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Leokris [45]
3 years ago
10

How can an element be identified using its emission spectrum?

Physics
1 answer:
posledela3 years ago
4 0
The correct choice is C .

Technically, choice-A is the process involved.  But it's already been done,
and widely published, and the spectrum of every element is easily available. 

For you to take the time and go to the trouble of constructing the spectra
would be like inventing wheels if you want to ride a bicycle. 
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Describe the energy transformation that occurs in a digital clock.
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the ball is accelerating at a rate of 10 m/s squared. the balls mass if 5 kg. what is the net force on the ball? help me men ple
nevsk [136]

Answer:

50N

Explanation:

F=ma\\F=(5kg)*(10m/s^{2})\\F=50N

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2 years ago
When cattle are processed, what is the most likely use for the animals' skin?
galben [10]

When cattle are processed, the most likely use for the animals' skin is in

the production of Leather.

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The commonly used animals in leather production is however cattle and

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3 0
3 years ago
A 2.0-kg projectile is fired with initial velocity components v0x = 30 m/s and v0y = 40 m/s from a point on the Earth's surface.
Alexxandr [17]

Answer:

Kinetic energy of the projectile at the vertex of the trajectory: 900\; {\rm J}.

Work done when firing this projectile: 2500\; {\rm J}.

Explanation:

Since the drag on this projectile is negligible, the horizontal velocity v_{x} of this projectile would stay the same (at 30\; {\rm m\cdot s^{-1}}) throughout the flight.

The vertical velocity v_{y} of this projectile would be 0\; {\rm m\cdot s^{-1}} at the vertex (highest point) of its trajectory. (Otherwise, if v_{y} > 0, this projectile would continue moving up and reach an even higher point. If v_{y} < 0, the projectile would be moving downwards, meaning that its previous location was higher than the current one.)

Overall, the velocity of this projectile would be v = 30\; {\rm m\cdot s^{-1}}\! when it is at the top of the trajectory. The kinetic energy \text{KE} of this projectile (mass m = 2.0\; {\rm kg}) at the vertex of its trajectory would be:

\begin{aligned} \text{KE} &= \frac{1}{2}\, m\, v^{2} \\ &= \frac{1}{2} \times 2.0\; {\rm kg} \times (30\; {\rm m\cdot s^{-1}})^{2} \\ &= 900\; {\rm J} \end{aligned}.

Apply the Pythagorean Theorem to find the initial speed of this projectile:

\begin{aligned}v &= \sqrt{(v_{x})^{2} + (v_{y})^{2}} \\ &= \left(\sqrt{900 + 1600}\right)\; {\rm m\cdot s^{-1}} \\ &= 50\; {\rm m\cdot s^{-1}}\end{aligned}.

Hence, the initial kinetic energy \text{KE} of this projectile would be:

\begin{aligned} \text{KE} &= \frac{1}{2}\, m\, v^{2} \\ &= \frac{1}{2} \times 2.0\; {\rm kg} \times (50\; {\rm m\cdot s^{-1}})^{2} \\ &=2500\; {\rm J} \end{aligned}.

All that energy was from the work done in launching this projectile. Hence, the (useful) work done in launching this projectile would be 2500\; {\rm J}.

7 0
2 years ago
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