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4 years ago

Science and math people where you at

Physics

1 answer:

PilotLPTM [1.2K]4 years ago

8 0

For the second question you’re solving for resistance. resistance= voltage/ current. 120/0.5= 240. the answer is 240 ohms
for the third question you would do 2*4 since it’s asking for voltage, the answer is 8 volts :)

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Find the total electric charge of 1.7 kg of electrons. me=9.11×10−31kg, e=1.60×10−19C.

Gelneren [198K]

Answer:

$2.99\cdot 10^{11}C$

Explanation:

The mass of one electron is

$m_e = 9.11\cdot 10^{-31}kg$

So the number of electrons contained in M=1.7 kg of mass is

$N=\frac{M}{m_e}=\frac{1.7 kg}{9.11\cdot 10^{-31}kg}=1.87\cdot 10^{30}$

The charge of one electron is

$e=1.60\cdot 10^{-19} C$

So, the total charge of these electrons is equal to the charge of one electron times the number of electrons:

$Q=Ne=(1.87\cdot 10^{30})(1.6\cdot 10^{-19}C)=2.99\cdot 10^{11}C$

8 0

3 years ago

Compute the expected shell-model quadrupole moment of 209Bi () and compare with the experimental value, - 0.37 b

Over [174]

Answer:

0.22 b

Explanation:

Quadrupole moment of the nucleon is,

$Q=-\frac{2j-1}{2(j+1)}\frac{3}{5}R^{2}$

And also,

$R^{2}=R^{2} _{0}A^{\frac{2}{3} }$

And, $R _{0}=1.2\times 10^{-15}m$

Now,

$Q=-\frac{2j-1}{2(j+1)}\frac{3}{5}R^{2} _{0}A^{\frac{2}{3} }$

For Bismuth $j=\frac{9}{2}$ and A is 209.

$Q=-\frac{2\frac{9}{2} -1}{2(\frac{9}{2} +1)}\frac{3}{5}(1.2\times 10^{-15}) ^{2}(209)^{\frac{2}{3} }\\Q=0.628\times 35.28\times 10^{-30} \\Q=22.15\times 10^{-30} m^{2} \\Q=0.2215\times 10^{-28} m^{2} \\Q=0.22 barn$

Therefore, the expected value of quadrupole is 0.22 b which is quite related with experimental value which is 0.37 b

3 0

3 years ago

Explain how a step-down transformer makes electricity safe for use in homes.

Savatey [412]

Answer:

it lowers the voltage in homes

Explanation:

this is due to safety reasons as a human cant survive power line voltages which are between 150k and 760k and if you plugged anything short of a piece of 1 inch thick wire not counting insulation it would immediately start smoking and burn the cable.

3 0

3 years ago

Read 2 more answers

Anyone know the answer ?

nata0808 [166]

Answer:

Explanation:

4 0

3 years ago

Read 2 more answers

] A new coal-fired 750 MWe power plant with a thermal efficiency of 42% burns 9000 Btu/lb coal, which contains 1.1% sulfur. a. I

Valentin [98]

Answer:

The net emissions rate of sulfur is 1861 lb/hr

Explanation:

Given that:

The power or the power plant = 750 MWe

Since the power plant with a thermal efficiency of 42% (i.e. 0.42) burns 9000 Btu/lb coal, Then the energy released per one lb of the coal can be computed as:

$\mathtt{=( 0.42\times 9000\times 1055.06) J}$

= 3988126.8 J

= 3.99 MJ

Also, The mass of the burned coal per sec can be calculated by dividing the molecular weight of the power plant by the energy released per one lb.

i.e.

The mass of the coal that is burned per sec $=\dfrac{750}{3.99}$

The mass of the coal that is burned per sec = 187.97 lb/s

The mass of sulfur burned $= \dfrac{1.1}{100} \times 187.97 \ lb/s$

= 2.067 lb/s

To hour; we have:

= 7444 lb/hr

However, If a scrubber with 75% removal efficiency is utilized,

Then; the net emissions rate of sulfur is (1 - 0.75) × 7444 lb/hr

= 0.25 × 7444 lb/hr

= 1861 lb/hr

Hence, the net emissions rate of sulfur is 1861 lb/hr

8 0

3 years ago