2. An electrical heater 200mm long and 15mm in diameter is inserted into a drilled hole normal to the surface of a large block of material having a thermal conductivity of 5W/m·K. Estimate the temperature reached by the heater when dissipating 25 W with the surface of the block at a temperature of 35 °C.

Answer:

0.0000000026 T

Explanation:

= Maximum electric field strength = 0.78 V/m

= Maximum magnetic field strength

c = Speed of light =

Relation between amplitudes of electric and magnetic fields is given by

The amplitude of the magnetic field is 0.0000000026 T

Answer:

Explanation:

vf=vi+at

vf=31 m/s

vi=0 m/s

a=g=9.8 m/s2

t=?

vf-vi=at

vf-vi/a=t

t=vf-vi/a

t=31 m/s-0/9.8

t=3.16 s

Answer:

v = 3×10^8 m/s

s= 384,400 km= 3.84×10^8 m/s

t = ?

v = s/t = 2s/t

t = 2s/v

t = (2×3.84×10^8) ÷ 3×10^8

t = 2.56 seconds

Explanation:

Earth's moon is the brightest object in our

night sky and the closest celestial body. Its

presence and proximity play a huge role in

making life possible here on Earth. The moon's gravitational pull stabilizes Earth's wobble on its axis, leading to a stable climate.

The moon's orbit around Earth is elliptical. At perigee — its closest approach — the moon comes as close as 225,623 miles (363,104 kilometers). At apogee — the farthest away it gets — the moon is 252,088 miles (405,696

km) from Earth. On average, the distance fromEarth to the moon is about 238,855 miles (384,400 km). According to NASA , "That means 30 Earth-sized planets could fit in between Earth and the moon."

Answer:

F'= 4F/9

Explanation:

Two small objects each with a net charge of +Q exert a force of magnitude F on each other. If r is the distance between them, then the force is given by :

...(1)

Now, if one of the objects with another whose net charge is + 4Q is replaced and also the distance between +Q and +4Q charges is increased 3 times as far apart as they were. New force is given by :

.....(2)

Dividing equation (1) and (2), we get :

Hence, the correct option is (d) i.e. " 4F/9"

Final answer:

The magnitude of the force on the +4Q charge, after replacing one of the original +Q charges and moving the charges three times farther apart, is calculated to be 4F/9 using Coulomb's Law. Therefore, the correct answer is D.

Explanation:

The magnitude of the electrostatic force between two charges can be described by Coulomb's Law, which states that F = k × (q1 × q2) / r^2, where F is the force between the charges, k is Coulomb's constant, q1 and q2 are the magnitudes of the charges, and r is the distance between the centers of the two charges. Originally, two objects each with charge +Q exert a force of magnitude F on each other. After one charge is replaced with a +4Q charge and they are moved to be three times as far apart, the force on the +4Q charge can be calculated using the modified version of Coulomb's Law that takes into account the new charges and distance.

Using the original scenario as a reference, where F = k × (Q × Q) / r^2, when the charge is replaced and the distance is tripled, the new force F' = k × (Q × 4Q) / (3r)^2 = 4kQ^2 / 9r^2. By comparing F' with F, we find that F' = (4/9)F. Thus, the magnitude of the force on the +4Q charge is 4F/9.

Answer:

a) h=3.16 m, b)  v_{cm }^ = 6.43 m / s

Explanation:

a) For this exercise we can use the conservation of mechanical energy

Starting point. Highest on the hill

           Em₀ = U = mg h

final point. Lowest point

            = K

Scientific energy has two parts, one of translation of center of mass (center of the sphere) and one of stationery, the sphere

           K = ½ m + ½ w²

angular and linear speed are related

           v = w r

           w = v / r

            K = ½ m v_{cm }^{2} + ½ I_{cm} v_{cm }^{2} / r²

            Em_{f} = ½ v_{cm }^{2} (m + I_{cm} / r2)

as there are no friction losses, mechanical energy is conserved

             Em₀ = Em_{f}

             mg h = ½ v_{cm }^{2} (m + I_{cm} / r²)         (1)

             h = ½ v_{cm }^{2} / g (1 + I_{cm} / mr²)

for the moment of inertia of a basketball we can approximate it to a spherical shell

             I_{cm} = ⅔ m r²

we substitute

            h = ½ v_{cm }^{2} / g (1 + ⅔ mr² / mr²)

            h = ½ v_{cm }^{2}/g    5/3

             h = 5/6 v_{cm }^{2} / g

let's calculate

           h = 5/6 6.1 2 / 9.8

           h = 3.16 m

b) this part of the exercise we solve the speed of equation 1

          v_{cm }^{2} = 2m gh / (1 + I_{cm} / r²)

in this case the object is a frozen juice container, which we can simulate a solid cylinder with moment of inertia

              I_{cm} = ½ m r²

we substitute

             v_{cm } = √ [2gh / (1 + ½)]

             v_{cm } = √(4/3 gh)

let's calculate

             v_{cm } = √ (4/3 9.8 3.16)

             v_{cm }^ = 6.43 m / s

Answer:

a

b

Explanation:

From the question we are told that

   The radius is  

   The current it carries is  

The  magnetic flux of the coil is mathematically represented as

Where  B is the  magnetic field which is mathematically represented as

Where   is the magnetic field with a constant value  

substituting  value

The area A is mathematically evaluated as

substituting values

the magnetic flux is mathematically evaluated as    

The self-inductance is evaluated as

substituting values