All categories

3 years ago

Compare the temperature of water as ice begins to form with the temperature of ice as it begins to melt

1 answer:

3 0

They are the same temperature.

If you have 1 gram of ice at zero degrees C, you have to pump 334 joules
of heat energy into it in order to<em> just change it from solid to liquid</em> at the <em>same temperature. </em>

This is one of the highest figures known for any substance, and that's why
a piece of ice does such a good job cooling your drink.

You might be interested in

Which of the following has the most thermal energy?

Lelechka [254]

Answer:
2. A 1 litre mug of hot chocolate at 75 degrees.

Explanation:
Thermal energy is directly proportional to mass, so as the mass increases, the thermal energy of the substance increases as well.

8 0

3 years ago

Given the height of rod,length of shadow of tree and length of shadow of the rod, estimate the height of the tree? Given:height

diamong [38]

Answer:

1000 cm.

Explanation:

To obtain the estimated tree height :

(Height of rod / length of rod shadow) = (height of tree / length of tree shadow)

Substituting values into the formula :

(150cm / 120 cm) = (height of tree / 800 cm)

Using cross multiplication :

Height of tree * 120 = 150 * 800

Height of tree = (150 * 800) / 120

Height of tree = 120,000 / 120

Height of tree = 1000

Hence, estimate height of tree = 1000 cm

5 0

2 years ago

A swimming pool is 50 ft wide and 100 ft long and its bottom is an inclined plane, the shallow end having a depth of 4 ft and th

Nina [5.8K]

Explanation:

We define force as the product of mass and acceleration.

F = ma

It means that the object has zero net force when it is in rest state or it when it has no acceleration. However in the case of liquids. just like the above mentioned case, the water is at rest but it is still exerting a pressure on the walls of the swimming pool. That pressure exerted by the liquids in their rest state is known as hydro static force.

Given Data:

Width of the pool = w = 50 ft

length of the pool = l= 100 ft

Depth of the shallow end = h(s) = 4 ft

Depth of the deep end = h(d) = 10 ft.

weight density = ρg = 62.5 lb/ft

Solution:

a) Force on a shallow end:

$F = \frac{pgwh}{2} (2x_{1}+h)$

$F = \frac{(62.5)(50)(4)}{2}(2(0)+4)$

$F = 25000 lb$

b) Force on deep end:

$F = \frac{pgwh}{2} (2x_{1}+h)$

$F = \frac{(62.5)(50)(10)}{2} (2(0)+10)$

$F = 187500 lb$

c) Force on one of the sides:

As it is mentioned in the question that the bottom of the swimming pool is an inclined plane so sum of the forces on the rectangular part and triangular part will give us the force on one of the sides of the pool.

1) Force on the Rectangular part:

$F = \frac{pg(l.h)}{2}(2(x_{1} )+ h)$