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2 years ago

What is the frequency of a monochromatic light used in a diffraction experiment that has a wavelength of 6.38 ✕ 10e-07 m?

Physics

1 answer:

Lady_Fox [76]2 years ago

4 0

Answer:

$f=4.70\times 10^{14}\ Hz$

Explanation:

Given that,

The wavelength of light, $\lambda=6.38\times 10^{-7}\ m$

We need to find the frequency of the light. We know that,

$c=f\lambda\\\\f=\dfrac{c}{\lambda}\\\\f=\dfrac{3\times 10^8}{6.38\times 10^{-7}}\\\\f=4.70\times 10^{14}\ Hz$

So, the required frequency of light is equal to $4.70\times 10^{14}\ Hz$ .

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If the equation of an average velocity of a sport car was given by the equation V(t) = 3t^2 -6t +24. Determine its displacement

alisha [4.7K]

Answer:

Explanation:

The displacement of an object can be found from the velocity of the object by integrating the expression for the velocity.

In this problem, the velocity of the sport car is given by the expression

$v(t)=3t^2-6t+24$

In order to find the expression for the position of the car, we integrate this expression. We find:

$x(t)=\int v(t) dt=t^3-3t^2+24t+C$

where C is an arbitrary constant.

Here we want to find the displacement after 3 seconds. The position at t = 0 is

$x(0)=0^3-0+0+C=C$

While the position after t = 3 s is

$x(3)=3^3-3(3)^2+24(3)+C=72+C$

Therefore, the displacement of the car in 3 seconds is

$d=x(3)-x(0)=72+C-C=72$

7 0

2 years ago

How long will it take a ball to roll 10 meters along the floor at a speed of 0.5m/s

Julli [10]

Answer:

Explanation:

because m/s=s/t so,

0.5m/s=10m/t

t=10m/0.5m/s

t=5sec

5 0

7 months ago

You kick a ball with a speed of 14 m/s at an angle of 51°. How far away does the ball land?

In-s [12.5K]

-- The vertical component of the ball's velocity is 14 sin(51°) = 10.88 m/s

-- The acceleration of gravity is 9.8 m/s².

-- The ball rises for 10.88/9.8 seconds, then stops rising, and drops for the
same amount of time before it hits the ground.

-- Altogether, the ball is in the air for (2 x 10.88)/(9.8) = 2.22 seconds
==================================

-- The horizontal component of the ball's velocity is 14 cos(51°) = 8.81 m/s

-- At this speed, it covers a horizontal distance of (8.81) x (2.22) = 19.56 meters
before it hits the ground.

As usual when we're discussing this stuff, we completely ignore air resistance.

4 0

2 years ago

Read 2 more answers

The outer layer of cable on a cable reel is 16.2 cm from the center of the reel. The reel is initially stationary and can rotate

ahrayia [7]

Answer:

B. w=12.68rad/s

C. α=3.52rad/s^2

Explanation:

We can solve this problem by taking into account that (as in the uniformly accelerated motion)

$\theta=\omega_{0}t+\frac{1}{2}\alpha t^{2}\\\theta = \frac{s}{r}$ ( 1 )

where w0 is the initial angular speed, α is the angular acceleration, s is the arc length and r is the radius.

In this case s=3.7m, r=16.2cm=0.162m, t=3.6s and w0=0. Hence, by using the equations (1) we have

$\theta=\frac{3.7m}{0.162m}=22.83rad$

$22.83rad=\frac{1}{2}\alpha (3.6s)^2\\\\\alpha=2\frac{(22.83rad)}{3.6^2s}=3.52\frac{rad}{s^2}$

to calculate the angular speed w we can use $\alpha=\frac{\omega _{f}-\omega _{i}}{t _{f}-t _{i}}\\\\\omega_{f}=\alpha t_{f}=(3.52\frac{rad}{s^2})(3.6)=12.68\frac{rad}{s}$

Thus, wf=12.68rad/s

C) We can use our result in B)

$\alpha=3.52\frac{rad}{s^2}$

I hope this is useful for you

regards

3 0

2 years ago

Read 2 more answers

Due to efficiency considerations related to its bow wake, the supersonic transport aircraft must maintain a cruising speed that

givi [52]

Answer:

decreases.

Explanation:

When the aircraft is flies from the warm air into the colder air then its speed will be decreases.

as we know that

we know mach number is constant

so that here Mach number M is expressed as

M = $\frac{u}{v}$ .............................1

here u is Local flow velocity with respect to the boundarie and v is the speed of sound in the medium

If the aircraft flies from hot air to cold air, the speed of sound in the medium will decrease. But the Mach number remains constant. Therefore, the local flow velocity relative to the boundaries also decreases.

7 0

2 years ago