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Ghella [55]
3 years ago
6

The light intensity incident on a metallic surface with a work function of 1.88 eV produces photoelectrons with a maximum kineti

c energy of 1.2 eV.
if the frequency of the light is doubled, what is the maximum kinetic energy of the photoelectrons?
Answer in units of eV.

​
Physics
1 answer:
ivanzaharov [21]3 years ago
7 0

Answer:

KE = KE (incidental) - KE of emitted photons

or KE = h * f - Wf

So   h * f = KE + Wf = 1.2 + 1.88 = 3.08    incident energy

If you double the frequency then h * f = 6.16

KE = 6.16 - 1.2 = 4.96 eV

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A 1900 kg car rounds a curve of 55 m banked at an angle of 11 degrees? . If the car is traveling at 98 km/h, How much friction f
Nat2105 [25]

Answer:

22000 N

Explanation:

Convert velocity to SI units:

98 km/h × (1000 m / km) × (1 h / 3600 s) = 27.2 m/s

Draw a free body diagram.  There are three forces acting on the car.  Normal force perpendicular to the bank, gravity downwards, and friction parallel to the bank.

I'm going to assume the friction force is pointed down the bank.  If I get a negative answer, that'll just mean it's actually pointed up the bank.

Sum of the forces in the radial direction (+x):

∑F = ma

N sin θ + F cos θ = m v² / r

Sum of the forces in the y direction:

∑F = ma

N cos θ - F sin θ - W = 0

To solve the system of equations for F, first solve for N and substitute.

N = (W + F sin θ) / cos θ

Substituting:

((W + F sin θ) / cos θ) sin θ + F cos θ = m v² / r

(W + F sin θ) tan θ + F cos θ = m v² / r

W tan θ + F sin θ tan θ + F cos θ = m v² / r

W tan θ + F (sin θ tan θ + cos θ) = m v² / r

W tan θ + F sec θ = m v² / r

F sec θ = m v² / r - W tan θ

F = m v² cos θ / r - W sin θ

F = m (v² cos θ / r - g sin θ)

Given that m = 1900 kg, θ = 11°, v = 27.2 m/s, and r = 55 m:

F = 1900 ((27.2)² cos 11 / 55 - 9.8 sin 11)

F = 21577 N

Rounding to two sig-figs, you need at least 22000 N of friction force.

4 0
3 years ago
A projectile is fired from the origin (at y = 0 m) as shown in the diagram. The initial velocity
Viktor [21]

Answer:

-26 m/s.

Explanation:

Hello,

In this case, since the vertical initial velocity is 26 m/s and the vertical final velocity is 0 m/s at P, we compute the time to reach P:

t=\frac{0m/s-26m/s}{-9.8m/s^2} =2.65s

With which we compute the maximum height:

y=26m/s*2.65s-\frac{1}{2}*9.8m/s^2*(2.65s)^2 \\\\y=34.5m

Therefore, the final velocity until the floor, assuming P as the starting point (Voy=0m/s), turns out:

v_f=\sqrt{0m/s-(-9.8m/s^2)*2*34.5m}\\ \\v_f=-26m/s

Which is clearly negative since it the projectile is moving downwards the starting point.

Regards.

3 0
3 years ago
A regulation basketball has a 32 cm diameter
Rudik [331]

Answer:

1.8 s

Explanation:

Potential energy = kinetic energy + rotational energy

mgh = ½ mv² + ½ Iω²

For a thin spherical shell, I = ⅔ mr².

mgh = ½ mv² + ½ (⅔ mr²) ω²

mgh = ½ mv² + ⅓ mr²ω²

For rolling without slipping, v = ωr.

mgh = ½ mv² + ⅓ mv²

mgh = ⅚ mv²

gh = ⅚ v²

v = √(1.2gh)

v = √(1.2 × 9.81 m/s² × 4.8 m sin 39.4°)

v = 5.47 m/s

The acceleration down the incline is constant, so given:

Δx = 4.8 m

v₀ = 0 m/s

v = 5.47 m/s

Find: t

Δx = ½ (v + v₀) t

t = 2Δx / (v + v₀)

t = 2 (4.8 m) / (5.47 m/s + 0 m/s)

t = 1.76 s

Rounding to two significant figures, it takes 1.8 seconds.

3 0
3 years ago
a 28.7 KG sled is pulled forward with a 63 net force across the ground with MK equals 0.169 what is the acceleration of the sled
abruzzese [7]

Answer:

0.54\ \text{m/s}^2

Explanation:

F = Force on the sled = 63 N

m = Mass of sled = 28.7 kg

\mu_k = Coefficient of kinetic friction = 0.169

g = Acceleration due to gravity = 9.81\ \text{m/s}^2

The force balance of the system is given by

F-\mu_k mg=ma\\\Rightarrow a=\dfrac{F-\mu_k mg}{m}\\\Rightarrow a=\dfrac{63-0.169\times 28.7\times 9.81}{28.7}\\\Rightarrow a=0.54\ \text{m/s}^2

The acceleration of the sled is 0.54\ \text{m/s}^2.

5 0
2 years ago
Two cars are racing! Car A steadily speeds up from rest to 100 mph at the finish line. Car B is already moving 60 mph at the sta
Vilka [71]

Answer:

b

Explanation:

I did this test before xx

5 0
3 years ago
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