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GenaCL600 [577]
2 years ago
6

4.A photon of green light strikes an unknown metal and an electron is emitted. The voltage is set to 2 volts. The electron canno

t make the journey to the second plate. What can be said about a similar experiment done with violet light? A. An electron may or may not be emitted in the second experiment. It cannot be determined. B. An electron will be emitted in the second experiment, but it cannot be determined whether it will reach the second plate. C. An electron will be emitted in the second experiment, and it will make it to the second plate. D. An electron will not be emitted in the second experiment.
3.Green light shines on an unknown piece of metal, and the lab instruments detect emitted electrons. Which of the following will be true if the light is changed to violet?
A. Violet light will cause electrons to be emitted at greater velocities than those removed by green light.
B. Violet light will cause electrons to be emitted at lower velocities than those removed by green light.
C. Violet light will cause electrons to be emitted only if the light is as intense as the green light.
D. It cannot be determined whether violet will cause electrons to be emitted or not.
Physics
1 answer:
Anarel [89]2 years ago
3 0
4) The correct answer is:
<span>B. An electron will be emitted in the second experiment, but it cannot be determined whether it will reach the second plate. 

In fact, violet light has higher frequency than green light. This means that photons of violet light carry more energy than photons of green light (remember that the energy of a photon is proportional to its frequency: </span>E=hf)<span>, so when they hit the surface of the metal, more energy is transferred to the electrons. The electron was already emitted with green light, so it must be emitted also with violet light, given the more energy transferred. The electron will also have more kinetic energy when hit by violet light, however, we cannot determine if it will reach the second plate, since we don't know how much energy has been used to extract the electron from the metal (in fact, we don't know the work function of the metal, i.e. the energy needed to extract the electron)


3)  The correct answer is
</span><span>A. Violet light will cause electrons to be emitted at greater velocities than those removed by green light.

In fact, </span>violet light has higher frequency than green light. This means that photons of violet light carry more energy than photons of green light (remember that the energy of a photon is proportional to its frequency: E=hf), so when they hit the surface of the metal, more energy is transferred to the electrons. Therefore, the emitted electrons will have on average greater energy (and so, greater velocity) than those removed by green light.
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A cylinder rotating about its axis with a constant angular acceleration of 1.6 rad/s2 starts from rest at t = 0. At the instant
OverLord2011 [107]

Answer:

The magnitude of the total linear acceleration is 0.27 m/s²

b. 0.27 m/s²

Explanation:

The total linear acceleration is the vector sum of the tangential acceleration and radial acceleration.

The radial acceleration is given by;

a_t = ar

where;

a is the angular acceleration and

r is the radius of the circular path

a_t = ar\\\\a_t = 1.6 *0.13\\\\a_t = 0.208 \ m/s^2

Determine time of the rotation;

\theta = \frac{1}{2} at^2\\\\0.4 = \frac{1}{2} (1.6)t^2\\\\t^2 = 0.5\\\\t = \sqrt{0.5} \\\\t = 0.707 \ s\\\\

Determine angular velocity

ω = at

ω = 1.6 x 0.707

ω = 1.131 rad/s

Now, determine the radial acceleration

a_r = \omega ^2r\\\\a_r = 1.131^2 (0.13)\\\\a_r = 0.166 \ m/s^2

The magnitude of total linear acceleration is given by;

a = \sqrt{a_t^2 + a_r^2} \\\\a = \sqrt{0.208^2 + 0.166^2} \\\\a = 0.266 \ m/s^2\\\\a = 0.27  \ m/s^2

Therefore, the magnitude of the total linear acceleration is 0.27 m/s²

b. 0.27 m/s²

5 0
2 years ago
A 1.25 in. by 3 in. rectangular steel bar is used as a diagonal tension member in a bridge truss. the diagonal member is 20 ft l
pentagon [3]

Answer:

axial stress in the diagonal bar =36,000 psi

Explanation:

Assuming we have to find axial stress

Given:

width of steel bar: 1.25 in.

height of the steel bar: 3 in

Length of the diagonal member = 20ft

modulus of elasticity E= 30,000,000 psi

strain in the diagonal member ε = 0.001200 in/in

Therefore, axial stress in the diagonal bar σ = E×ε

=  30,000,000 psi×  0.001200 in/in =36,000 psi

5 0
2 years ago
THE RELATIVE ANGLE AT THE KNEE CHANGES FROM 0O TO 85O DURING THE KNEE FLEXION PHASE OF A SQUAT EXERCISE. IF 10 COMPLETE SQUATS A
Mice21 [21]

Answer: angular distance = 1700° and 29.7 rad

      also the angular displacement = 0

Explanation:

To explain this, i will give a breakdown of this works.

we are asked to find both the angular distance and displacement the knee undergo.

Ok to get the distance of the knee, we would first take note that for one to squat down and get back up, the knee would travel through 85° of flexion to gpo down, and also through another 85° of extension to return standing (upright). So, the actual angular distance of the squat is 170°.

taking ten squats, the knee would have to go through 170° motion times 10 i.e;

10 * 170° = 1700°

Therefore the angulaar distance is 1700°

now converting this distance to radians since we will be required to have our answer in both degree and rad.

Given that 2pi = 360°, it means that one degree will give 57.3°;

∅ (rad) = ∅ (deg) * 2π/360°

∅ (rad) = 1170° * 2π/360°  = 29.7 rad

∅ (rad) = 29.7 rad

b. For the other part, let us remember that angular displacement is equal to angular distance divided by time, so the angular displacement displacement of the knee will be zero, because the knee's position at the final third will be the same as the initial position.

cheers i hope this helps!!!!

π

5 0
2 years ago
An investigation was done with an electromagnetic system made from a battery and wire wrapped around a nail. Different sizes of
alex41 [277]
Dependant variable is something which you MEASURE during an experiment

So your answer would be : B
5 0
2 years ago
jesse is swinging miguel in a circle at a tangential speed of 3.50 m/s. if the radius of the circle is 0.600 m and miguel has a
Morgarella [4.7K]
Centripetal acceleration = (speed)² / (radius) .

Force = (mass) · (acceleration)

Centripetal force = (mass) · (speed)² / (radius) .

                             = (11 kg) · (3.5 m/s)² / (0.6 m)

                             = (11 kg) · (12.25 m²/s²) / (0.6 m)

                             =  (11 · 12.25) / 0.6  kg-m/s²

                             =      224.58 newtons.    (about 50.5 pounds)

That's the tension in Miguel's arm or leg or whatever part of his body
Jesse is swinging him by.  It's the centripetal force that's needed in
order to swing 11 kg in a circle with a radius of 0.6 meter, at 3.5
meters/second.  If the force is less than that, then the mass has to
either swing slower or else move out to follow a bigger circle.
6 0
2 years ago
Read 2 more answers
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