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LuckyWell [14K]

4 years ago

10

A microwave oven operates at 2.00 ghz . what is the wavelength of the radiation produced by this appliance? express the waveleng

th numerically in nanometers.

1 answer:

yan [13]4 years ago

7 0

<span>Since frequency and wavelength have inverse relationship. It can be expressed by the equation: Î˝.Î» = c Where, v = frequency of the electromagnetic wave. Î» = it's wavelength c = the speed of light in a vacuum. v = 2.00 Ghz x 10^9 Hz / 1 Ghz = 2.00 x 10^9 Hz that means that in one second it covers 2.00 x 10^9 cycles. Î» = 3.10^8 m/s / 2.00 x 10^9 /s = 1.25E-10 nanometers</span>

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Suppose a diving board with no one on it bounces up and down in a simple harmonic motion with a frequency of 6.90 Hz. The board

Kobotan [32]

Answer:

The frequency of the simple harmonic motion of a 67.6 kg diver on the board = 2.48 Hz

Explanation:

The frequency in simple harmonic motion is related to spring constant and mass causing the motion through the relation

f = (1/2π) √(k/m)

When mass = 10 kg, f = 6.90 Hz,

6.9 = (1/2π) √(k/10)

(√(k/10) = 6.9×2π = 43.354

k/10 = 43.354² = 1879.57

K = 18795.7 N/m

When a diver of mass 67 kg climbs the diving board, the total mass on the diving board now becomes (10+67.6) = 77.6 kg

Spring constant of the diving board doesn't change,

So, the frequency is then given by

f = (1/2π) √(18795.7/77.6)

f = 2.48 Hz

8 0

3 years ago

Read 2 more answers

Which statement best describes who the period and frequency of electromagnetic was change between gamma rays and microwaves?

nalin [4]

Answer : <em> The period increases and the frequency decreases</em>.

Explanation :

Gamma rays are the rays having frequency( $\nu$ ) greater than $10^{19}\ Hz$ and frequency of microwave lie between 300 MHz to 300 GHz.

Since, $E=h\nu$

So, as we move from gamma rays to microwaves the energy increases because frequency is increasing.

Also, Time period $T=\dfrac{1}{\nu}$

We can say that as the time period increases, frequency decreases.

6 0

3 years ago

Read 2 more answers

John throws a rock down with speed 14 m/s from the top of a 30 m tower. If air resistance is negligible, what is the rock's spee

astra-53 [7]

The final velocity of the rock before it touches the ground is 28 m/s.

Answer:

Explanation:

As the rock is thrown down, this means the acceleration due to gravity will be exerting on the rock. So the rock will be exhibiting a free fall motion. Thus, the acceleration of the rock will be equal to the magnitude of acceleration due to gravity. Then using the third equation of motion, we can determine the final velocity of the rock provided the values for initial velocity, displacement and acceleration is given in the problem itself.

So the acceleration is equal to 9.8 m/s² due to its free fall motion and displacement will be equal to the height of the tower which is given as 30 m. And the initial speed of the rock is stated as 14 m/s. The initial speed is represented as u, final speed is represented as v, displacement is represented as s and acceleration is represented as a.

$2as=v^{2}-u^{2}$

Then, 2 × 9.8 × 30 = v²-(14)²

v²=784

v= 28 m/s

So the final velocity of the rock before it touches the ground is 28 m/s.

8 0

4 years ago

A parachutist bails out and freely falls 63 m. Then the parachute opens, and thereafter she decelerates at 1.5 m/s2. She reaches

yaroslaw [1]

Answer:

(a) The parachutist spent 24.84 secs in air

(b) The height the fall begins is 472 m

Explanation:

Here is the complete question:

A parachutist bails out and freely falls 63 m. Then the parachute opens, and thereafter she decelerates at 1.5 m/s2. She reaches the ground with a speed of 3.3 m/s. (a) How long is the parachutist in the air (b) At what height does the fall begin?

Explanation:

From one of the equations of kinematics for free fall

$H = ut - \frac{1}{2}gt^{2}$

Where H is the height

u is the initial velocity

t is the time

and g is the acceleration due to gravity (Take g = 9.8 m/s2)

Now, we can find the time spent before the parachute opens.

u = 0 m/s (we assume the parachutist starts from rest)

H = - 63 m

∴ $-63 = 0(t) - \frac{1}{2}(0.98) t^{2} \\-63 = -4.9 t\\t^{2} = \frac{63}{4.9} \\t^{2} = 12.86\\t = \sqrt{12.86} \\t = 3.59 s$

This the time spent before the parachute opens

Also from one of the equations of kinematics for free fall

$v = u - gt$

where v is the final velocity

We can determine the final velocity before the parachute opens and she starts to decelerate

∴ $v = - 9.8(3.59)\\v = - 35.18m/s$

Now, we will calculate the time spent after the parachute opens

From one of the equation of kinematics for linear motion,

$v = u + at$

Here, the initial velocity will be the final velocity just before the parachute opens, that is

$u = - 35.18 m/s$

From the question,

$v = - 3.3 m/s$

$a = 1.5 m/s^{2}$

We then get

$-3.3 = - 35.18 + (1.5)t$

$-3.3 + 35.18 = 1.5t\\31.88 = 1.5t$

$t = \frac{31.88}{1.5}$

$t = 21.25 secs$

(a) To determine how long the parachutist is in the air,

That is sum of the time used when falling freely and the time used after the parachute opens

= 3.59 secs + 21.25 secs

24.84 secs

Hence, the parachutist spent 24.84 secs in air

(b) To determine what height the fall begins

First, we will calculate the height from which the parachute opens

From one of the equation of kinematics for linear motion,

$x = ut + \frac{1}{2}at^{2} \\$

$x = -35.18(21.25) + \frac{1}{2}(1.5)(21.25)^{2} \\x = -747.58 + 338.67\\x = -408.91m\\$

$x$ ≅ $- 409m$

Hence, the height the fall begins is 63m + 409m

= 472 m

3 0

3 years ago

Fix this problem please.

Vinil7 [7]

Answer:

In my knowledge....

Option E.........

Explanation:

I have no doubts it is option E 350N

4 0

3 years ago