Places with a constant temperature are most likely to be located

Help me please!!!!!..............

qaws [65]

Answer:

<em>The car took 3 hours to travel 180 miles</em>

Explanation:

<u>Constant Speed Motion</u>

An object travels at constant speed if the ratio of the distance traveled by the time taken is constant.

Expressed in a simple equation, we have:

$\displaystyle v=\frac{d}{t}$

Where

v = Speed of the object

d = Distance traveled

t = Time taken to travel d.

From the equation above, we can solve for t:

$\displaystyle t=\frac{d}{v}$

The car travels at v=60 mi/h for a distance of d=180 miles. The time taken is:

$\displaystyle t=\frac{180\ mi}{60\ mi/h}$

t = 3 hours

The car took 3 hours to travel 180 miles

6 0

3 years ago

Does the earth behave like a magnet how do you prove it

Arlecino [84]

Take any item that's magnetized. With a sharpie or a drop of nail polish, make a small mark on one end of the item. Then, hang it up by a thread, in such a way that its two ends are at the same height off the floor, and it's free to turn on the thread.

After it settles down and stops turning and wiggling, it'll come to rest with the marked end pointing in the same direction EVERY TIME you try this.

The only thing that makes a magnet point in the same direction every time is another magnet.

5 0

4 years ago

Which circuit would have the most electrical power-will mark brainiest if correct

jolli1 [7]

Answer:

c

Explanation:

6 0

3 years ago

Read 2 more answers

The problem with your knees being over your toes is _______.

FrozenT [24]

Answer:

D. All of the above

Explanation:

7 0

2 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