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5 days ago

Consider steady-state conditions for one-dimensional conduction in a plane wall having a thermal conductivity k = 50 W/m · K and

a thickness L = 0.35 m, with no internal heat generation. Determine the heat flux and the unknown quantity for each case and sketch the temperature distribution, indicating the direction of the heat flux.

Physics

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If you secure a refrigerator magnet about 2mmfrom the metallic surface of a refrigerator door and then move the magnet sideways,

Yuliya22 [3333]

Response:

(A) 4* 6 ^ ⁻6 T m² (B) 2 * 10 ^ ⁻6 v

Clarification:

Solution

Given that:

A refrigerator magnet with a depth of approximately 2 mm

The estimated magnetic field strength of the magnet is = 5 m T

The Area = 8 cm²

Now,

(A) The magnetic flux ΦB = BA

Therefore,

ΦB = (5 * 10^⁻ 3) ( 4 * 10 ^⁻2) * ( 2 * 10^ ⁻2) Tm²

Thus,

ΦB = 4* 6 ^ ⁻6 T m²

(B) By employing Faraday's Law, the subsequent equation applies:

Ε = Bℓυ

Where,

ℓ = 2 cm equals 2 * 10 ^⁻2 m

B = 5 m T = 5 * 10 ^ ⁻3 T

υ = 2 cm/s = 2 * 10 ^ ⁻2 m/s

Therefore,

Ε = (5 * 10 ^ ⁻3 T) * (2 * 10 ^ ⁻2) (2 * 10 ^ ⁻2) v

E =2 * 10 ^ ⁻6 v

7 0

1 month ago

A pressure cooker is a pot whose lid can be tightly sealed to prevent gas from entering or escaping. even without knowing how bi

Sav [3153]

Even if we lack details about the size of the pressure cooker or the altitude of its operation, we can reliably assess the force on the lid based on prior knowledge because, similar to boiling water, the pressure buildup inside the cooker increases in line with the rising temperature.

5 0

2 months ago

Three cars (car F, car G, and car H) are moving with the same velocity when the driver suddenly slams on the brakes, locking the

inna [3103]

Answer:

Car F.

Explanation:

Because, when skidding, the longer distance is associated with greater inertia. A larger mass results in increased inertia.

6 0

1 month ago

In a novel from 1866 the author describes a spaceship that is blasted out of a cannon with a speed of about 11.000 m/s. The spac

Sav [3153]

Answer:

$a=0.284\ m/s^2$

Explanation:

We start with the fact that

Initially, the spacecraft was at rest, u = 0

The final velocity of the rocket is given as v = 11 m/s

The distance that the rocket covers during acceleration is given as d = 213 m

We seek to determine the acceleration that the occupants of the spacecraft experience during launch, which is derived from the principles of kinematics. By applying the

third motion equation we can find the acceleration.

$v^2-u^2=2ad\\\\a=\dfrac{v^2-u^2}{2d}\\\\a=\dfrac{(11)^2-(0)^2}{2\times 213}\\\\a=0.284\ m/s^2$