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kkurt [141]
3 years ago
9

Silver has a work function of 4.5 eV . Part A What is the longest wavelength of light that will release an electron from a silve

r surface
Physics
1 answer:
uranmaximum [27]3 years ago
7 0

Answer:

λ = 2.7608 x 10⁻⁷ m = 276.08 nm

Explanation:

The work function of a metallic surface is the minimum amount of photon energy required to release the photo-electrons from the surface of metal. The work function is given by the following formula:

Work Function = hc/λ

where,

Work Function = (4.5 eV)(1.6 x 10⁻¹⁹ J/1 eV) = 7.2 x 10⁻¹⁹ J

h = Plank's Constant = 6.626 x 10⁻³⁴ J.s

c = speed of light = 3 x 10⁸ m/s

λ = longest wavelength capable of releasing electron.

Therefore,

7.2 x 10⁻¹⁹ J = (6.626 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/λ

λ = (6.626 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/(7.2 x 10⁻¹⁹ J)

<u>λ = 2.7608 x 10⁻⁷ m = 276.08 nm</u>

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Answer:

Explanation:

Given

mass of crane m=1400\ kg

distance moved d=40\ m

Since it is moving with a constant velocity therefore net force on it is zero

Tension force=weight

T=mg

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W_T=T\cdot d

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W_T=548.8\ KJ

Work done by Gravity will be equal in magnitude but opposite in sign and can be obtained by work energy theorem which states that change in kinetic energy of object is equal to work done by all the forces

W_T+W_g=0

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3 years ago
The weights in atwoods machine, starting at rest, attain a velocity of 2ft/sec in one sec. Find the ratio of the masses
Orlov [11]
Refer to the figure shown below.
Let m₁ and m₂ e the two masses.
Let a = the acceleration.
Let T =  tension over the frictionless pulley.

Write the equations of motion.
m₂g - T = m₂a            (1)
T - m₁g = m₁a            (2)

Add equations (1) and (2).
m₂g - T + T - m₁g = (m₁ + m₂)a
(m₂ - m₁)g = (m₁ + m₂)a

Divide through by m₁.
(m₂/m₁ - 1)g = (1 + m₂/m₁)a

Define r = m₂/m₁ as the ratio of the two masses. Then
(r - 1)g = (1 +r)a
r(g-a) = a + g
r = (g - a)/(g + a)

With  = 2 ft/s from rest, the acceleration is
a = 2/32.2 = 0.062 ft/s²
Therefore
r = (32.2 - 0.062)/(32.2 + 0.062) = 0.9962

Answer:
The ratio of masses is 0.9962 (heavier mass divided by the lighter mass).

8 0
3 years ago
In Physics, work depends on two factors. What are those two<br> factors?
Aloiza [94]

Answer:

Force and displacement.

Explanation:

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5 0
3 years ago
A force acts on a 9.90 kg mobile object that moves from an initial position of to a final position of in 5.40 s. Find (a) the wo
horrorfan [7]

Given that,

Mass of object = 9.90 kg

Time =5.40 s

Suppose the force is (2.00i + 9.00j + 5.30k) N, initial position is (2.70i - 2.90j + 5.50k) m and final position is (-4.10i + 3.30j + 5.40k) m.

We need to calculate the displacement

Using formula of displacement

s=r_{2}-r_{1}

Where, r_{1} = initial position

r_{2} = final position

Put the value into the formula

s= (-4.10i + 3.30j + 5.40k)-(2.70i - 2.90j + 5.50k)

s= -6.80i+6.20j-0.1k

(a). We need to calculate the work done on the object

Using formula of work done

W=F\cdot s

Put the value into the formula

W=(2.00i + 9.00j + 5.30k)\cdot (-6.80i+6.20j-0.1k)

W=-13.6+55.8-0.53

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(b). We need to calculate the average power due to the force during that interval

Using formula of power

P=\dfrac{W}{t}

Where, P = power

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Put the value into the formula

P=\dfrac{41.67}{5.40}

P=7.71\ Watt

(c). We need to calculate the angle between vectors

Using formula of angle

\theta=\cos^{-1}(\dfrac{r_{1}r_{2}}{|r_{1}||r_{2}|})

Put the value into the formula

\theta=\cos^{-1}\dfrac{(-4.10i + 3.30j + 5.40k)\cdot(2.70i - 2.90j + 5.50k)}{7.54\times6.778})

\theta=79.7^{\circ}

Hence, (a). The work done on the object by the force in the 5.40 s interval is 41.67 J.

(b). The average power due to the force during that interval is 7.71 Watt.

(c).  The angle between vectors is 79.7°

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