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liubo4ka [24]
3 years ago
15

An element emits a spectrum that____________.a)is evenly spaced.b)is the same as all other elements.c)is unique to that element.

d)is indistinguishable from most other elements.e)is evenly spaced and unique to that element.
Physics
1 answer:
Fittoniya [83]3 years ago
5 0

Answer:

c)is unique to that element

Explanation:

 The spectrum emitted by a element shows the internal electronic structure of that element.We know that all element is having different internal electronic structure that is why all element emits different spectrum.And the spacing between the spectrum is uneven .

Therefore the answer will be option C.

c)is unique to that element

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this is a link to a web sight with a diagram to help you

https://www.google.com/url?sa=i&source=images&cd=&ved=2ahUKEwjw4v7knNDgAhVyUN8KHWgWD-wQjRx6BAgBEAU&url=%2Furl%3Fsa%3Di%26source%3Dimages%26cd%3D%26ved%3D%26url%3Dhttps%253A%252F%252Ffuturism.com%252Fwhere-do-all-the-elements-come-from%26psig%3DAOvVaw19_FOCuWs_nMsyY1YT0Da-%26ust%3D1550955269844922&psig=AOvVaw19_FOCuWs_nMsyY1YT0Da-&ust=1550955269844922

7 0
4 years ago
What conditions are required for a solar eclipse?
IgorLugansk [536]
The phase of the Moon must be new, and the nodes of the Moon's orbit must be nearly aligned with Earth and the Sun.
5 0
3 years ago
You and your friends are doing physics experiments on a frozen pond that serves as a frictionless, horizontal surface. Sam, with
Misha Larkins [42]

Answer:

Sam's and Abigail speeds before colliding were a. 12.34 m/s and b. 2.86 m/s, respectively. Their total kinetic energy was diminished by c. 1484.42 J, approximately

Explanation:

By conservation of momentum, we have

\Delta \vec{p} = 0\\\\m\vec{v}_i = m\vec{v}_f\\m_S v_S\hat{i} + m_A v_A\hat{j} = m_S \times 7 \cos{(40)} \hat{i} + m_S \times 7 \sin{(40)} \hat{j} + m_A \times 9.4 \cos{(18)} \hat{i} - m_A \times 9.4 \sin{(18)} \hat{j}.

Writing for each direction at a time,

m_S v_S = 7m_S \cos{(40)} +9.4 m_A \cos{(18)}\\v_S = 7 \cos{(40)} + 9.4 \frac{m_A}{m_S} \cos{(18)} = 7 \cos{(40)} + 9.4\times\frac{57}{73} \cos{(18)} \approx \mathbf{12.34}\\m_A v_A = 7m_S\sin{(40)} - 9.4m_A\sin{(18)}\\v_A = 7\frac{m_S}{m_A}\sin{(40)} - 9.4\sin{(18)} = 7\times\frac{73}{57}\sin{(40)} - 9.4\sin{(18)} \approx \mathbf{2.86}.

Their kinetic energy changed by

K_f - K_i = \left( \frac{1}{2}m_s v_{fs}^2 + \frac{1}{2}m_a v_{fa}^2 \right) - \left( \frac{1}{2}m_s v_{is}^2 + \frac{1}{2}m_a v_{ia}^2 \right) =  \left( \frac{1}{2}\times 73 \times 7^2 + \frac{1}{2}\times 57 \times 9.4^2 \right) - \left( \frac{1}{2}\times 73 \times 12.34^2  + \frac{1}{2}\times 57 \times 2.86^2 \right) \approx \mathbf{-1484.418 J}.

3 0
3 years ago
Find the polar coordinates corresponding to a point located at (−6.48, 12.58) in Cartesian coordinates
barxatty [35]

Answer:

Explanation:

r = √(-6.48² + 12.58²) = 14.1508586311... 14.15

θ = arctan(12.58 / -6.48) = 117.25314692... 117.25

(r, θ) = (14.15, 117.25)

5 0
3 years ago
A fireboat is to fight fires at coastal areas by drawing seawater with a density of 1030 kg/m3 through a 10-cm-diameter pipe at
GaryK [48]

Answer:

50.93 m/s

199.5 kW

Explanation:

From the question, the nozzle exit diameter = 5 cm, Radius= diameter/2= 5cm/2= 2.5cm. we can convert it to metre for unit consistency= (2.5×0.01)=

0.025m

We can calculate the The cross sectional area of the nozzle as

A= πr^2

A= π ×0.025^2

= 1.9635 ×10^- ³ m²

From the question, the water is moving through the pipe at a rate of 0.1 m /s , then for the water to move through it at a seconds, it must move at

(0.1 / 1.9635 ×10^- ³ m²)

= 50.93 m/s

During the Operation of the pump, the Dynamic energy of the water= potential energy provided there is no loss during the Operation

mgh = 1/2mv²

We can make "h" subject of the formula, which is the height of required head of water

h = (1/2mv²)/mg

h= v² / 2g

h = 50.93² / (2 ×9.81)

h = 132.21m

From the question;

The total irreversible head loss of the system = 3 m,

the given position of nozzle = 3 m

the total head the pump needed=(The total irreversible head loss of the system + the position of the nozzle + required head of water )

=(3 + 3 + 132.21m)

=138.21m

mass of water pumped in a seconds can be calculated since we know that mass is a product of volume and density

Volume= 0.1m³

Density of sea water=1030 kg/m

(0.1 m^3× 1030)

= 103kg

We can calculate the Potential enegry, which is = mgh

= (103 ×9.81 × 138.21)

= 139651.5 Watts

= 139.65kW

To determine required shaft power input to the pump and the water discharge velocity

Energy= efficiency × power

But we are given efficiency of 70 percent, then

139651.5 Watts = 0.7P

=199502.18 Watts

P=199.5 kW

Therefore, the required shaft power input to the pump and the water discharge velocity is 199.5 kW

5 0
3 years ago
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