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VladimirAG [237]
3 years ago
8

When one "throws” a punch, the forearm applies force to the fist. Consider a(n) 0.75 kg fist that goes from rest to a

Physics
1 answer:
deff fn [24]3 years ago
6 0
Formula for acceleration

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If the distance between the first order maximum and the tenth order maximum of a double-slit pattern is 18 mm and the slits are
Setler79 [48]

Answer:

Wavelength of light is 600 nm

Explanation:

Given

Distance between the first order maximum and the tenth order maximum of a double-slit pattern = 18 mm

Separation between the slits = 0.15 mm

Distance of screen from the slits = 50 cm

Wavelength

= \frac{18*10^{-3} * 0.15 *10^{-3}}{0.50*9} \\= 6 *10^{-7}\\= 600nm

4 0
3 years ago
What is the maximum wavelength of incident light that can produce photoelectrons from silver? The work function for silver is Φ=
Oksi-84 [34.3K]

Answer:

The maximum wavelength of incident light that can produce photoelectrons from silver is 423.5 nm.

Explanation:

Given;

work function of silver, Φ = 2.93 eV = 2.93 x 1.602 x 10⁻¹⁹ J = 4.6939 x 10⁻¹⁹ J

Apply Einstein Photo electric effect;

E = K.E + Ф

Where;

E is the energy of the incident light

K.E is the kinetic of electron

Ф is the work function of silver surface

For the incident light to have maximum wavelength, the kinetic energy of the electron will be zero.

E = Ф

hf = Ф

h\frac{c}{\lambda} = \phi

where;

c is speed of light = 3 x 10⁸ m/s

h is Planck's constant, = 6.626 x 10⁻³⁴ J/s

λ is the wavelength of the incident light

\lambda = \frac{hc}{\phi}\\\\\lambda =\frac{6.626*10^{-34} *3*10^8}{4.6939*10^{-19}} \\\\\lambda = 4.235 *10^{-7} \ m\\\\\lambda = 423.5 *10^{-9} \ m\\\\\lambda = 423.5 \ nm

Therefore, the maximum wavelength of incident light that can produce photoelectrons from silver is 423.5 nm.

8 0
3 years ago
A sound wave of the form s = sm cos(kx - ?t + f) travels at 343 m/s through air in a long horizontal tube. At one instant, air m
Naddik [55]

Answer:

960.24 Hz

Explanation:

Here is the complete question

A sound wave of the form

S=Smcos(kx−ωt+Φ)

travels at 343 m/s through air in a long horizontal tube. At one instant, air molecule A at x = 2.000 m is at its maximum positive displacement of 6.00 nm and air molecule B at x = 2.070 m is at a positive displacement of 2.00 nm. All the molecules between A and B are at intermediate displacements. What is the frequency of the wave?

Solution

Given x₁ = 2.0 m, x₂ = 2.070 m, maximum positive displacement s = 6.00 nm at x₁, positive displacement s = 2.00 nm at x₂, velocity of wave v = 343 m/s, maximum positive displacement s₀ = 6.00 nm

Let t₀ = 0 at x₁ = 2.0 m for maximum displacement.

So, s = s₀cos(kx−ωt+Φ)

     6 = 6cos(2k - 0 + Φ) = 6cos(2k + Φ)⇒ cos(2k + Φ) = 6/6 = 1

cos(2k + Φ) = 1 ⇒ (2k + Φ) = cos⁻¹ (1) = 0 ⇒ 2k + Φ = 0

Let t₀ = 0 at x₂ = 2.070 m for displacement s = 2.00 nm.

So, s = s₀cos(kx−ωt+Φ)

     2 = 6cos(2.070k - 0 + Φ) = 6cos(2.070k + Φ)⇒ cos(2.070k + Φ) = 2/6 = 1/3

cos(2.070k + Φ) = 1/3 ⇒ (2.070k + Φ) = cos⁻¹ (1/3) = 70.53 ⇒ 2.070k + Φ = 70.53.

We now have two simultaneous equations.

2k + Φ = 0   (1)

2.070k + Φ = 70.53.    (2)

Subtracting (2) -(1)

2.070k - 2k = 70.53

0.070k = 70.53

k = 70.53/0.070 = 1007.554π/180 rad/m = 17.59 rad/m

k = 2π/λ ⇒ λ = 2π/k

and frequency, f = v/λ = v/2π/k = kv/2π = 17.59 × 343/2π = 960.24 Hz

4 0
3 years ago
If an antelope can run 14.1 km in 39 min, how fast is it km/h(for reference , 110 km/h is 65 mph )
shtirl [24]
It might make more sense putting it another way but this is basically it. you just take the minutes and divide them by 60 to convert them to hours. then simplify the ratio

8 0
3 years ago
Read 2 more answers
There are ________ main groups in the modern periodic table of elements.
patriot [66]
The Correct answer: 18

They are also known as families because they are in the same column.
3 0
3 years ago
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