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3 years ago

Pls help guys this is my 7th time posting XD

Physics

2 answers:

borishaifa [10]3 years ago

5 0

Answer:

Explanation:

initial velocity u = 54 km/h = 15 m/s

final velocity v = 18 km/h = 5 m/s

distance s = 10 m

1. v^2 = u^2 + 2as

5^2 = 15^2 + 2a × 10

25 = 225 + 20a

25 - 225 = 20a

20 a = -200

a = -200/2

a = -100m/s^2

∴Deacceleration = -100m/s^2

2. v = u + at

5 = 15 -100t

5-15 = -100t

-10 = -100t

t = 100 / 10

∴t = 10 sec

Total distance covered by the car 10 m

zysi [14]3 years ago

5 0

To me, the hardest part of this whole thing is keeping the units straight. We're starting out with information given to us in kilometers, hours, and meters, and we have to come up with answers in m/s² , seconds, and meters.

When I worked this problem, I jumped right in without thinking, and I immediately got bogged down when I had to go off to the side and convert some units.

Now I know better. THIS time, before we get all tangled up trying to solve anything, let's get clever and change everything to m/s right now !

(54 km/hour) · (1,000m/km) · (1 hour/3600 sec) = 15 meters/second

(18 km/hour) · (1,000 m/km) · (1 hour/3600 sec) = 5 meters/second

NOW I'll betcha it's gonna be about 70% faster and easier !

i). Acceleration = (change in speed / time for the change)

We know the distance, but not the time. I know there's a formula for it, but I've learned so many formulas during my lifetime that I can't remember ALL of them. So I just memorize some of them, and I work things out from the formulas that I know. Here's how I do time:)

Average speed during the given slow-down = (1/2)·(15+5) = 10 m/s

Distance covered during the given slow-down = 10 m.

Time = (distance) / (average speed) = (10m) / (10 m/s) = 1 second

Acceleration = (change in speed) / (time for the change)

Acceleration = (5 m/s - 15 m/s) / (1 second)

Acceleration = (-10 m/s) / (1 second)

Acceleration = -10 m/s² (or 'Deceleration' = 10 m/s² )

_____________________________________________

For parts ii). and iii)., there's a big shift in the question.

It only gave you the slow-down from 54 to 18 km/hr for the purpose of calculating the deceleration. NOW, for the rest of the answers, it's talking about a complete stop ... 0 m/s .

____________________________________________

ii). Time required to stop = (initial speed) / (deceleration)

Time to stop from 54 km/hr = (15 m/s) / (10 m/s²)

Time to stop = 1.5 seconds

iii). Distance covered = (average speed) · (time to stop)

Distance covered = (7.5 m/s) · (1.5 sec)

Distance covered = 11.25 meters

OR ... use the official formula:

Distance = (1/2) · (acceleration) · (time² )

Distance = (1/2) · (10 m/s²) · (1.5 sec)²

Distance = 11.25 meters yay !

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2 years ago

A rigid tank initially contains 3kg of carbon dioxide (CO2) at a pressure of 3bar.The tank is connected by a valve to a friction

marissa [1.9K]

Answer:

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Part b: The total amount of energy transfer by the heat is 54.81 kJ

Explanation:

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the gas constant is given as

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Final mass is given as

$m_2=\frac{P_2V}{RT}\\m_2=\frac{200\times 0.5481}{0.189\times 290}\\m_2=2 kg$

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The initial mass in the cylinder is given as

$m_{(cyl)_1}=\frac{P_{(cyl)_1}V_1}{RT}\\m_{(cyl)_1}=\frac{200\times 0.5}{0.189 \times 290}\\m_{(cyl)_1}=1.82 kg$

The mass after the process is

$m_{(cyl)_2}=m_{(cyl)_1}+m\\m_{(cyl)_2}=1.82+1\\m_{(cyl)_2}=2.82\\$

Now the volume 2 of the cylinder is given as

$V_{(cyl)_2}=\frac{m_{(cyl)_2}RT}{P_2}\\m_{(cyl)_2}=\frac{2.82\times 0.189\times 290}{200}\\m_{(cyl)_1}=0.774 m^3$

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So the Work done is given as

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The total amount of energy transfer by the work done is 54.81 kJ.

Part b:

The total energy transfer by heat is given as

$Q=\Delta U+W\\Q=0+W\\Q=54.81 kJ$

As the temperature is constant thus change in internal energy is 0.

The total amount of energy transfer by the heat is 54.81 kJ

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3 years ago

Pls help guys this is my 7th time posting XD​

Pls help guys this is my 7th time posting XD