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3 years ago

15

If the Sun were to mysteriously turn into a black hole (don't worry . . . it won't!) and retain its current mass, what would hap

pen to Earth?

1 answer:

kap26 [50]3 years ago

8 0

The Earth would continue to orbit it
as usual.(But it would be very cold and dark around here.)

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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

Ice skaters often end their performances with spin turns, where they spin very fast about their center of mass with their arms f

vampirchik [111]

Answer:

$\large \boxed{\text{30 rev/s}}$

Explanation:

This question is based on the Law of Conservation of Angular Momentum.

Angular momentum (L) equals the moment of inertia (I) times the angular speed (ω).

L = Iω

If momentum is conserved,

I₁ω₁ = I₂ω₂

Data:

I₁ = 3.5 kg·m²s⁻¹

ω₁ = 6.0 rev·s⁻¹

I₂ = 0.70 kg·m²s⁻¹

Calculation:

$\begin{array}{rcl}I_{1}\omega_{1} &= &I_{2}\omega_{2}\\\text{3.5 kg$\cdot$m$^{2}$}\times \text{6.0 rev/s} &= &\text{0.70 kg$\cdot$m$^{2}$}\times\omega_{2}\\\text{21 rev/s} &= &0.70\omega_{2}\\\omega_{2} & = & \dfrac{\text{21 rev/s}}{0.70}\\\\&=&\textbf{30 rev/s}\\\end{array}\\\text{The skater's final rotational speed is $\large \boxed{\textbf{30 rev/s}}$}$

8 0

4 years ago

The Greek philosopher Democritus coined what word for a tiny piece of matter that cannot be divided?

Afina-wow [57]

Word for a tiny piece of matter that cannot be divided is B - Atom

5 0

3 years ago

Which of the following are ways in which energy can be transferred to or from a substance?

Gnesinka [82]

Answer:

a

Explanation:

the answer is a because it is feasible

3 0

3 years ago

If the work required to speed up a car from 11 km/h to 21 km/h is 6.0×103 J , what would be the work required to increase the ca

goblinko [34]

Explanation:

We need convert the velocities first to m/s and we get the following:

v2 = 21 km/hr = 5.8 m/s

v1 = 11 km/hr = 3.1 m/s

We need to find the mass of the car also for later use do using the work-energy theorem:

$delta \: w = \frac{1}{2} m(v \frac{2}{2} - v\frac{2}{1} )$

6.0x10^3 J = (0.5) m [(5.8)^2 - (3.1)^2]

or

m = 499.4 kg

Now we determine work needed delta W to change its velocity from 21 km/hr to 33 km/hr

v2 = 33 km/hr = 9.2 m/s

v1 = 21 km/hr = 5.8 m/s

delta W = (0.5)(499.4)[(9.2)^2 - (5.8)^2]

= 1.3 x 10^4 J

6 0

3 years ago