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Lapatulllka [165]
4 years ago
14

When the heat source is removed from a fluid, convection currents in the fluid will eventually ____________then stop.

Physics
1 answer:
Blizzard [7]4 years ago
5 0

When the heat source is removed from a fluid, convection currents in the fluid will eventually distribute heat uniformly throughout the fluid.  When all of the fluid is at the same temperature, convection currents will stop.

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choose the correct answer 14: which of the following force follows the inverse square law of distance A: gravitaional force B: e
Delicious77 [7]

Answer:

C. both A and B

Explanation:

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8 0
3 years ago
Let's say you go on a run and start to sweat because you are heating up.
Jobisdone [24]

Answer:

kinetic to thermal

Explanation:

4 0
4 years ago
Considerando que los coeficientes de dilatación de los siguientes metales son: hierro 11.7 x 10-6; plomo 27.3 x 10-6; cobre 16.7
Rasek [7]

Answer:

el plomo será el más largo

Explanation:

Dado que;

longitud inicial (l1) = 4m

Longitud final l2

aumento de temperatura (θ) = 10 ° C

Coeficiente de expansión lineal α

Ahora para el hierro;

α = 11,7 x 10-6

Desde;

l2-l / l1θ = α

l2 = α l1θ + l1

l2 = l1 (αθ + 1)

l2 = 4 ((11,7 x 10-6 * 10) + 1)

l2 = 4.00044 m

Para el plomo

l2 = 4 ((27,3 x 10-6 * 10) + 1)

l2 = 4,00109 m

Para cobre

l2 = 4 ((16,7 x 10-6 * 10) + 1)

l2 = 4.000668 m

Por lo tanto, el plomo será el más largo

7 0
3 years ago
For a particular casting setup, the top of the sprue has a diameter of 0.030 m, and its length is 0.200 m. The volume flow rate
faust18 [17]

Answer with Explanation:

We are given that

Diameter=0.030 m

Length of sprue=h_1=0.200 m

Metal volume flow  rate,Q=0.03m^3/min

Q=\frac{0.03}{60}=5\times 10^{-4}m^3/s because 1 minute=60 seconds

Let 1 for the top and 2 for the bottom

d_=0.030 m

h_2=0

A_1=\frac{\pi d^2}{4}=\frac{3.14\times (0.030)^2}{4}

A_1=7.065\times 10^{-4} m^2

v_1=\frac{Q}{A_1}=\frac{5\times 10^{-4}}{7.065\times 10^{-4}}

v_1=0.708 m/s

Pressure at the top and bottom of the sprue is atmospheric

h_1+\frac{v^2_1}{2g}=h_2+\frac{v^2_2}{2g}

Substitute the values

0.2+\frac{(0.708)^2}{2\cdot 9.8}=0+\frac[v^2_2}{2\cdot 9.8}

v^2_2=2\cdot 9.8\cdot \frac{0.2\cdot 9.8\cdot 2+0.501264}{2\cdot 9.8}=4.421264

v_2=\sqrt{4.421264}=2.1 m/s

Q=A_2v_2

5\times 10^{-4}=A_2\times 2.1

A_2=\frac{5\times 10^{-4}}{2.1}=2.381\times 10^{-4} m^2

Reynolds number=\frac{v_2D\rho}{\eta}

\eta=0.004 N.s/m^2

\rho=2700 kg/m^3

Substitute the values then we get

Reynolds number=\frac{2.1\times 0.03\times 2700}{0.004}

Reynolds number=42525

The Reynolds number is greater than 4000 .Therefore, the flow is turbulent.

8 0
3 years ago
Jane, looking for tarzan, is running at top speed (6.8 m/s) and grabs a vine hanging vertically from a tall tree in the jungle.
erica [24]
Jane's mechanical energy at any time is
E=U+K
where U=mgh is the potential energy, while K= \frac{1}{2} mv^2 is the kinetic energy.

Initially, Jane is on the ground, so the altitude is h=0 and the potential energy is zero: U=0. She's running with speed v, so she has kinetic energy only:
E=K= \frac{1}{2} mv^2
Then she grabs the vine, and when she reaches the maximum height h, her speed is zero: v=0, and so the kinetic energy becomes zero: K=0. So now her mechanical energy is just potential energy:
E=U=mgh

But E must be conserved, so the initial kinetic energy must be equal to the final potential energy:
\frac{1}{2}mv^2=mgh
from which we can find h, the maximum height Jane can reach:
h= \frac{v^2}{2g}= \frac{(6.8 m/s)^2}{2\cdot 9.81 m/s^2}=2.36 m
6 0
3 years ago
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