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