{2(0.024)}=-101 m/s^2

where the negative sign means the baseball is slowing down.

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4 0
3 years ago
Sedimentary rock turns into magnum through which process
alexandr402 [8]
Sedimentary rocks are formed when sediment is deposited out of air, ice, wind, gravity, or water flows carrying the particles in suspension. This sediment is often formed when weathering and erosion break down a rock into loose material in a source area.
4 0
3 years ago
A car moves with constant acceieration of -0.s m/s? on a straight portion of the road. Att- 0s the car has a velocity of 69 mph,
Crazy boy [7]

Answer:

b) d = 0.71 Km

Explanation:

Car kinematics

Car 1 moves with uniformly accelerated movement

v_f^2=v_o^2+2a*d Formula (1)

d: displacement in meters (m)

v₀: initial speed in m/s

vf: final speed in m/s

a: acceleration in m/s²

Equivalences:

1mile = 1609.34 meters

1 hour = 3600s

1km = 1000m

Known data

v_o = 69\frac{mile}{hour} *1609.34\frac{meter}{mile} *\frac{1}{3600}\frac{hour}{s}=30.8 \frac{m}{s}

v_f= \frac{69}{2} \frac{mile}{hour} =34.5\frac{mile}{hour}=15.4 \frac{m}{s}

a = -0.5 m/s²

Distance calculation

We replace data in the Formula (1)

15.4^2 = 30.8^2+2(-0.5)*d

2(0.5)*d = 30.8^2 - 15.4^2

d =\frac{ 30.8^2 - 15.4^2}{2(0.5) }= 717.6m

d = 717.6 m\frac{1km}{1000m} = 0.7176Km

8 0
3 years ago
2)
earnstyle [38]

Answer:

0.0312J

Explanation:

Let x be the distance the staple moves:

x=0.150m-0.115m=0.035m

And spring constant is k=51.0N/m

PE=0.5kx^2\\=0.5\times 51.0\times 0.035^2\\\\=0.312

Hence, the potential energy is 0.0312J

8 0
3 years ago
What is the force of gravity (from the Earth) on the 700kg satellite if it’s 10km above the Earths surface?
joja [24]

Answer:

The force of gravity on a 700 kg satellite if its 10 km above Earth's surface is given by

    = {\frac{(6.674\times 10^{-11}N. m^2/kg^2)(5.97\times 10^{24}kg) }{(10\times10^3)^2} = 3984378 m / s^{2}

Explanation:

The force of gravity on a 700 kg satellite if its 10 km above Earth's surface is given by

    = {\frac{(6.674\times 10^{-11}N. m^2/kg^2)(5.97\times 10^{24}kg) }{(10\times10^3)^2} = 3984378 m / s^{2}

3 0
3 years ago
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