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vampirchik [111]
4 years ago
7

What is the jet stream and how does it separate cold air to warm air?

Physics
1 answer:
docker41 [41]4 years ago
4 0
Jet stream<span>ˈjet ˈˌstrēm/</span>noun1.a narrow, variable band of very strong, predominantly westerly air currents encircling the globe several miles above the earth. There are typically two or three jet streams in each of the northern and southern hemispheres.
(credit to google) 

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A wooden block has a mass of 20.0 kg and a specific heat of 1700 J/kg °C. Find the change in thermal energy of the block as it w
andrew-mc [135]
M = 20.0kg, the mass of the block.
c = 1700 J/(kg-°C), the specific heat
ΔT = 25 - 15 = 10 °C = 10 K, the change in temperature.

The change in thermal energy is
Q = (20.0 \, kg)*(1700 \,  \frac{J}{kg-C} )*(10 \, C) = 340 \times 10^{3} \, J

Answer: 340 kJ (or 340,000 J)

6 0
3 years ago
How do satellites stay in orbit?
Vinil7 [7]

Answer:A satellite maintains its orbit by balancing two factors its velocity the speed it takes to travel in a straight line and the gravitational pull that Earth has on it. A satellite orbiting closer to the Earth requires more velocity to resist the stronger gravitational pull.

Explanation:

6 0
3 years ago
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A parallel-plate vacuum capacitor has 8.38 J of energy stored in it. The separation between the plates is 2.30 mm. If the separa
Elanso [62]

Answer:

Explanation:

plate separation = 2.3 x 10⁻³ m

capacity C₁ = ε A / d

= ε A / 2.3 x 10⁻³

C₂ = ε A / 1.15 x 10⁻³

\frac{C_2}{C_1} = \frac{2.3}{1.15}

a ) when charge remains constant

energy = \frac{q^2}{2C}

q is charge and C is capacity

energy stored initially E₁= \frac{q^2}{2C_1}

energy stored finally E₂ = \frac{q^2}{2C_2}

\frac{E_1}{E_2} = \frac{C_2}{C_1} = \frac{2.3}{1.15}

E_2 = \frac{1.15}{2.3 } \times E_1

= \frac{1.15}{2.3 } \times 8.38

= 4.19 J

b )

In this case potential diff remains constant

energy of capacitor = 1/2 C V²

energy is proportional to capacity as V is constant .

\frac{E_2}{E_1} = \frac{C_2}{C_1}

\frac{E_2}{8.38} = \frac{2.3}{1.15}

E_2 = 16.76 .

8 0
3 years ago
A lightbulb is rated by the power that it dissipates when connected to a given voltage. For a lightbulb connected to 120 V house
zhenek [66]

1. increase, 2. decrease

3 0
3 years ago
Please I need help........
LenKa [72]

Answer:

7.46 J/kg/K

Explanation:

The heat absorbed or lost is:

q = mCΔT

where m is the mass, C is the heat capacity, and ΔT is the change in temperature.

Given q = 15.0 J, m = 0.201 kg, and ΔT = 10.0 °C:

15.0 J = (0.201 kg) C (10.0 °C)

C = 7.46 J/kg/°C

Which is the same as 7.46 J/kg/K.

7 0
3 years ago
Read 2 more answers
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