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Alenkasestr [34]
3 years ago
6

AgNO3(aq)+NaCl(aq)—>|NaN03(aq)+AgCl(s)

Physics
1 answer:
ASHA 777 [7]3 years ago
8 0

the answer to the question is (D)

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Does the sun always shine directly overhead at the equator at noon
Nikitich [7]
The sun always shines directly overhead at noon. This is because the equator always gets the equivalent amount of sunlight. The area always get 12 hours of sunlight, because it's 0 degrees north and south and it's at the center of the Earth.
4 0
3 years ago
A 10-turn coil of wire having a diameter of 1.0 cm and a resistance of 0.50 Ω is in a 1.0 mT magnetic field, with the coil orien
n200080 [17]

Answer:

The voltage across the capacitor is 1.57 V.

Explanation:

Given that,

Number of turns = 10

Diameter = 1.0 cm

Resistance = 0.50 Ω

Capacitor = 1.0μ F

Magnetic field = 1.0 mT

We need to calculate the flux

Using formula of flux

\phi=NBA

Put the value into the formula

\phi=10\times1.0\times10^{-3}\times\pi\times(0.5\times10^{-2})^2

\phi=7.85\times10^{-7}\ Tm^2

We need to calculate the induced emf

Using formula of induced emf

\epsilon=\dfrac{d\phi}{dt}

Put the value into the formula

\epsilon=\dfrac{7.85\times10^{-7}}{dt}

Put the value of emf from ohm's law

\epsilon =IR

IR=\dfrac{7.85\times10^{-7}}{dt}

Idt=\dfrac{7.85\times10^{-7}}{R}

Idt=\dfrac{7.85\times10^{-7}}{0.50}

Idt=0.00000157=1.57\times10^{-6}\ C

We know that,

Idt=dq

dq=1.57\times10^{-6}\ C

We need to calculate the voltage across the capacitor

Using formula of charge

dq=C dV

dV=\dfrac{dq}{C}

Put the value into the formula

dV=\dfrac{1.57\times10^{-6}}{1.0\times10^{-6}}

dV=1.57\ V

Hence, The voltage across the capacitor is 1.57 V.

5 0
3 years ago
Suppose 1 kg of Hydrogen is converted into Helium. a) What is the mass of the He produced? b) How much energy is released in thi
morpeh [17]

Answer:

a) m = 993 g

b) E = 6.50 × 10¹⁴ J

Explanation:

atomic mass of hydrogen = 1.00794

4 hydrogen atom will make a helium atom = 4 × 1.00794 = 4.03176

we know atomic mass of helium = 4.002602

difference in the atomic mass of helium = 4.03176-4.002602 = 0.029158

fraction of mass lost = \dfrac{0.029158}{4.03176}= 0.00723

loss of mass for 1000 g = 1000 × 0.00723 = 7.23

a) mass of helium produced = 1000-7.23 = 993 g (approx.)

b) energy released in the process

E = m c²

E = 0.00723 × (3× 10⁸)²

E = 6.50 × 10¹⁴ J

4 0
3 years ago
Read 2 more answers
A constant tone is being applied to a speaker. The voltage across the speaker is 5 volts. The voltage across the speaker is incr
lubasha [3.4K]

Answer:

decibel gain is 10 db

Explanation:

given data

voltage v1 =  5 volts

voltage v2 = 15 volts

to find out

decibel gain

solution

we have given that 5 volt voltage that is increase to 15 volts

so

we know here decibel gain formula that is

decibel gain = 20 log(v2/v1)   ....................1

put here value of v1 and v2  in equation 1

decibel gain = 20 log(15/5)

decibel gain = 20 log3

decibel gain = 20 × 0.4771

decibel gain = 9.54 db = approx 10 db

so decibel gain is 10 db

5 0
4 years ago
Read 2 more answers
Dez pours water (n 1.333) into a container made of crown glass (n 1.52). The light ray in ner made of crown glass (n = 1.52). Th
siniylev [52]

Answer:

The angle of the corresponding refracted ray is 34.84°

Explanation:

Given that,

Refractive index of water n= 1.33

Refractive index of glass n= 1.52

Incident angle = 30.0°

We need to calculate the refracted angle

Using formula of Snell's law

n_{i}\sin i=n_{r}\sin r

Put the value into the formula

\sin r=\dfrac{n_{i}\sin i}{n_{r}}

\sin r=\dfrac{1.52\times\sin30}{1.33}

\sin r=0.5714

r=sin^{-1}0.5714

r = 34.84^{\circ}

Hence, The angle of the corresponding refracted ray is 34.84°

8 0
4 years ago
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