How many centimeters are in 24 meters

True or false? Once the sun explodes, the whole solar system will go with it.

liberstina [14]

True. That is how powerful the sun is. But it will explode in only about 5 billion years.

6 0

4 years ago

A parallel-plate capacitor has plates of area 0.40 m2 and plate separation of 0.20 mm. The capacitor is connected to a 9.0 V bat

mafiozo [28]

Answer:

a) E = 4.5*10⁴ V/m

b) C= 17.7 nF

c) Q = 159. 3 nC

Explanation:

a)

By definition, the electric field is the electrostatic force per unit charge, and since the potential difference between plates is just the work done by the field, divided by the charge, assuming a uniform electric field, if V is the potential difference between plates, and d is the separation between plates, the electric field can be expressed as follows:

$E = \frac{V}{d} = \frac{9.0V}{2*10-4m} =4.5 * 10e4 V/m (1)$

b)

For a parallel-plate capacitor, applying the definition of capacitance as the quotient between the charge on one of the plates and the potential difference between them, and assuming a uniform surface charge density σ, we get:

$Q = \sigma* A (2)$

From (1), we know that V = E*d, but at the same time, applying Gauss'

Law at a closed surface half within the plate, half outside it , it can be

showed than E= σ/ε₀, so finally we get:

$C = \frac{Q}{V} =\frac{\sigma*A}{E*d} = \frac{\sigma*A}{\frac{\sigma}{\epsilon_{o} } d} =\frac{\epsilon_{0}*A}{d} = \frac{8.85e-12F/m*0.4m2}{2e-4m} = 17.7 nF (3)$

c)

From (3) we can solve for Q as follows:

$Q = C* V = 17.7 nF * 9.0 V = 159.3 nC (4)$

6 0

3 years ago

The masses of the earth and moon are 5.98 x 1024 and 7.35 x 1022 kg, respectively. Identical amounts of charge are placed on eac

dybincka [34]

Answer:

The magnitude of charge on each is $5.707\times 10^{13} C$

Solution:

As per the question:

Mass of Earth, $M_{E} = 5.98\times 10^{24} kg$

Mass of Moon, $M_{M} = 7.35\times 10^{22} kg$

Now,

The gravitational force of attraction between the earth and the moon, if 'd' be the separation distance between them is:

$F_{G} = \frac{GM_{E}M_{M}}{d^{2}}$ (1)

Now,

If an identical charge 'Q' be placed on each, then the Electro static repulsive force is given by:

$F_{E} = \frac{1}{4\pi\epsilon_{o}}\frac{Q^{2}}{d^{2}}$ (2)

Now, when the net gravitational force is zero, the both the gravitational force and electro static force mut be equal:

Equating eqn (1) and (2):

$\frac{GM_{E}M_{M}}{d^{2}} = \frac{1}{4\pi\epsilon_{o}}\frac{Q^{2}}{d^{2}}$

$(6.67\times 10^{- 11})\times (5.98\times 10^{24})\times (7.35\times 10^{22}) = (9\times 10^{9}){Q^{2}}$

$\sqrt{\farc{(6.67\times 10^{- 11})\times (5.98\times 10^{24})\times (7.35\times 10^{22})}{9\times 10^{9}}} = Q$

Q = $\pm 5.707\times 10^{13} C$

7 0

3 years ago

Which happens to the magnetic field of a wire when you change the direction of the current in the wire?

SVEN [57.7K]

Nothing will happen to the magnetic field.
BUT If you're talking about the magnetic force , then when changing the current direction , the direction of the magnetic force will be reversed.

(You'ld like to know :
-Changing direction of current ONLY , reverses the direction of magnetic force.
-Changing direction of Magnetic field only (poles of magnet) , reverses the direction of magnetic force.
-Changing the direction of both , magnetic field and current , doesn't effect direction of magnetic force.)

N.B : Magnetic Force is experienced only when direction of current is PARALLEL to direction of magnetic field .

--> Flemming Left Hand Rule is used to know direction of magnetic force.

Hope This Helps.

6 0

3 years ago

Read 2 more answers

Since this object is traveling at terminal velocity, what must be the value of Fd? (HINT: how to the lengths of both arrows comp

Juli2301 [7.4K]

The magnitude of the air drag when the object is traveling at terminal velocity is C. 850 Newtons

Explanation:

There are only two forces acting on the object here:

- The force of gravity, of magnitude $F_g = 850 N$ , acting downward