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3 years ago

6

In a young's double-slit experiment, the angle that locates the second dark fringe on either side of the central bright fringe i

s 6.5 degrees. find the ratio d/λ of the slit separation d to the wavelength λ of the light.

1 answer:

dem82 [27]3 years ago

6 0

Answer:

(d/λ) = 17.67

Explanation:

Equation for double-slit:

d*sinθ = m*λ

Rearrange the equation

(d/λ) = m/(sinθ)

Since it is at the second dark fringe, m = 2

Therefore:

(d/λ) = 2/(sin(6.5)

(d/λ) = 17.67

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Answer:

B. Tomatos reflect red light

Explanation:

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3 years ago

Read 2 more answers

Consider a car travelling at 60 km/hr. If the radius of a tire is 25 cm, calculate the angular speed of a point on the outer edg

vlabodo [156]

To solve this problem it is necessary to apply the concepts given in the kinematic equations of movement description.

From the perspective of angular movement, we find the relationship with the tangential movement of velocity through

$\omega = \frac{v}{R}$

Where,

$\omega =$ Angular velocity

v = Lineal Velocity

R = Radius

At the same time we know that the acceleration is given as the change of speed in a fraction of the time, that is

$\alpha = \frac{\omega}{t}$

Where

$\alpha =$ Angular acceleration

$\omega =$ Angular velocity

t = Time

Our values are

$v = 60\frac{km}{h} (\frac{1h}{3600s})(\frac{1000m}{1km})$

$v = 16.67m/s$

$r = 0.25m$

$t=6s$

Replacing at the previous equation we have that the angular velocity is

$\omega = \frac{v}{R}$

$\omega = \frac{ 16.67}{0.25}$

$\omega = 66.67rad/s$

Therefore the angular speed of a point on the outer edge of the tires is 66.67rad/s

At the same time the angular acceleration would be

$\alpha = \frac{\omega}{t}$

$\alpha = \frac{66.67}{6}$

$\alpha = 11.11rad/s^2$

Therefore the angular acceleration of a point on the outer edge of the tires is $11.11rad/s^2$

5 0

2 years ago

An automobile engine delivers 55.0 hp. How much time will it take for the engine to do 6.22 × 105 J of work? One horsepower is e

Gennadij [26K]

Answer:

15.2 s

Explanation:

Convert hp to W:

55.0 hp × 746 W/hp = 41,030 W

Power = energy / time

41030 W = 6.22×10⁵ J / t

t = 15.2 s

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2 years ago

A boat travels at 30 mph for a huge, solid cliff that is about 3,000 meters away. When the horn on the boat makes a toot, you ca

AleksandrR [38]

Answer:The frequency of the echo is slightly decreased

Explanation:

Given

speed of boat $=30\ mph$

cliff is $3000\ m$ away

when boat is still , suppose t is the time taken by the echo to reach observer on the boat

But as soon as boat starts moving the distance between cliff and boat decreasing and time for echo to reach observer also decreases

and we know $time \propto \frac{1}{frequency}$

therefore frequency of the echo slightly decreased.

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3 years ago

12. A rocket, initially at rest on the ground, accelerates vertically. It accelerates uniformly until it

vodomira [7]

Answer:

We kindly invite you to read carefully the explanation and check the image attached below.

Explanation:

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$v_{f} = v_{o}+a\cdot (t-t_{o})$ (1)

Where:

$v_{o}$ - Initial velocity, measured in meters per second.

$v_{f}$ - Final velocity, measured in meters per second.

$a$ - Acceleration, measured in meters per square second.

$t_{o}$ - Initial time, measured in seconds.

$t$ - Final time, measured in seconds.

Now we obtain the kinematic equations for thrust and free fall stages:

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$v = 30\cdot t$ (2)

Free fall ( $v_{o} = 900\,\frac{m}{s}$ , $a = -9.807\,\frac{m}{s}$ , $t_{o} = 30\,s$ , $30\,s \le t \le 120\,s$ )

$v = 900-9.81\cdot (t-30)$ (3)

Now we created the graph speed-time, which can be seen below.

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2 years ago