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Goshia [24]
3 years ago
13

When using the spectroscope to view the spectra of the discharge lamps, what observations are recorded?.

Physics
1 answer:
natima [27]3 years ago
8 0

When using the spectroscope to view the spectra of the discharge lamps, what observations are recorded by the wavelengths and colors of all bands.

<h3>Spectra of discharge lamps</h3>

The spectra of discharge lamp from spectroscope shows the different electromagnetic spectrum, they include the following;

  • Ultraviolet rays
  • Visible lights
  • Infra-red, etc

The electromagnetic spectrum consists of radiations of different wavelengths which are indicated by different color bands.

Thus, when using the spectroscope to view the spectra of the discharge lamps, what observations are recorded by the wavelengths and colors of all bands.

Learn more about spectroscope here: brainly.com/question/2114494

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A bowling ball with a momentum of 18kg-m/s strikes a stationary bowling pin. After the collision, the ball has a momentum of 13k
Veronika [31]

Answer:

14.98\ \text{kg m/s}

45.26^{\circ}

Explanation:

P_1 = Initial momentum of the pin = 13 kg m/s

P_i = Initial momentum of the ball = 18 kg m/s

P_2 = Momentum of the ball after hit

55^{\circ} = Angle ball makes with the horizontal after hitting the pin

\theta = Angle the pin makes with the horizotal after getting hit by the ball

Momentum in the x direction

P_i=P_1\cos55^{\circ}+P_2\cos\theta\\\Rightarrow P_2\cos\theta=P_i-P_1\cos55^{\circ}\\\Rightarrow P_2\cos\theta=18-13\cos55^{\circ}\\\Rightarrow P_2\cos\theta=10.54\ \text{kg m/s}

Momentum in the y direction

P_1\sin55=P_2\sin\theta\\\Rightarrow P_2\sin\theta=13\sin55^{\circ}\\\Rightarrow P_2\sin\theta=10.64\ \text{kg m/s}

(P_2\cos\theta)^2+(P_2\sin\theta)^2=P_2^2\\\Rightarrow P_2=\sqrt{10.54^2+10.64^2}\\\Rightarrow P_2=14.98\ \text{kg m/s}

The pin's resultant velocity is 14.98\ \text{kg m/s}

P_2\sin\theta=10.64\\\Rightarrow \theta=sin^{-1}\dfrac{10.64}{14.98}\\\Rightarrow \theta=45.26^{\circ}

The pin's resultant direction is 45.26^{\circ} below the horizontal or to the right.

4 0
3 years ago
According to Newton's<br> Ist law, what will an<br> object in motion tend<br> to do?
SCORPION-xisa [38]

Answer:

According to <em>Newton's first law of motion:</em>

<u>An object in motion tends to remain in motion unless an external force acts upon it.</u>

<u>It stays in motion with the same speed and goes in the same direction.</u>

<u></u>

<em>Hope this helped </em>

<em>:)</em>

4 0
3 years ago
Read 2 more answers
A hypothesis may also be called a(n) _____.
gavmur [86]
This is the answer to what a hypothesis may also be called

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3 years ago
4. Johnny exerts a 3.55 N rightward force on a 0.200-kg box to accelerate it across a low-friction track. If the total resistanc
Anon25 [30]

a) 15.2 m/s^2

b) 1.96 N

c) 1.96 N

Explanation:

a)

To find the box's acceleration, we have to find first the net force acting on the box in the horizontal direction.

We have:

- The forward force of 3.55 N

- The backward, resistive force of 0.52 N

So, the net force forward is

\sum F=3.55-0.52=3.03 N

Now we can find the acceleration by using Newton's second law of motion, which states that:

\sum F=ma

where

m = 0.200 kg is the mass of the box

a is its acceleration

And solving for a, we find the acceleration:

a=\frac{\sum F}{m}=\frac{3.03}{0.200}=15.2 m/s^2

b)

The gravitational force on an object is the force with which the object is pulled towards the ground by the Earth.

It is given by

W=mg

where

m is the mass of the object

g is the gravitational field strength

In this problem we have

m = 0.200 kg is the mass of the box

g=9.8 m/s^2 is the gravitational field strength

So, the gravitational force on the box is

W=(0.200)(9.8)=1.96 N

c)

The normal force is the reaction force exerted by the floor on the box, in the upward direction.

In order to find the magnitude of this force, we apply Newton's second law of motion along the vertical direction.

We have two forces in this direction:

- The gravitational force, W, downward

- The normal force, N, upward

So the net force is

\sum F=N-W

According to Newton's second law,

\sum F=ma

However, the box is at rest in the vertical direction, so the vertical acceleration is zero:

a=0

This means that the net force is zero:

\sum F=0

And so, we can find the normal force:

N-W=0\\N=W=1.96 N

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The Earth’s orbit around the Sun is slightly elliptical. At Earth's closest approach to the Sun (perihelion) the orbital radiu
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Answer:

1.25

Explanation:

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