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ipn [44]
3 years ago
13

Activity that uses 150 calories of energy per day, or 1,000 calories per week, describes ___________.

Physics
1 answer:
serious [3.7K]3 years ago
8 0
Moderate physical activity
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True or False:
bekas [8.4K]
I think it false. Sorry if i'm wrong.

5 0
2 years ago
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How many electrons are in the third energy level?<br><br> 2<br> 8<br> 18<br> 20
Blababa [14]

Answer:

B)8

Explanation:

In the first energy level you can have, at most, 2. Every energy level after that wants to have 8 electrons. Valence electrons I believe.

Hope this helps, have a nice day! (^-^)

6 0
3 years ago
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What is step one of a star called?​
yKpoI14uk [10]

Answer: Stage 1- Stars are born in a region of high density Nebula, and condenses into a huge globule of gas and dust and contracts under its own gravity. This image shows the Orion Nebula or M42 . Stage 2 - A region of condensing matter will begin to heat up and start to glow forming Protostars.

Explanation:

3 0
3 years ago
A student used apparatus as shown above. The beaker held 750 g of a liquid. The current from the power supply was 1.8 A and the
liq [111]

Answer:

2365.71\ \text{J/kg}^{\circ}\text{C}

Explanation:

V = Voltage = 230 V

I = Current = 1.8 A

\Delta T = Temperature change = (40-12)^{\circ}\text{C}

t = Time taken = 2 minutes

m = Mass of liquid = 750 g

c = Specific heat capacity of the liquid

Energy required to heat the water is equal to the heat released due to current passing

mc\Delta T=VIt\\\Rightarrow c=\dfrac{VIt}{m\Delta T}\\\Rightarrow c=\dfrac{230\times 1.8\times 2\times 60}{0.75\times (40-12)}\\\Rightarrow c=2365.71\ \text{J/kg}^{\circ}\text{C}

The specific heat capacity of the liquid is 2365.71\ \text{J/kg}^{\circ}\text{C}

6 0
3 years ago
A charge of 1.0 × 10-6 μC is located inside a sphere, 1.25 cm from its center. What is the electric flux through the sphere due
Korvikt [17]

Answer:

\Phi_E=0.11\frac{N\cdot m^2}{C}

Explanation:

According to Gauss's Law, the electric flux of a charged sphere is the electric field multiplied by the area of ​​the spherical surface:

\Phi_E=EA\\\Phi_E=E(4\pi R^2)\\\Phi_E=\frac{q}{4\pi \epsilon_0 R^2}(4\pi R^2)\\\Phi_E=\frac{q}{\epsilon_0}

This is identical to the electric flux of a point charge located in the center of the sphere.

\Phi_E=\frac{1*10^{-12}C}{8.85*10^{-12} \frac{C^2}{N\cdot m^2}}\\\Phi_E=0.11\frac{N\cdot m^2}{C}

8 0
3 years ago
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