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2 years ago

The wavelengths of visible light vary from about 300 nm to 700 nm. What is the range of frequencies of visible light in a vacuum

Physics

1 answer:

exis [7]2 years ago

6 0

The range of frequencies of visible light in a vacuum is mathematically given as

Fmin=4.19*10^14Hz to Fmax=1*10^15Hz

<h3>What is the range of frequencies of visible light in a vacuum?</h3>

Question Parameters:

The wavelengths of visible light vary from about 300 nm to 700 nm.

Generally, the equation for the frequency is mathematically given as

F=C/\lambda

Therefore

For Fmax

$Fmax=\frac{300*10^8}{3*10^9}$

Fmax=1*10^15Hz

Where

$Fmin=\frac{3*10^8}{700*10^9}$

Fmin=4.19*10^14Hz

For more information on Wave

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An egg of mass 0.060 kg is dropped from the top of a building. Just before it reaches the ground, it has a total kinetic energy

3241004551 [841]

The velocity is 14 m/s

The parameters given on the question are

mass= 0.060 kg

kinetic energy= 5.9 joules

K.E= 1/2mv²

5.9= 1/2 × 0.060 × v²

5.9= 0.5 × 0.060v²

5.9= 003v²

v²= 5.9/0.03

v²= 196.66

v= √196.66

v= 14 m/s

Hence the velocity of the egg before it strikes the ground is 14 m/s

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3 years ago

A pitcher throws a 0.140 kg baseball, and it approaches the bat at a speed of 35.0 m/s. The bat does Wnc = 75.0 J of work on the

Eva8 [605]

Answer:

The speed of the ball is 42.5 m/s

Explanation:

The initial kinetic energy of the ball is:

$K_1=\frac{1}{2} m v_0^2=\frac{1}{2}*0.140 kg*(35.0 m/s)^2$ = 85.75 J

The speed of the ball after leaving the bat is:

$K_2=K_1+W_{nc}\\ \frac{1}{2}mV^2= 85.75 J + 75 J\\ (\frac{1}{2}mV^2)2=( 160.75 J)2\\ mV^2= 321.5 J\\ V^2= \frac{321.5 J}{0.140kg} \\ V=\sqrt{\frac{321.5 J}{0.140kg}}$

V=47.92 m/s

Using kinematic equation we can find the speed of the ball after being 25 m above the point of collision:

$V_f^2-V^2=-2gh$

$V_f^2-(47.92 m/s)^2=-2*9.81m/s^2*25m$

$V_f^2=-2*9.81m/s^2*25m+(47.92 m/s)^2$

$V_f=\sqrt{-2*9.81m/s^2*25m+(47.92 m/s)^2}$

$V_f=42.5m/s$

3 0

3 years ago

A simple harmonic oscillator consists of a block (m = 0.50 kg) attached to a spring (k = 128 N/m). The block is pulled a certain

Zinaida [17]

Answer:

0.5 m

14.00595

8 m/s, 0.0625 s

5.71314 m/s

Explanation:

k = Spring constant = 128 N/m

A = Amplitude

E = Energy in spring = 16 J

Energy in spring is given by

$E=\dfrac{1}{2}kA^2\\\Rightarrow A=\sqrt{\dfrac{2E}{k}}\\\Rightarrow A=\sqrt{\dfrac{2\times 16}{128}}\\\Rightarrow A=0.5\ m$

The amplitude is 0.5 m

Time period is given by

$T=2\pi\sqrt{\dfrac{m}{k}}\\\Rightarrow T=2\pi\sqrt{\dfrac{0.5}{128}}\\\Rightarrow T=0.39269\ s$

Number of oscillations is given by

$N=\dfrac{5.5}{0.39269}\\\Rightarrow N=14.00595$

The number of oscillations is 14.00595

For maximum speed

$\dfrac{1}{2}mv^2=16\\\Rightarrow v=\sqrt{\dfrac{16\times 2}{0.5}}\\\Rightarrow v=8\ m/s$

The maximum speed is 8 m/s

For a distance of 0.5 m which is the amplitude

$Time=\dfrac{Distance}{Speed}\\\Rightarrow Time=\dfrac{0.5}{8}\\\Rightarrow Time=0.0625\ s$

The time taken would be 0.0625 s

The maximum kinetic energy is equal to the mechanical energy

$\dfrac{1}{2}mv^2+\dfrac{1}{2}kx^2=16$

At x = 0.35 m

$v=\sqrt{\dfrac{16-\dfrac{1}{2}kx^2}{\dfrac{1}{2}m}}\\\Rightarrow v=\sqrt{\dfrac{16-\dfrac{1}{2}128\times 0.35^2}{\dfrac{1}{2}0.5}}\\\Rightarrow v=5.71314\ m/s$

The speed of the block is 5.71314 m/s

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3 years ago

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Answer: im trying to find the same answer too

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2 years ago

Which of the following describes the relationship between work and power?

Vinvika [58]

Answer: The answer is B

Explanation:

Work can be defined as the energy that is required to apply a force to an object in order to move it from one point to another. In physics, work = force x distance travelled. On the other hand, Power is the work done per time. In other words, it the rate at which work is done and is determined by using the formula, Power = Work/time. In these relationships, it can be seen that power is directly proportional to the amount of work done, hence as power increases, more work is done.

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3 years ago