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ehidna [41]
3 years ago
7

gayle cooks a roast in her microwave oven. the klystron tube in the oven emits photons whose energy is 1.20 x 10^-3 ev. what are

the wavelengths of these photons
Physics
1 answer:
AveGali [126]3 years ago
7 0

Answer:

\lambda=1.03\times 10^{-3}\ m

Explanation:

Given that,

The energy of the microwave oven is 1.2\times 10^{-3}\ eV.

We need to find the wavelength of these photons.

1.2\times 10^{-3}\ eV=1.2\times 10^{-3}\times 1.6\times 10^{-19}\\\\=1.92\times 10^{-22}\ J

The energy of a wave is given by :

E=\dfrac{hc}{\lambda}\\\\\lambda=\dfrac{hc}{E}

Put all the values,

\lambda=\dfrac{6.63\times 10^{-34}\times 3\times 10^8}{1.92\times 10^{-22}}\\\\\lambda=1.03\times 10^{-3}\ m

So, the wavelength of these photon is 1.03\times 10^{-3}\ m.

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An important safety consideration with an electromagnet is to disconnect the wires when not in use because they can get ________
Rashid [163]

Answer:

Hot (A)

Explanation:

because there are electrons going through there, the friction between the metallic nuclei and electrons creates vibrations known as heat.

if you touch it the electrons will run through your body and caus damadge.

5 0
3 years ago
The value of the normal force exerted on a 2,000 gram block on a table
kobusy [5.1K]

Answer:

N = 19.6 N

Explanation:

Given that,

Mass of a block, m = 2000 g

1 kg = 1000 g

It means, 2000 g = 2kg

We need to find the value of normal force on the block on a table. Normal force is balanced by the weight of the block as follows :

N = mg, g is acceleration due to gravity

N = 2 kg × 9.8 m/s²

N = 19.6 N

So, the normal force acting on the block is 19.6 N.

4 0
3 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
A motorboat traveling on a straight course slows
never [62]

Answer:

2.572 m/s²

Explanation:

Convert the given initial velocity and final velocity rates to m/s:

  • 65 km/h → 18.0556 m/s
  • 35 km/h → 9.72222 m/s

The motorboat's displacement is 45 m during this time.

We are trying to find the acceleration of the boat.

We have the variables v₀, v, a, and Δx. Find the constant acceleration equation that contains all four of these variables.

  • v² = v₀² + 2aΔx

Substitute the known values into the equation.

  • (9.72222)² = (18.0556)² + 2a(45)
  • 94.52156173 = 326.0046914 + 90a
  • -231.4831296 = 90a
  • a = -2.572

The magnitude of the boat's acceleration is |-2.572| = 2.572 m/s².

3 0
3 years ago
A.to reduce<br> b.to dispose<br> c.to prevent<br> d.to help
kenny6666 [7]

Answer:

to prevent cuz now for my mind when you prevent it it will not happen to you or nothing happened to you

5 0
3 years ago
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