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mariarad [96]
3 years ago
7

A Young's double-slit experiment is performed using light that has a wavelength of 631 nm. The separation between the slits is 5

.25E-5 m. Calculate the angle that locates the first-order bright fringes on the screen.
Physics
1 answer:
Sergeu [11.5K]3 years ago
7 0

Answer:

 θ = 1.21  10⁻² rad

Explanation:

The double slit experiment is described for constructive interference

           d sin θ = m λ

where d is the separation between the slits, λ the wavelength used, m a stereo that represents the order of interference

The first order of interference corresponds to m = 1 and the separation of the slits is d = 5.2 10⁻⁵ m

        sin θ = m λ / d

        θ = sin⁻¹ (m λ / d)

     

let's calculate

          θ = sin⁻¹ (1 631 10⁻⁹ / 5.2 10⁻⁵)

          θ = 1.21  10⁻² rad

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A 75 kg football player is gliding forward across very smooth ice at 4.6 m/s. He throws a 0.47 kg football straight forward. A)
lord [1]

Answer:

4.53482 m/s

4.506 m/s

Explanation:

m_1 = Mass of player = 75 kg

v_1 = Initial velocity of player = 4.6 m/s

m_2 = Mass of ball = 0.47 kg

v_1 = Initial velocity of ball = 15 m/s

The linear momentum of the system is conserved

(m_1+m_2)v_1=m_1v+m_2v_2\\\Rightarrow v=\dfrac{(m_1+m_2)v_1-m_2v_2}{m_1}\\\Rightarrow v=\dfrac{(75+0.47)4.6-0.47\times 15}{75}\\\Rightarrow v=4.53482\ m/s

The player's speed is 4.53482 m/s

In the second case the equation of momentum is

(m_1+m_2)v_1=m_1v+m_2(v_2+v_1)\\\Rightarrow v=\dfrac{(m_1+m_2)v_1-m_2(v_2+v_1)}{m_1}\\\Rightarrow v=\dfrac{(75+0.47)4.6-0.47\times (15+4.6)}{75}\\\Rightarrow v=4.506\ m/s

The player's speed is 4.506 m/s

4 0
4 years ago
Physics class 9 chapter 8 please tell​ please
natali 33 [55]

Answer:

(a) The motion is uniform

(b)  11.11 m/s

Explanation:

(a)

From the table below, the motion of the bus is uniform.

(b)

Speed(s) = Δd/Δt

s = Δd/Δt............. Equation 1

From the table,

Given: Δd = 10 km = 10000 m, Δt = 15 minutes = (15×60) = 900 seconds

Substitute these values into equation 1

s = 10000/900

s = 11.11 m/s

7 0
3 years ago
A body of mass 400 kg is suspended at a lower end of a light vertical chain and is being pulled up vertically. Initially the bod
yKpoI14uk [10]

Answer:31.62 m/s

Explanation:

Given

mass of body m=400 kg

Pull on chain is F_1=6000g N=60 kN

Pull get smaller at the rate of F_2=360g N/m

Net Upward Force F=6000 g-360 g\times 10=24 kN

net acceleration a=\frac{F}{m}

a=\frac{24\times 1000}{m}

a=\frac{24000}{400}

a=60 m/s^2

but g is acting downward

a_{net}=a-g=60-10=50 m/s^2

using v^2-u^2=2 as

here initial velocity is zero

v^2=2\times 50\times 10

v=31.62 m/s

7 0
4 years ago
600 kg elephant runs at 5 m/s and jumpts onto a 100kg cart. If the coefficient of friction is .04 how far will the cart travel o
Minchanka [31]

Answer:

Eleven seconds.

Explanation:

Two keys are needed to solve this problem. First, the conservation of momentum: allowing you to calculate the cart's speed after the elephant jumped onto it. It holds that:

m_ev_e+m_c\cdot0=(m_e+m_c)v_0\implies \\v_0=\frac{m_ev_e}{m_e+m_c}=\frac{600kg\cdot 5\frac{m}{s}}{700kg}=4.29\frac{m}{s}

So, once loaded with an elephant, the cart was moving with a speed of 4.29m/s.

The second key is the kinematic equation for accelerated motion. There is one force acting on the cart, namely friction. The friction acts in the opposite direction to the horizontal direction of the velocity v0, its magnitude and the corresponding deceleration are:

F_r = 0.04\cdot (m_e+m_c)g\implies a_r = 0.04\cdot g= 0.04 \cdot 9.8 \frac{m}{s^2}=0.39\frac{m}{s^2}

The kinematic equation describing the decelerated motion is:

v = -a_r t+v_0\\0 = -a_rt+v_0\implies \\t = \frac{v_0}{a_r}=\frac{4.29\frac{m}{s}}{0.39\frac{m}{s^2}}=11s

It takes 11 seconds for the comical elephant-cart system to come to a halt.

7 0
3 years ago
20 Points! Please Help!
Katyanochek1 [597]

Answer: tu dia es INCORRECTO

Explanation:

4 0
3 years ago
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