Ask question

Ask question

All categories

3 years ago

9

Help please an observer at the North Pole sees the moon phase shown below. What Moon phase will be observed approximately one we

ek later?

1 answer:

ziro4ka [17]3 years ago

4 0

Answer:

it should be a new moon

Explanation:

You might be interested in

How far will a car travel from a standing start if it accelerates at 4.0m/s2 for 9.0 seconds

ehidna [41]

Answer:

4*9 = 36 m/sec at 9 seconds. The average speed is (0 + 36)/2 = 18 .

Explanation:

5 0

3 years ago

Oxygen makes up more than 90 percent of the volume of the Earth's crust. Is oxygen found as a solid, liquid, or gas?

kipiarov [429]

Answer:

Oxygen is an element that can be a solid, liquid or gas depending on its temperature and pressure. In the atmosphere it is found as a gas, more specifically, a diatomic gas. This means that two oxygen atoms are connected together in a covalent double bond.

5 0

3 years ago

A sphere is originally at a temperature of 500°c. The sphere is melted and recast, without loss of mass, into a cube with the sa

aleksandr82 [10.1K]

Answer: The value of the celsius temperature of the cube is 472.2°c.

Explanation:

The expression for the power radiated is as follows;

$P=A\epsilon\sigma T^{4}$

Here, A is the area, $\sigma$ is the stefan's constant, $\epsilon$ is the emissivity and T is the temperature.

It is given in the problem that A sphere is originally at a temperature of 500°c. The sphere is melted and recast, without loss of mass, into a cube with the same emissivity as the sphere.

Then the expression for the radiated power for the cube and the sphere can be expressed as;

$A_{1}\epsilon \e\sigma T_{1}^{4}=A_{2}\epsilon \e\sigma T_{2}^{4}$

Here, $A_{1}$ is the area of the sphere, $A_{2}$ is the area of the cube, $T_{1}$ is the temperature of the sphere and $T_{2}$ is the temperature of the cube.

The radiated powers and emissivity of the cube and the sphere are same.

$A_{1}T_{1}^{4}=A_{2}T_{2}^{4}$

The area of the sphere is $A_{1}=4\pi \times r^{2}$ .

Here, r is the radius of the sphere.

The area of the cube is $A_{2}=6\times a^{2}$ .

Here, a is the edge of the cube.

Put $A_{1}=4\pi \times r^{2}$ and $A_{2}=6\times a^{2}$ .

$T_{2}=T_{1}(\frac{2\pi }{3}\times (\frac{r}{a})^{2})^{\frac{1}{4}}$ ....(1)

The masses and the densities of the sphere and the cube are same. Then the volumes are also same.

$V_{1}=V_{2}$

Here, $V_{1},V_{1}$ are the volumes of the sphere and the cube.

$\frac{4}{3}\pi r^{3}=a^{3}$

$\frac{r}{a}=(\frac{3}{4\pi })^{\frac{1}{3}}$

Put this value in the equation (1).

$T_{2}=T_{1}(\frac{2\pi }{3}\times (\frac{r}{a})^{2})^{\frac{1}{4}}$ $T_{2}=T_{1}(\frac{2\pi }{3}\times ((\frac{3}{4\pi })^{\frac{1}{3}})^{2})^{\frac{1}{4}}$

Put T_{1}=500°c.

$T_{2}=(500)(\frac{2\pi }{3}\times (\frac{3}{4\pi })^{\frac{2}{3}})^{\frac{1}{4}}$

$T_{2}=472.2^{\circ}c$

Therefore, the value of the celsius temperature of the cube is 472.7°c.

5 0

3 years ago

An object has a mass of 120 kg on the moon. What is the force of gravity acting on the object on the moon?

Art [367]

$m-mass \\ g-gravitational \ acceleration \\ G-the \ force \ of \ gravity \\\\ G=m*g \\\\ G=120*1.62 \\\\ \boxed{G=194.4N}$

5 0

3 years ago

Read 2 more answers

Jamal plugged his radio into the wall. The radio's plug had copper wires surrounded by rubber. The rubber protects Jamal from___

Mariana [72]

The rubber protects him from being electrocuted by the flow of current going through the plug.
Hope this helped!!

3 0

4 years ago

Read 2 more answers