The method of heat transfer by which the sun energy reaches the earth is

A solid conducting sphere of radius 2 cm has a charge of 8microCoulomb. A conducting spherical shell of inner radius 4 cm andout

nika2105 [10]

Answer:

C) 7.35*10⁶ N/C radially outward

Explanation:

If we apply the Gauss'law, to a spherical gaussian surface with radius r=7 cm, due to the symmetry, the electric field must be normal to the surface, and equal at all points along it.
So, we can write the following equation:

$E*A = \frac{Q_{enc} }{\epsilon_{0}} (1)$

As the electric field must be zero inside the conducting spherical shell, this means that the charge enclosed by a spherical gaussian surface of a radius between 4 and 5 cm, must be zero too.
So, the +8 μC charge of the solid conducting sphere of radius 2cm, must be compensated by an equal and opposite charge on the inner surface of the conducting shell of total charge -4 μC.
So, on the outer surface of the shell there must be a charge that be the difference between them:

$Q_{enc} = - 4e-6 C - (-8e-6 C) = + 4 e-6 C$

Replacing in (1) A = 4*π*ε₀, and Qenc = +4 μC, we can find the value of E, as follows:

$E = \frac{1}{4*\pi*\epsilon_{0} } *\frac{Q_{enc} }{r^{2} } = \frac{9e9 N*m2/C2*4e-6C}{(0.07m)^{2} } = 7.35e6 N/C$

As the charge that produces this electric field is positive, and the electric field has the same direction as the one taken by a positive test charge under the influence of this field, the direction of the field is radially outward, away from the positive charge.

6 0

4 years ago

What type of change occurs when water change from a solid to a iquid?

Oksanka [162]

The type of change is melting

6 0

3 years ago

A technician wearing a brass bracelet enclosing area 0.00500 m2 places her hand in a solenoid whose magnetic field is 3.10 T dir

aliya0001 [1]

Explanation:

Given that,

Area enclosed by a brass bracelet, $A=0.005\ m^2$

Initial magnetic field, $B_i=3.1\ T$

The electrical resistance around the circumference of the bracelet is, R = 0.02 ohms

Final magnetic field, $B_f=0.93\ T$

Time, $t=16\ ms=16\times 10^{-3}\ s$

The expression for the induced emf is given by :

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

$\phi$ = magnetic flux

$\epsilon=-\dfrac{d(BA)}{dt}$

$\epsilon=-A\dfrac{d(B)}{dt}$

$\epsilon=-A\dfrac{B_f-B_i}{t}$

$\epsilon=-0.005 \times \dfrac{0.93-3.1}{16\times 10^{-3}}$

$\epsilon=0.678\ volts$

So, the induced emf in the bracelet is 0.678 volts.

Using ohm's law to find the induced current as :

V = IR

$I=\dfrac{V}{R}$

$I=\dfrac{0.678}{0.02}$

I = 33.9 A

or

I = 34 A

So, the induced current in the bracelet is 34 A. Hence, this is the required solution.

5 0

3 years ago

An object is pulled with a 85 N force inclined at 27° along a horizontal surface ABC

OLEGan [10]

Answer:

m = 15.15 kg

Explanation:

Newton's Second Law of motion states that when an unbalanced force is applied on a body, an acceleration is produced in it in the direction of force. The component of force along the horizontal direction here, will be given by the Newton's Second Law as:

Fx = ma

F Cosθ = ma

where,

F = Magnitude of Force = 85 N

θ = Angle with horizontal = 27°

m = mass of object = ?

a = acceleration of object = 5 m/s²

Therefore,

85 N Cos 27° = m(5 m/s²)

m = 75.73 N/5 m/s²

6 0

3 years ago

In this diagram, which force is represted by L?

Dmitry [639]

Answer is a) the force caused by the wend nag encountering wind and air.

8 0

3 years ago