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balandron [24]
3 years ago
14

Infrared radiation has wavelengths ranging from about 800 nm to 1 mm. what is the frequency of radiation of wavelength 960 nm? a

nswer in units of s−1 .
Physics
1 answer:
Xelga [282]3 years ago
3 0
The correct answer to your question and how to solve it is
 
The relation between wavelength (λ)and the frequency of electromagnetic oscillation (f) is described by the following expression: λ=c/f, where c–is the speed of light in vacuum = 3*10^8 m/s
Derive f from above: f = c/λ.How to Calculate: λ=890nm = 890*10^-9m = 8.9*10^-7m
f =3*10^8m/s Divided by 8.9*10^-7m = 0.34*10^15 s-1=3.4*10^14 s-1
So your Answer is: The frequency of radiation of wavelength 890 nm is 3.4*10^14s-1
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A ball is thrown straight upward and rises to a maximum
Leviafan [203]

Answer:

Explanation:

As we know that the ball is projected upwards so that it will reach to maximum height of 16 m

so we have

v_f^2 - v_i^2 = 2 a d

here we know that

v_f = 0

also we have

a = -9.81 m/s^2

so we have

0 - v_i^2 = 2(-9.81)(16)

v_i = 17.72 m/s

Now we need to find the height where its speed becomes half of initial value

so we have

v_f = 0.5 v_i

now we have

v_f^2 - v_i^2 = 2 a d

(0.5v_i)^2 - v_i^2 = 2(-9.81)h

-0.75v_i^2 = -19.62 h

0.75(17.72)^2 = 19.62 h

h = 12 m

3 0
3 years ago
Wire sizes are often reported using the AWG (American Wire Gauge) system in which smaller diameter wires are said to be of highe
NISA [10]

Answer:

Explanation:

24 - gauge wire , diameter = .51 mm .

Resistivity of copper ρ = 1.72 x 10⁻⁸ ohm-m

R = ρ l / s

1.72x 10⁻⁸ / [3.14 x( .51/2)² x 10⁻⁶ ]

= 8.42 x 10⁻² ohm

= .084 ohm

B )  Current required through this wire

= 12 / .084 A

= 142.85 A

C )

Let required length be l

resistance = .084 l

2 = 12 / .084 l

l = 12 / (2 x .084)

= 71.42 m

6 0
3 years ago
Read 2 more answers
What is the density of an object that has a mass of 30 g and a volume of 20cm cubed/ to the third power?
Ksenya-84 [330]

Answer:

d= 1.5 g/cm3

Explanation:

datos

m= 30g

v= 20cm3

d=?

formula

d= m / v

solución

d= 30g / 20cm3 = 1.5g/cm3

6 0
3 years ago
A 45 kg bear slides, from rest, 11 m down a lodgepole pine tree, moving with a speed of 5.8 m/s just before hitting the ground.
Viefleur [7K]

Answer:

Explanation:

a )

change  in the gravitational potential energy of the bear-Earth system during the slide  = mgh

= 45 x 9.8 x 11

= 4851 J

b )

kinetic energy of bear just before hitting the ground

= 1/2 m v²

= .5 x 45 x 5.8²

= 756.9 J

c ) If  the average frictional force that acts on the sliding bear be F

negative work done by friction

= F x 11 J

then ,

4851 J -  F x 11 =  756.9 J

F x 11 = 4851 J -   756.9 J

= 4094.1 J

F = 4094.1 / 11

= 372.2 N  

4 0
3 years ago
Suppose that a freely falling object were somehow equipped with an odometer. Would the readings of distance fallen each second i
maks197457 [2]

Answer:

1  greater distances fallen in successive seconds

Explanation:

When a body falls freely it is subjected to the action of the force of gravity, which gives an acceleration of 9.8 m / s2, consequently, we are in an accelerated movement

If we use the kinematic formula we can find the position of the body

       Y = Vo t + ½ to t2

Where the initial velocity is zero or constant and the acceleration is the acceleration of gravity

Y = - ½ g t2 = - ½ 9.8 t2 = -4.9 t2

Let's look for the position for successive times

t (s)      Y (m)

  1          -4.9

  2         -19.6

   3        -43.2

The sign indicates that the positive sense is up

It can be clearly seen that the distance is greatly increased every second that passes

3 0
3 years ago
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