All categories

2 years ago

Which freezes faster, hot water or cold water?

2 answers:

8 0

Hot water seems to freeze faster than cold water, known as the Mpemba effect. The effect was named after the Tanzanian student who in 1963 noticed that hot ice cream mix freezes faster than a cold one. hope this helped

Marina CMI [18]2 years ago

5 0

Hot water freezes faster

You might be interested in

What is the coldest place on earth

exis [7]

The continent of Antartica is located at the bottom of the world. the South Pole is at its center. Antarctica is the coldest and windiest place on earth. It is covered with ice up to 3 miles thick. Very few plants and animals can survive here, but penguins, fish, and seals live on the coast and in the seas. No people live on Antarctica permanently, but scientists and tourists visit.

6 0

3 years ago

PLZ HELP

lianna [129]

The air movements toward the equator are called trade winds, which are warm, steady breezes that blowalmost continuously. The Coriolis Effect makes the trade winds appear to be curving to the west, whether they are traveling to the equator from the south or north. Answer trade wind

7 0

2 years ago

A ball of mass M collides with a stick with moment of inertia I = βml2 (relative to its center, which is its center of mass). Th

ZanzabumX [31]

Answer:

Part a)

$v_2 = \frac{\frac{2\beta mL^2v_o}{d}}{(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})}$

Part b)

$v_1 = v_0 - \frac{m}{M}(\frac{\frac{2\beta mL^2v_o}{d}}{(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})})$

Explanation:

Since the ball and rod is an isolated system and there is no external force on it so by momentum conservation we will have

$Mv_o = M v_1 + m v_2$

here we also use angular momentum conservation

so we have

$M v_o d = M v_1 d + \beta mL^2 \omega$

also we know that the collision is elastic collision so we have

$v_o = (v_2 + d\omega) - v_1$

so we have

$\omega = \frac{v_o + v_1 - v_2}{d}$

also we know

$M v_o d - M v_1 d = \beta mL^2(\frac{v_o + v_1 - v_2}{d})$

also we know

$v_1 = v_o - \frac{m}{M}v_2$

so we have

$M v_o d - M(v_o - \frac{m}{M}v_2)d = \beta mL^2(\frac{v_o + v_o - \frac{m}{M}v_2 - v_2}{d})$

$mv_2 d = \beta mL^2\frac{2v_o}{d} - \beta mL^2(1 + \frac{m}{M})\frac{v_2}{d}$

now we have

$(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})v_2 = \frac{2\beta mL^2v_o}{d}$

$v_2 = \frac{\frac{2\beta mL^2v_o}{d}}{(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})}$

Part b)

Now we know that speed of the ball after collision is given as

$v_1 = v_o - \frac{m}{M}v_2$

so it is given as

$v_1 = v_0 - \frac{m}{M}(\frac{\frac{2\beta mL^2v_o}{d}}{(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})})$

3 0

2 years ago

What type of heat transfer is used in a restaurant that uses ovens with fans inside?

Marina CMI [18]

convection

please mark brainliest any other problems or questions feel free to ask

5 0

2 years ago

Bodies A and B have equal mass. Body B is initially at rest. Body A collides with body B in a one-dimensional elastic collision.

jek_recluse [69]

According to the statement " Collision between two bodies in which the total kinetic energy of the two bodies after the collision is equal to their total kinetic energy before the collision."
The best answer is :
Option A " BODY A COMES TO REST BODY B STARTS MOVING WITH INITIAL VELOCITY OF BODY A "

4 0

2 years ago

Read 2 more answers