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4 years ago

6

The velocity of Desmond’s car changes from 20 meters/second to 10 meters/second. The change happens in 5 seconds. What’s the acc

eleration of the car? Use .

2 answers:

Sphinxa [80]4 years ago

8 0

V-V₀=at
a=(V-V₀)/t
a=(10-20)/5=-2 m/s²
Acceleration of the car is -2 m/s²

Vikentia [17]4 years ago

4 0

I would say its -2 m/s<span>2 ....hope this help..</span>

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A slider Crank Mechanism creates linear and rotary motion

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In lab you find that a 1-kg rock suspended above water weighs 10 N. When the rock is suspended beneath the surface of the water,

pav-90 [236]

Answer:

a. 2N

b. 12N

c. 20N

Explanation:

Dtaa given,

Weight of rock in air=10N

weight of rock in water =8N

a. When the rock is suspend in water, it will displace some water from thr container since Buoyant force is the acts upward and equals in magnitude to the volume of water displays,

The buoyant force is calculated as 10N-8N=2N

b. The scale reading will be the sum of the buoyant force and the weight of the container

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3 years ago

This food chain is incomplete. Two groups of organisms are missing. One group of organisms that is missing are the producers, pl

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Decomposers that recycle matter back into the food chain.

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4 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:

Part a: <em>The total amount of energy transfer by the work done is 54.81 kJ.</em>

Part b: <em>The total amount of energy transfer by the heat is 54.81 kJ</em>

Explanation:

Mass of Carbon Dioxide is given as m1=3 kg

Pressure is given as P1=3 bar =300 kPA

Volume is given as V1=0.5 m^3

Pressure in tank 2 is given as P2=2 bar=200 kPa

T=290 K

Now the Molecular weight of $CO_2$ is given as

M=44 kg/kmol

the gas constant is given as

$R=\frac{\bar{R}}{M}\\R=\frac{8.314}{44}\\R=0.189 kJ/kg.K$

Volume of the tank is given as

$V=\frac{mRT}{P_1}\\V=\frac{3 \times 0.189 \times 290}{300 }\\V=0.5481 m^3$

Final mass is given as

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

Mass of the CO2 moved to the cylinder

$m=m_1-m_3\\m=3-2=1 kg$

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

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

Part a:

So the Work done is given as

$W=P(V_2-V_1)\\W=200(0.774-0.5)\\W=54.81 kJ$

<em>The total amount of energy transfer by the work done is 54.81 kJ.</em>

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.

<em>The total amount of energy transfer by the heat is 54.81 kJ</em>

7 0

3 years ago