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3 years ago

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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.

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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Umar has two copper pans, each containing 500cm3 of water. Pan A has a mass of 750g and pan B has a mass of 1.5kg. Which pan wil

Olin [163]

Answer:

heat required in pan B is more than pan A

Explanation:

Heat required to raise the temperature of the substance is given by the formula

$Q = ms\Delta T$

now we know that both pan contains same volume of water while the mass of pan is different

So here heat required to raise the temperature of water in Pan A is given as

$Q_1 = (m_w s_w + m_ps_p)\delta T$

$Q_1 = (0.5(4186) + 0.750(s))\Delta T$

Now similarly for other pan we have

$Q_2 = (m_w s_w + m_ps_p)\delta T$

$Q_2 = (0.5(4186) + 1.50(s))\Delta T$

So here by comparing the two equations we can say that heat required in pan B is more than pan A

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An electron is released from rest in a uniform electric field and accelerates to the east at a rate of 4x106m/s2. What is the ma

Jet001 [13]

Answer:

Explanation:

Force on electron in an electric field E = eE where E is electric field .

acceleration = eE / m where m is mass of electron .

Putting the values

4 x 10⁶ = 1.6 x 10⁻¹⁹ x E / 9.1 x 10⁻³¹

E = 22.75 x 10⁻⁶ N/C

The direction of electric field will be towards west ( opposite to east )

because of negative charge on electron .

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3 years ago

Suppose that an asteroid traveling straight toward the center of the earth were to collide with our planet at the equator and bu

vlada-n [284]

Answer:

$\frac{1}{10}$ M

Explanation:

To apply the concept of <u>angular momentum conservation</u>, there should be no external torque before and after

As the <u>asteroid is travelling directly towards the center of the Earth</u>, after impact ,it <u>does not impose any torque on earth's rotation,</u> So angular momentum of earth is conserved

⇒ $I_{1} \times W_{1} =I_{2} \times W_{2}$

$I_{1}$ is the moment of interia of earth before impact
$W_{1}$ is the angular velocity of earth about an axis passing through the center of earth before impact
$I_{2}$ is moment of interia of earth and asteroid system
$W_{2}$ is the angular velocity of earth and asteroid system about the same axis

let $W_{1}=W$

since $\text{Time period of rotation}∝$ $\frac{1}{\text{Angular velocity}}$

⇒ if time period is to increase by 25%, which is $\frac{5}{4}$ times, the angular velocity decreases 25% which is $\frac{4}{5}$ times

therefore $W_{1}$ $=$ $\frac{4}{5} \times W_{1}$

$I_{1}=\frac{2}{5} \times M\times R^{2}$ (moment of inertia of solid sphere)

where M is mass of earth

R is radius of earth

$I_{2}=\frac{2}{5} \times M\times R^{2}+M_{1}\times R^{2}$

(As given asteroid is very small compared to earth, we assume it be a particle compared to earth, therefore by parallel axis theorem we find its moment of inertia with respect to axis)

where $M_{1}$ is mass of asteroid

⇒ $\frac{2}{5} \times M\times R^{2} \times W_{1}=}(\frac{2}{5} \times M\times R^{2}$ $+$ $M_{1}\times R^{2})\times(\frac{4}{5} \times W_{1})$

$\frac{1}{2} \times M\times R^{2}$ = $(\frac{2}{5} \times M\times R^{2}$ + $M_{1}\times R^{2})$

$M_{1}\times R^{2}=$ $\frac{1}{10} \times M\times R^{2}$

⇒ $M_{1}=}$ $\frac{1}{10} \times M$

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Which statement is true of a glass lens that diverges light in air?

Arisa [49]

C it is uniformly thick

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During a circus act, an elderly performer thrills the crowd by catching a cannon ball shot at him. The cannon ball has a mass of

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Answer:

m v1 = (m + M) v2

v2 = m v1 / (m + M)

v2 = 7 * 74 / (74 + 65)

3.73 m/s

74 kg is too heavy for the cannonball (over 150 lbs)

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3 years ago