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Otrada [13]
4 years ago
13

Which of these elements has the largest atom?

Physics
1 answer:
notka56 [123]4 years ago
7 0
The largest atommmmmmmmmmm
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A capacitor stores 7.6 × 10–11 C when the electric potential difference between the plates is 5.8 V. What is the electric potent
N76 [4]

Energy stored in the capacitor= 2.2 x 10⁻¹⁰ J

Explanation:

The energy stored in the capacitor is given by

E= 1/2 q V

E= energy stored

q= charge= 7.6 x 10⁻¹¹ C

V= potential difference=5.8 V

so energy stored= E= 1/2 (7.6 x 10⁻¹¹) (5.8)

Energy stored= 2.2 x 10⁻¹⁰ J

5 0
4 years ago
53. A recently discovered planet has a mass four times as great as Earth's and a radius twice as large as Earth's. What will be
marusya05 [52]

Answer:

g' = g = 9.81 m/s^2

so gravity will be same as that of surface of earth

Explanation:

As we know that the acceleration due to gravity is given as

g = \frac{GM}{R^2}

here we have

M = 4M_e

R = 2R_e

we know that for earth we have

g = 9.81 = \frac{GM_e}{R_e^2}

now if the radius and mass is given as above

g' = \frac{G(4M_e)}{(2R_e)^2}

g' = \frac{GM_e}{R_e^2}

g' = g = 9.81 m/s^2

so gravity will be same as that of surface of earth

6 0
3 years ago
Which number of planets are in oreader towards the sun?
Masja [62]
C. the order from closest to the sun to furthest is Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, Neptune so C. would be the opposite heading towards the sun. A good mnemonic to use is MVEMJSUN: My Very Educated Mother Just Served Us Nectarines
4 0
4 years ago
Read 2 more answers
True or False?
lana66690 [7]
The answer is A. True 
3 0
4 years ago
Read 2 more answers
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
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