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galina1969 [7]
3 years ago
11

A wave of infrared light has a speed of 6 m/s and a wavelength of 12 m. What is the frequency of this wave?

Physics
2 answers:
givi [52]3 years ago
5 0
I'll be happy to solve the problem using the information that
you gave in the question, but I have to tell you that this wave
is not infrared light.

If it was a wave of infrared, then its speed would be close
to 300,000,000 m/s, not 6 m/s, and its wavelength would be
less than 0.001 meter, not 12 meters.

For the wave you described . . .

             Frequency  =  (speed)  /  (wavelength)

                                 =  (6 m/s)  /  (12 m)

                                 =      0.5 / sec

                                 =      0.5 Hz .  

(If it were an infrared wave, then its frequency would be
greater than 300,000,000,000 Hz.)
Anna [14]3 years ago
4 0
Frequency(ν) = speed of infrared light( c )/wavelength( λ ) = 6/12 = 1/2 = 0.5 s^-1 . 
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Answer:

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Explanation:

6 0
4 years ago
Some birds migrate 20,000 miles. If 1 mile equals 1.6 kilometers, calculate the distance these birds fly in kilometers.
faust18 [17]
Set up a proportion

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6 0
4 years ago
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A weight lifter is trying to do a bicep curl with a weight of 300 N. At the "sticking point", the moment arm of this weight is 3
lesantik [10]

Answer:

The weight lifter would not get past this sticking point.

Explanation:

Generally torque applied on the weight is mathematically represented as

             T =  F z

To obtain Elbow torque we substitute 4000 N for F (the force ) and 2cm = \frac{2}{100} = 0.02m for z the perpendicular distance

So Elbow Torque is   T_e= 4000 * 0.02

                                   = 80Nm

 To obtain the torque required we substitute 300 N for F and 30cm =\frac{30}{100} = 0.3 m

  So the Required Torque is T_R = 300 *0.3

                                                     =90Nm

Now since   T_e < T_R it mean that the weight lifter would not get past this sticking point

                                   

7 0
3 years ago
A charge of 7.2 × 10-5 C is placed in an electric field with a strength of 4.8 × 105 StartFraction N over C EndFraction. If the
barxatty [35]

Answer:

2.2 meters

Explanation:

Potential energy, PE created by a charge, q at a radius r from the charge source, Q,  is expressed as:

KE=\frac{kQq}{r}\     \ \ \ \ \ \ ...i

k is Coulomb's constant.

#The electric field,E at radius r is expressed as:

E=\frac{kQ}{r^2}\ \ \ \ \ \ \ \ \ \ ...ii

From i and ii, we have:

KE=Eqr

r=(KE)/Eq

#Substitute actual values in our equation:

r=\frac{75J}{(7.2\times 10^{-5}C)(4.8\times 10^5 V/m)}\\\\=2.1701\approx2.2\ m

Hence, the distance between the charge and the source of the electric field is 2.2 meters

7 0
3 years ago
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A 5 kg block is being pushed horizontally across a level surface at a constant velocity. What is the magnitude of the Normal for
luda_lava [24]

Answer:

50 N.

Explanation:

On top of a horizontal surface, the normal force acting on an object is equivalent to the force of gravity acting on the object. That is:

\displaystyle \begin{aligned} F_N = F_g & = ma \\ & = mg\end{aligned}

The mass of the block is 5 kg and the given force due to gravity is 10 N/kg. Substitute and evaluate:

\displaystyle F_N = F_g = (5\text{ kg})(10 \text{ N/kg}) = 50 \text{ N}

In conclusion, the normal force acting on the block is 50 N.

5 0
2 years ago
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