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andrew-mc [135]
2 years ago
14

The 3rd bright fringe of a double slit interference pattern is 30.0 cm above the central bright fringe. If the angle from the ho

rizontal to this 3rd bright fringe is 12.0 degrees, what is the distance (in meters) between the double slits and the viewing screen
Physics
1 answer:
Anastasy [175]2 years ago
5 0

Answer:

Explanation:

12 degree = (π / 180) x 12 radian

= .2093 radian

position of third bright fringe = 3λ D/ d where λ is wave length of light , D is screen distance and d is slit separation

given

3λ D/ d = 30 x 10⁻²

angular separation

3λ / d = .2093

from the two equation

.2093 D= 30 x 10⁻²

D = 30 x 10⁻² / .2093

= 1.43 m

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netineya [11]

Answer:

The activation energy for this reaction, Ea = 159.98 kJ/mol

Explanation:

Using the Arrhenius equation as:

ln\frac {K_2}{K_1}=-\frac {E_a}{R}\times (\frac {1}{T_2}-\frac {1}{T_1})

Where, Ea is the activation energy.

R is the gas constant having value 8.314 J/K.mol

K₂ and K₁ are the rate constants

T₂ and T₁ are the temperature values in kelvin.

Given:

K₂ = 8.66×10⁻⁷ s⁻¹ , T₂ = 425 K

K₁ = 3.61×10⁻¹⁵ s⁻¹ , T₁ = 298 K

Applying in the equation as:

ln\frac {8.66\times 10^{-7}}{3.61\times 10^{-15}}=-\frac {E_a}{8.314}\times (\frac {1}{425}-\frac {1}{298})

Solving for Ea as:

Ea = 159982.23 J /mol

1 J/mol = 10⁻³ kJ/mol

Ea = 159.98 kJ/mol

7 0
2 years ago
An amusement park ride spins you around in a circle of radius 2.5 m with a speed of 8.5 m/s. If your mass is 75 kg, what is the
zhannawk [14.2K]
B is the answer to this truly did the math
3 0
2 years ago
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An ac source of period T and maximum voltage V is connected to a single unknown ideal element that is either a resistor, and ind
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The answer is d.a capacitor
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Keisha looks out the window from a tall building at her friend Monique standing on the ground, 8.3 m away from the side of the b
Salsk061 [2.6K]

Answer:

Explanation:

GIVEN DATA:

Distance between keisha and her friend 8.3 m

angle made by keisha toside building 30 degree

height of her friend monique is 1.5 m

from the figure

\Delta ACB

tan 30 = \frac{8.3}{h}

h= \frac{8.3}{tan 30} = 14.376 m

therefore

height of keisha is = h  + 1.5 m

                               = 14.376 + 1.5

= 15.876 \simeq 16 m

therefore option c is correct

5 0
2 years ago
1. Do alto de uma plataforma com 15m de altura, é lançado horizontalmente um projéctil. Pretende-se atingir um alvo localizado n
sveta [45]

Answer:

(a). The initial velocity is 28.58m/s

(b). The speed when touching the ground is 33.3m/s.

Explanation:

The equations governing the position of the projectile are

(1).\: x =v_0t

(2).\: y= 15m-\dfrac{1}{2}gt^2

where v_0 is the initial velocity.

(a).

When the projectile hits the 50m mark, y=0; therefore,

0=15-\dfrac{1}{2}gt^2

solving for t we get:

t= 1.75s.

Thus, the projectile must hit the 50m mark in 1.75s, and this condition demands from equation (1) that

50m = v_0(1.75s)

which gives

\boxed{v_0 = 28.58m/s.}

(b).

The horizontal velocity remains unchanged just before the projectile touches the ground because gravity acts only along the vertical direction; therefore,

v_x = 28.58m/s.

the vertical component of the velocity is

v_y = gt \\v_y = (9.8m/s^2)(1.75s)\\\\{v_y = 17.15m/s.

which gives a speed v of

v = \sqrt{v_x^2+v_y^2}

\boxed{v =33.3m/s.}

4 0
2 years ago
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