Which of the answers shows three examples of a physical change

Water is leaking out of an inverted conical tank at a rate of 1.5 cm3 /min at the same time that water is being pumped into the

iris [78.8K]

Answer:

a) Check Explanation

b) Check Explanation

c) The rate at which water is being pumped into the tank = 2.631 cm³/min

Explanation:

Let the rate of flow of water into the tank be k cm³/min

a) The image of the conical tank is presented in the attached image

Note, the radius and height of a cone are related through the similar triangles principle.

As shown in the attached image, it is evident that

r/h = 3/10

r = 3h/10 = 0.3 h

b) The quantities given in the problem.

- Shape of the tank, conical tank, Hence volume of the tank = πr²h/3

- total height of the tank, H = 10 cm

- Radius of the tank at the top, R = D/2 = 6/2 = 3 cm

- rate at which water is leaking from the tank = 1.5 cm³/min

- water is being pumped into the tank at constant rate of k cm³/min

- As at height of water, h = 2 cm, the rate of rise in water level = 1 cm/min

c) volume of the tank at any time = πr²h/3

Rate of change in the volume of water in the tank = (rate of flow into the tank) - (Rate of water flow out of the tank)

dV/dt = k - 1.5

V = πr²h/3 and r = 0.3 h, r² = 0.09 h²

V = 0.03πh³

dV/dt = (dV/dh) × (dh/dt)

dV/dh = 0.09π h²

dV/dt = 0.09π h² (dh/dt)

dV/dt = k - 1.5

0.09π h² (dh/dt) = k - 1.5

But at h = 2 cm, (dh/dt) = 1.0 cm/min

0.09π h² (dh/dt) = k - 1.5

0.09π 2² (1) = k - 1.5

k - 1.5 = 1.131

k = 1.5 + 1.131 = 2.631 cm³/min

5 0

3 years ago

In a second order lever system the force ratio is 2.5, the load is at the distance of 0.5m from the fulcrum find distance of eff

Fynjy0 [20]

Answer:

1.25 m

Explanation:

From the question given above, the following data were obtained:

Force ratio = 2.5

Distance of load from the fulcrum = 0.5 m

Distance of effort =.?

The distance of the effort from the fulcrum can be obtained as illustrated below:

Force ratio = Distance of effort / Distance of load

2.5 = Distance of effort / 0.5

Cross multiply

Distance of effort = 2.5 × 0.5

Distance of effort = 1.25 m

Therefore, the distance of the effort from the fulcrum is 1.25 m

8 0

3 years ago

¿Qué nombre recibe un electrón que abandona el espacio interno del átomo dirigiéndose al exterior del átomo?

Alenkinab [10]

Answer:

i dont speack spanish sorry

Explanation: agian sorry

7 0

2 years ago

A block of ice with mass 2.00 kg slides 0.750 m down an inclined plane that slopes downward at an angle of 36.9 degrees below th

zhannawk [14.2K]

Answer: $V_{f}=2.96m/s$

Firstly we have to draw the Free Body Diagram (FBD) as shown in the figure attached.

Where the weight $w$ of the block has an x-component and y-component:

$w_{x}=wsin(\theta)$ (1)

$w_{y}=wcos(\theta)$ (2)

As well as the Normal Force $N$ :

$N_{x}=Nsin(\theta)$ (3)

$N_{y}=Ncos(\theta)$ (4)

In addition, we know $N=w$ , then $\sum F_{y}=0$

In the X-component:

$\sum F_{x}=m.a$

$m.a=w_{x}$ (5)

Substituting (1) in (5):

$wsin(\theta)=m.a$ (6)

In addition, we know $w=m.g$ , where $m$ is the mass of the block and $g$ the gravity acceleration, which is equal to $9.8m/{s}^{2}$

So:

$m.g.sin(\theta)=m.a$ (7)

$a=g.sin(\theta)$ (8)

$a=5.88m/{s}^{2}$ (9) >>>>This is the acceleration of the block

On the other hand, we have the following equation that expresses a relation between the distance $d$ with the acceleration $a$ and time $t$ :

$d=\frac{1}{2}a{t}^{2}$ (10)

We already know the value of $d$ and calculated $a$ , we have to find $t$ :

$t=\sqrt{\frac{2d}{a}}$ (11)

$t=\sqrt{\frac{2(0.75m)}{5.88m/{s}^{2}}}$ (12)

$t=0.50s$ (13) >>>This is the time it takes to the block to go from the initial velocity $V_{o}$ to its final velocity $V_{f}$

If the acceleration is the variation of the velocity in time, we can use the following equation to find $V_{f}$ :

$V_{f}-V_{o}=a.t$ (13)

If $V_{o}=0$

$V_{f}=a.t$ (14)

$V_{f}=(5.88m/{s}^{2})(0.50s)$ (15)

Finally we get the value of the Final Velocity of the block:

$V_{f}=2.96m/s$

6 0

3 years ago

Most interstellar clouds are: Most interstellar clouds are: much bigger than our solar system. similar in size to clouds in Eart

cupoosta [38]

Answer:

Most interstellar clouds are much bigger than our solar system.

Explanation:

An interstellar cloud refers:

It is generally an accumulation of gas, plasma, and dust in our and other galaxies.
It is basically a denser-than-average region of the interstellar medium (ISM).

Interstellar clouds can be large up to 106 solar masses

It is also often said to be the most massive entities in the galaxy.

Hence

we can say about Interstellar clouds,

They are much bigger than our solar system.

learn more about interstellar clouds here:

brainly.com/question/14726563

#SPJ4

4 0

2 years ago