If you applied a force of 200 Newtons to lift a box 2.20 meters above the floor, how much work would you be doing?

When you boil potatoes, will your cooking time be reduced with vigorously boiling water instead of gently boiling water? (Direct

lara [203]

Answer:

The cooling time will not be reduced.

Explanation:

The time to cook is virtually the same in both types, vigorously and gently boiling water.

The reason cooking of spaghetti calls for vigorously boiling water is to keep the pasta agitated so that they do not stick to one another.

The temperature of boiling water is the same for both vigorously boiling water and gently boiling water, therefore there will be little time difference in when the potatoes will cook when it is done with vigorously boiling water than when it is cooked with gently boiling water.

However cooking potatoes in vigorously boiling water may cause the water to dry up on time and the potatoes get burnt.

8 0

2 years ago

How far will a man travel in 15 min driving his car down the highway at an average speed of 24 m/?s

emmasim [6.3K]

Answer:

21.6 km

Explanation:

15×60×24 = 21600 m

$\frac{21600}{1000}$ = 21.6 km

7 0

3 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:

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