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3 years ago

Which of the following devices is used in this experiment to separate the different wavelengths emitted by the light source?

Physics

1 answer:

adelina 88 [10]3 years ago

7 0

Answer:

A. Prism

Explanation:

In the experiment we are able to isolate different wavelengths by shining the light through a prism, which separates the different wavelengths through refraction.

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A star is held together by:gravitynuclear fissionnuclear fusionacceleration

alina1380 [7]

Nuclear fusion in the core tries to blow the star apart. Gravity holds it together. Whoever designed that system really knew what he was doing. I'm kinda grateful to him.

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3 years ago

Real springs have mass. How will the true period andfrequency

Ad libitum [116K]

Explanation:

An perfect mass less spring, attached at one end and with a free mass attached at the other end, will have a distinct frequency of oscillation depending on its constant spring and mass. On the other hand, a spring with mass along its length will not have a characteristic frequency of oscillation.

Alternatively, based on its spring constant and mass per length, it will now have a wave Speed. It would be possible to use all wavelengths and frequencies, as long as the component fλ= S, where S is the spring wave size. If that sounds like longitudinal waves, like solid sound waves.

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4 years ago

How do you comprehend the fourth dimension?

sergij07 [2.7K]

The fourth dimension is technically time. the fourth dimension that you are talking about is actually impossible to comprehend.

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3 years ago

Estimate the wavelength corresponding to maximum emission from each of the following surfaces: the sun, a tungsten filament at 2

igomit [66]

Answer

Applying Wein's displacement

$\lamda_{max}\ T = 2898 \mu_mK$

1) for sun T = 5800 K

$\lambda_{max} = \dfrac{2898}{5800}$

$\lambda_{max} = 0.5 \mu_m$

2) for tungsten T = 2500 K

$\lambda_{max} = \dfrac{2898}{2500}$

$\lambda_{max} = 1.16 \mu_m$

3) for heated metal T = 1500 K

$\lambda_{max} = \dfrac{2898}{1500}$

$\lambda_{max} = 1.93 \mu_m$

4) for human skin T = 305 K

$\lambda_{max} = \dfrac{2898}{305}$

$\lambda_{max} = 9.50 \mu_m$

5) for cryogenically cooled metal T = 60 K

$\lambda_{max} = \dfrac{2898}{60}$

$\lambda_{max} = 48.3 \mu_m$

range of different spectrum

UV ----0.01-0.4

visible----0.4-0.7

infrared------0.7-100

for sun T = 5800

λ 0.01 0.4 0.7 100

λT 58 2320 4060 5.8 x 10⁵

F 0 0.125 0.491 1

fractions

for UV = 0.125

for visible = 0.441-0.125 = 0.366

for infrared = 1 -0.491 = 0.509

8 0

4 years ago

2. Two toy cars are involved in a race. Car A has mass m while car B has mass 2m. a. The two cars have the same force applied to

ArbitrLikvidat [17]

Answer:

a) The kinetic energy of the two cars is the same

the moment of car 2 is greater than the moment of car 1

b) the kinetic energy of car 1 is greater than that of car 2

the moment of the two cars is the same

Explanation:

a) to know the kinetic energy of each car, we must find the speed, use Newton's second law to find the acceleration

Car 1

F = m a

a = F / m

Let's use kinematics to find the velocity after x = 1 m

v² = v₀² + 2 a x

The initial speed is zero

v = √ (2 F/m x)

For the distance of x = 1 m

v₁ = √ (2 F / m)

Car 2

F = 2m a

a = F / 2m

v² = 2 a x

v = √ (F/m x)

For x = 1 m

v₂ = √(F / m)

Let's calculate the kinetic energy of each car

Car 1

K₁ = ½ m v₁²

K₁ = ½ m 2F / m

K₁ = F

Car 2

K₂ = ½ 2m v₂²

K₂ = ½ 2m F / m

K₂ = F

The kinetic energy of the two cars is the same

Let's calculate the moment

Car 1

P₁ = m v₁

P₁ = m √ (2F / m)

Car 2

P₂ = 2m v²

P₂ = 2m √(F / m)

We see that the moment of car 2 is greater than the moment of car 1

b) in this part the force is applied by t = 10 s

Acceleration is the same, let's find the speed

Car1

v = v₀ + a t

v = F / m t

v₁ = F / m 10

Car 2

v₂ = F / 2m 10

v₂ = F / m 5

Let's calculate the kinetic energy of each car

Car 1

K₁ = ½ m v₁²

K₁ = ½ m (F / m 10)²

K₁ = 50 F² / m

Car2

K₂ = ½ 2m v₂²

K₂ = m (F / m 5)²

K₂ = 25 F² / m

In this case we see that the kinetic energy of car 1 is greater than that of car 2

Let's calculate the moment

Car 1

P₁ = m v₁

P₁ = m F / m 10

P₁ = 10 F

Car 2

P₂ = 2m v₂

P₂ = 2m F / m 5

P₂ = 10 F

In this case the moment of the two cars is the same

7 0

3 years ago