Was Chantal's snack nutritious? What could she have eaten instead to boost

The heat flux for a given wall is in the x-direction and given as q^n = 11 W/m^2, the walls thermal conductivity is 1.7 W/mK and

MrMuchimi

Answer:

$\frac{dT}{dx} = 6.47 ^oC/m$

Also as we can see the equation that heat flux directly depends on the temperature gradient so more is the temperature gradient then more will be the heat flux.

For positive temperature gradient the heat will flow outwards while for negative temperature gradient the heat will flow inwards

Explanation:

As we know that heat flux is given by the formula

$q^n = K\frac{dT}{dx}$

here we know that

K = thermal conductivity

$\frac{dT}{dx}$ = temperature gradient

now we know that

$q^n = 11 W/m^2$

also we know that

K = 1.7 W/mK

now we have

$11 = 1.7 \frac{dT}{dx}$

so temperature gradient is given as

$\frac{dT}{dx} = \frac{11}{1.7} = 6.47 K/m$

also in other unit it will be same

$\frac{dT}{dx} = 6.47 ^oC/m$

Also as we can see the equation that heat flux directly depends on the temperature gradient so more is the temperature gradient then more will be the heat flux.

For positive temperature gradient the heat will flow outwards while for negative temperature gradient the heat will flow inwards

5 0

3 years ago

23. Lunar Gravity Compare the force holding a

romanna [79]

Explanation:

Given

mass of the rock is 10 kg

Force requires to hold the rock is equal to its weight

Weight is given by the product of mass and acceleration due to gravity

Weight on the earth surface

$\Rightarrow W_e=10\times 9.8\\\Rightarrow W_e=98\ N$

Weight on the moon surface

$\Rightarrow W_m=1.6\times 10\\\Rightarrow W_m=16\ N$

So, the force holding the rock on earth is approximately 6 times the force on the moon.

3 0

3 years ago

If a 0.15 kg ball falls and has a KE of 20 J just before striking the ground, from what height did it fall. A. 1.36m B. 3m C. 13

RUDIKE [14]

According to the conservation of mechanical energy, the kinetic energy just before the ball strikes the ground is equal to the potential energy just before it fell.

Therefore, we can say KE = PE
We know that PE = m·g·h

Which means KE = m·g·h

We can solve for h:

h = KE / m·g
= 20 / (0.15 · 9.8)
= 13.6m

The correct answer is: the ball has fallen from a height of 13.6m.

5 0

3 years ago

Which of the following statements is true?

icang [17]

i think b and d are true

5 0

3 years ago

A 73-kg person in a moving car stops during a car collision in a distance of 0.80 m . The stopping force that the air bag exerts

fredd [130]

Answer:

v₀ = 13.24 m / s

Explanation:

Let's use Newton's second law to find the average acceleration during the crash

F = m a

. a = F / m

a = 8000/73

a = 109.59 m / s²

Now we can use the kinematic equations to find the initial velocity, since when the velocity stops it is zero (v = 0)

v² = v₀² - 2 a x

v₀² = 2 a x

v₀ = √ 2 a x

v₀ = √ (2 109.59 0.80)

v₀ = 13.24 m / s

3 0

4 years ago

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