The working equation to be used here is written below:

Q = kA(T₁ - T₂)/Δx
where
Q is the rate of heat transfer
k is the heat transfer coefficient
A is the cross-sectional area of the wall
T₁ - T₂ is the temperature difference between the sides of the wall
Δx is the thickness of the wall

The solution is as follows:

Q = (0.69 W/m²·°C)(5 m × 6 m)(50°C - 20°C)/(30 cm * 1 m/100 cm)
Q = 2,070 W/m

4 0

3 years ago

The moons of Mars, Phobos (Fear) and Deimos (Terror), are very close to the planet compared to Earth's Moon. Their orbital radii

tankabanditka [31]

Answer:

$0.2528$

Explanation:

To calculate the period we need the formula:

$T=\frac{2\pi r^{3/2}}{\sqrt{GM}}$

Where $r$ is the radius of the moon, $G$ is the universal constant of gravitation and $M$ is the mass of mars.

The period of Phobos:

$T_{p}=\frac{2\pi r_{p}^{3/2}}{\sqrt{GM}}$

The period of Deimos:

$T_{D}=\frac{2\pi r_{D}^{3/2}}{\sqrt{GM}}$

The ratio of the period of Phobos and Deimos:

$\frac{T_{p}}{T_{D}}=\frac{\frac{2\pi r_{p}^{3/2}}{\sqrt{GM}}}{\frac{2\pi r_{D}^{3/2}}{\sqrt{GM}}}$

$\frac{T_{p}}{T_{D}}=\frac{\sqrt{GM}2\pi r_{p}^{3/2}}{\sqrt{GM}2\pi r_{D}^{3/2}}$

Most terms get canceled and we have:

$\frac{T_{p}}{T_{D}}=\frac{r_{p}^{3/2}}{r_{D}^{3/2}}$

According to the problem

$r_{p}=9,378km\\r_{D}=23,459km$

so the ratio will be:

$\frac{T_{p}}{T_{D}}=\frac{(9,378)^{3/2}}{(23,459)^{3/2}}=\frac{908166.22}{3593058.125}=0.25275$ ≈ $0.2528$

the ratio of the period of revolution of Phobos to that of Deimos is 0.2528

8 0

3 years ago

A 1.5 kg frictionless pendulum is released from point A at an angle θ of 35 degrees. The speed of the pendulum at point C is 3.4

sergeinik [125]

The length of the pendulum is 3.3 m.

The given parameters:

Mass, m = 1.5 kg
Angle, θ = 35⁰
Speed, v = 3.4 m/s

<h3>What is principle of conservation of energy?</h3>

The principle of conservation of energy states that, the total energy of a system is always conserved.

P.E = K.E

mgh = ¹/₂mv²

gh = ¹/₂v²

g(L - Lcosθ) = ¹/₂v²

gL(1 - cosθ) = ¹/₂v²

$L = \frac{v^2}{2g(1- cos\theta)} \\\\L = \frac{(3.4)^2}{2\times 9.8(1 - cos35)} \\\\L = 3.28 \ m\\\\L = 3.3 \ m$

Thus, the length of the pendulum is 3.3 m.

Learn more about length of pendulum here: brainly.com/question/8168512

3 0

3 years ago

In a free body diagram the weight is

torisob [31]

Answer:

Explanation:

a. WEIGHT: As we have seen, weight is the gravitational force exerted on an object by the Earth (or any other celestial body). If an object is near the Earth's surface and has mass, then the object has a weight. The magnitude of its weight is w = mg and its direction is toward the center of the Earth.

7 0

3 years ago