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3 years ago

What are TWO benefits of scientific using a diagram to model the water cycle? [PLS HELP ASAP]~pick 2 letters~

Physics

1 answer:

LenKa [72]3 years ago

8 0

Answer:

A.) it can show changes that occur in many different parts of earth at the same time

D.) it can be used to show how the parts of the cycle relate to one another

Explanation:

When a scientific diagram is used as a model, it explains the phenomena described in the research.

Scientific models describing water cycle help students understand that a group of parts which relate can combine to form a larger system. The different parts of a water cycle such as evaporation and condensation all work together to facilitate continuity of the water cycle. Additionally, the models highlights the individual processes contributed by the stages of water cycle showing changes and resulting processes.

Option B and C are incorrect because the model diagrams show much details including individual processes contributed by each of the parts leading to functioning of the whole water cycle process.

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A 1.5 kilogram car is moving at 10 meters per second east. A braking force acts on the car for 5.0 seconds, reducing its velocit

Arada [10]

Answer:

I don't know

Explanation:

8 0

3 years ago

platform diving in the olympic games takes place at two heights: 5 meters and 10 meters. What is the velocity of a diver enterin

posledela

1) Velocity: 9.9 m/s and 14 m/s

The motion of the diver is a free-fall motion, so it is a uniform accelerated motion. Choosing downward as positive direction, the final velocity can be found by using the following suvat equation:

$v^2-u^2=2as$

where

v is the final velocity

u = 0 is the initial velocity (the diver starts from rest)

$a=g=9.8 m/s^2$ is the acceleration of gravity

s is the displacement

For the diver jumping from 5 m, s = 5 m, so

$v=\sqrt{2as}=\sqrt{2(9.8)(5)}=9.9 m/s$

For the diver jumping from 10 m, s = 10 m, so

$v=\sqrt{2as}=\sqrt{2(9.8)(10)}=14 m/s$

2) Time: 1.01 s and 1.43 s

The time of flight of each diver can be found by using the other suvat equation

$s=ut+\frac{1}{2}at^2$

And since u = 0, it can be reduced to

$s=\frac{1}{2}at^2$

For the diver jumping from 5 m, s = 5 m, so we find

$t=\sqrt{\frac{2s}{a}}=\sqrt{\frac{2(5)}{9.8}}=1.01 s$

For the diver jumping from 10 m, s = 10 m, so we find

$t=\sqrt{\frac{2s}{a}}=\sqrt{\frac{2(10)}{9.8}}=1.43 s$

5 0

3 years ago

A 23 kg body is moving through space in the positive direction of an x axis with a speed of 130 m/s when, due to an internal exp

babymother [125]

Answer:

a) Vx = 1088m/s

b) Vy = -162.93m/s

c) 5246745J

Explanation:

Mass of unbroken body = 23kg

Its velocity along +ve X-axis = 130m/s

Mass of first broken body, m1= 9.4kg

Its velocity along +ve X-axis = 130m/s

Nass of 2nd broken body, m2 = 6.1kg

Its velocity long-lived X - axis = -550m/s

Mass of 3rd broken body = ?

m3 = (23 - 9.4 - 6.1)kg

m3 = 7.5kg

Let velocity along the x-axis = Vx

Let the velocity along the x-axis = Vy

Applying law of conservation of momentum along x-axis

a) m1×0 + m2×(-550) + m3×(Vx) =M × 130

9.4 × 0 + 6.1× (-550) + 7.5(Vx) = 23 ×130

0 + (-5170) + 7.5Vx = 2990

2990 + 5170 = 7.5Vx

8160 = 7.5Vx

Vx = 8160/7.5

Vx = 1088m/s

b) Aplying conservation of momentum along the x-axis

(m1×130) + (m2 × 0) + (m3× Vy) = 0

(9.4 × 130) + (6.1 ×550) + 7.5Vy = 0

1222 + 0 + 7.5Vy = 0

1222 = -7.5Vy

Vy = 1222/(-7.5)

Vy = -262.93m/s

c) The energy released or change in KE is given by:

1/2[(m1v1^2) + (m2v2^2) +(m3Vx^2) ]= MV^2

Change in KE = 1/2[ 9.4× 130^2 + 6.1 × 550^2 + 7.5 × 1088^2 ] - 1/2(23 × 130^2)

Change in KE = 1/2[158860 + 1845250 + 8878080] - 1/2[388700]

Change in KE = 5441095 - 194350

Change in KE = 5246745J

4 0

3 years ago

Fill in the blanks for the following:

storchak [24]

Answer:

a. 4.21 moles

b. 478.6 m/s

c. 1.5 times the root mean square velocity of the nitrogen gas outside the tank

Explanation:

Volume of container = 100.0 L

Temperature = 293 K

pressure = 1 atm = 1.01325 bar

number of moles n = ?

using the gas equation PV = nRT

n = PV/RT

R = 0.08206 L-atm- $mol^{-1}$ $K^{-1}$

Therefore,

n = (1.01325 x 100)/(0.08206 x 293)

n = 101.325/24.04 = 4.21 moles

The equation for root mean square velocity is

Vrms = $\sqrt{\frac{3RT}{M} }$

R = 8.314 J/mol-K

where M is the molar mass of oxygen gas = 31.9 g/mol = 0.0319 kg/mol

Vrms = $\sqrt{\frac{3*8.314*293}{0.0319} }$ = 478.6 m/s

For Nitrogen in thermal equilibrium with the oxygen, the root mean square velocity of the nitrogen will be proportional to the root mean square velocity of the oxygen by the relationship

$\frac{Voxy}{Vnit}$ = $\sqrt{\frac{Mnit}{Moxy} }$