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Mazyrski [523]
3 years ago
7

In your own words who discovered gadolinium

Physics
2 answers:
scoundrel [369]3 years ago
6 0

Answer:

Who Discovered Gadolinium Gadolinium was only discovered in 1880 by a Swiss chemist called Jean Charles Galissard de Marignac.

Explanation:

When analyzing samples of two minerals which contain small amounts of gadolinium, didymium and gadolinite, he discovered spectroscopic lines that were unrecognisable. Every single element has distinctive spectroscopic lines so he knew he had discovered a new element, and called it gadolinium after Finnish chemist Johan Gadolin.

77julia77 [94]3 years ago
3 0

Answer:

Paul-Émile Lecoq de Boisbaudran

Jean Charles Galissard de Marignac

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A quick USB charger claims its output current is 1.97Amp. We know that the standard USB output voltage is 5V. What is the output
Arte-miy333 [17]

Answer:

Output power of the charger is 9.85 watts.

Explanation:

It is given that,

Output current of the USB charger, I = 1.97 A

The standard USB output voltage, V = 5 V

We need to find the output power of the charger. It can be determined using the following formula as :

P = V × I

P=5\ V\times 1.97\ A

P = 9.85 watts

The output power of the charger is 9.85 watts. Hence, this is the required solution.

3 0
3 years ago
Find the electron and hole mobilities, and the resistivity of intrinsic silicon at 300K. Is intrinsic silicon a semiconductor
tino4ka555 [31]

Answer:

Resistivity = 231.481 K Ohm

Yes, Intrinsic Silicon is the semiconductor.

Explanation:

Solution:

At 300K:

Let suppose mobility of electron in intrinsic semiconductor = M_{e}

Mobility of electron in intrinsic semiconductor is:

M_{e}  = 1300 cm^{2}/volt.sec

Let suppose mobility of hole in intrinsic semiconductor = M_{h}

M_{h} = 500 cm^{2}/volt.sec

We know that, intrinsic silicon semiconductor has equal number of holes and electrons. So,

At 300 K

Intrinsic Carrier Concentration = 1.5 x 10^{10}/cm^{3} = C

And,

Conductivity of intrinsic Silicon is:

σ = C x (M_{h} + M_{e}) e

e = 1.6 x 10^{-19} C

So, plugging in the values, we get:

σ = C x (M_{h} + M_{e}) e

σ = 1.5 x 10^{10} x (500 + 1300) x 1.6 x 10^{-19}

σ = 4.32 x 10^{-6}

So, now we can find the resistivity.

Resistivity = 1/σ

Resistivity = 1/ 4.32 x 10^{-6}

Resistivity = 231.481 K Ohm

Yes, Intrinsic Silicon is the semiconductor.

7 0
3 years ago
car 1 drives 45 mph to the west and car 2 drives 30mph to the east . from the frame of reference of car 1, what is the velocity
Naya [18.7K]
Distance between the two cars is increasing at the rate of 85 mph. 

A passenger in Car-1 says that he is at rest in his own frame of reference,
and Car-2 is moving away from him at 85 mph, toward the west.


6 0
3 years ago
Read 2 more answers
Two charges, each of 2.9 microC are placed at two corners of a square 50cm on a side, If the charges are on one side of the squa
anyanavicka [17]

Answer:

The magnitude of the electric field and direction of electric field are 146.03\times10^{3}\ N/C and 75.36°.

Explanation:

Given that,

First charge q_{1}= 2.9\mu C

Second chargeq_{2}= 2.9\mu C

Distance between two corners r= 50 cm

We need to calculate the electric field due to other charges at one corner

For E₁

Using formula of electric field

E_{1}=\dfrac{kq}{r'^2}

Put the value into the formula

E_{1}=\dfrac{9\times10^{9}\times2.9\times10^{-6}}{(50\sqrt{2}\times10^{-2})^2}

E_{1}=52200=52.2\times10^{3}\ N/C

For E₂,

Using formula of electric field

E_{1}=\dfrac{kq}{r^2}

Put the value into the formula

E_{2}=\dfrac{9\times10^{9}\times2.9\times10^{-6}}{(50\times10^{-2})^2}

E_{2}=104400=104.4\times10^{3}\ N/C

We need to calculate the horizontal electric field

E_{x}=E_{1}\cos\theta

E_{x}=52.2\times10^{3}\times\cos45

E_{x}=36910.97=36.9\times10^{3}\ N/C

We need to calculate the vertical electric field

E_{y}=E_{2}+E_{1}\sin\theta

E_{y}=104.4\times10^{3}+52.2\times10^{3}\sin45

E_{y}=141310.97=141.3\times10^{3}\ N/C

We need to calculate the net electric field

E_{net}=\sqrt{E_{x}^2+E_{y}^2}

Put the value into the formula

E_{net}=\sqrt{(36.9\times10^{3})^2+(141.3\times10^{3})^2}

E_{net}=146038.69\ N/C

E_{net}=146.03\times10^{3}\ N/C

We need to calculate the direction of electric field

Using formula of direction

\tan\theta=\dfrac{141.3\times10^{3}}{36.9\times10^{3}}

\theta=\tan^{-1}(\dfrac{141.3\times10^{3}}{36.9\times10^{3}})

\theta=75.36^{\circ}

Hence, The magnitude of the electric field and direction of electric field are 146.03\times10^{3}\ N/C and 75.36°.

4 0
3 years ago
According to Newton's first law, which characteristic of a moving object would remain constant if there were no other
suter [353]
In object in motion stays in motion; speed
8 0
3 years ago
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