If low CVP precipitates a suction alarm, rapid infusion of volume can remedy the situation after dropping the P-level.

What happens when ice melts, according to the kinetic theory?

lukranit [14]

The water molecules increase their speed.
This is because when an object is in its solid form, the molecules are packed closely together, restricting movement.
On this table;

Solid
Liquid
Gas

Liquid is the next step on the ladder, and going from up to down, the molecules speed up and desperate.

7 0

3 years ago

What is NOT one of the three primary resources that families have to reach financial goals?

Elodia [21]

What is NOT one of the three primary resources that families have to reach financial goals? It is c) education

8 0

3 years ago

In Fig. 4-41, a ball is thrown up onto a roof, landing 4.00 s later at height h ???? 20.0 m above the release level. The ball’s

Yanka [14]

"Fig is attacted with answer"

Answer:

a) d = 33.72 m

b) $v_{i}$ = 26 m/s

c) β = 71.08°

Explanation:

a)

When an object is thrown into the air under the effect of the gravitational force, the movement of the projectile is observed. Then it can be considered as two separate motions, horizontal motion and vertical motion. Both motions are different, so that they can be handled independently.

Given data:

time = t = 4.00 s

Height = h = 20 m

Angle = θ = 60°

Horizontal distance = d = ?

Using 2nd equation of motion

$h = v_{y_{f}}t + \frac{1}{2}gt^{2}$

-20 = $v_{y_{f}}$ (4) + 0.5(-9.8)(4)²

$v_{y_{f}}$ (4) = 58.4

$v_{y_{f}}$ = 14.6 m/s

This is vertical component of velocity when the ball is on the roof. To calculate the Final velocity and horizontal component, we use

$v_{f}$ = $v_{y_{f}}$ / sinθ

$v_{f}$ = 14.6 / sin 60

$v_{f}$ = 16.86 m/s

$v_{x_{f}}$ = $v_{f}$ cosθ

$v_{x_{f}}$ = 16.86 cos 60

$v_{x_{f}}$ = 8.43 m/s

To calculate the horizontal distance

d = $v_{y_{f}}$ t

d = (8.43)(4)

d = 33.72 m

b)

We know the values of Landing angle, height of roof, time of flight. In part a, We calculate the landing velocity of the ball and also its horizontal and vertical component. As the ball followed the projectile path, and we know that in projectile motion the horizontal component of the velocity remain constant throughout his motion. So there is no acceleration along horizontal path.

So,

$v_{x_{f}}$ = $v_{x_{i}}$