Give the formula for average velocity

Highlight two factors which shows that heat from the sun does reach the earth's surface by convection.

MAXImum [283]

radiation

when the suns radiation fall on the earth and its objects they receive heat energy and hence get heated. Thus the suns heat reaches the earth by. the process of radiation

3 0

2 years ago

A motorboat traveling with a current can go 160 km in 4 hours. against the current it takes 5 hours to go the same distance. Fin

MatroZZZ [7]

<h2>Speed of motorboat is 36 km/hr and speed of current is 4 km/hr.</h2>

Explanation:

Let speed of motor boat be m and speed of current be c.

A motorboat traveling with a current can go 160 km in 4 hours.

Distance = 160 km

Time = 4 hours

Speed = m + c

We have

Distance = Speed x Time

160 = (m+c) x 4

m + c = 40 --------------------- eqn 1

Against the current it takes 5 hours to go the same distance.

Distance = 160 km

Time = 5 hours

Speed = m - c

We have

Distance = Speed x Time

160 = (m-c) x 5

m - c = 32 --------------------- eqn 2

eqn 1 + eqn 2

2m = 40 + 32

m = 36 km/hr

Substituting in eqn 1

36 + c = 40

c = 4 km/hr

Speed of motorboat is 36 km/hr and speed of current is 4 km/hr.

3 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:

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