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ryzh [129]
3 years ago
12

Question 14

Physics
1 answer:
Nimfa-mama [501]3 years ago
5 0

Answer:

The answer is Dependent Variable

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In a bumper car arena, two cars of equal mass are heading straight toward each other. The orange one is traveling at a speed of
marshall27 [118]

Answer: D.

Explanation:   Orange, at 3 meters per second if you calculate the net force being applied to the system.

hope this helps! ✌

4 0
3 years ago
Calculate the kinetic energy of a car of mass 800kg moving at 100 kmph.
valentina_108 [34]

Answer:

KE = 308642.02469136 J

4 0
2 years ago
1. If you push a 40.0kg mass object with a force of 145.ON, what will be the object's
horrorfan [7]

Newton said . . . F = m a

Divide each side by  m . . .  a = F / m

Acceleration = (force) / (mass)

Acceleration = (145.O N) / (40.0 kg)

<em>Acceleration = 3.625 m/s²</em>

4 0
3 years ago
A cylinder with a piston contains 0.300 mol of oxygen at 2.50×105 Pa and 360 K . The oxygen may be treated as an ideal gas. The
alukav5142 [94]

Answer:

a) W =  900   J.  b) Q =  3142.8   J . c) ΔU =  2242.8   J. d) W = 0. e) Q =   2244.78   J.  g) Δ U  =  0.

Explanation:

(a) Work done by the gas during the initial expansion:

The work done W for a thermodynamic constant pressure process is given as;

W  =  p Δ V

where  

p  is the pressure and  Δ V  is the change in volume.

Here, Given;

P 1 = i n i t i a l  p r e s s u r e  =  2.5 × 10^ 5   P a

T 1 = i n i t i a l   t e m p e r a t u r e  =  360   K

n = n u m b er   o f   m o l e s  =  0.300  m o l  

The ideal gas equation is given by  

P V = nRT

where ,

p  =  absolute pressure of the gas  

V =  volume of the gas  

n  =  number of moles of the gas  

R  =  universal gas constant  =  8.314   K J / m o l   K

T  =  absolute temperature of the gas  

Now we will Calculate the initial volume of the gas using the above equation as follows;

PV  =  n R T

2.5 × 10 ^5 × V 1  =  0.3 × 8.314 × 360

V1 = 897.91 / 250000

V 1  =  0.0036   m ^3  = 3.6×10^-3 m^3

We are also given that

V 2  =  2× V 1

V2 =  2 × 0.0036

V2 =  0.0072   m^3  

Thus, work done is calculated as;

W  =  p Δ V  = p×(V2 - V1)

W =  ( 2.5 × 10 ^5 ) ×( 0.0072  −  0.0036 )

W =  900   J.

(b) Heat added to the gas during the initial expansion:

For a diatomic gas,

C p  =  7 /2 ×R

Cp =  7 /2 × 8.314

Cp =  29.1  J / mo l K  

For a constant pressure process,  

T 2 /T 1  =  V 2 /V 1

T 2  =  V 2 /V 1 × T 1

T 2  =  2 × T 1  = 2×360

T 2  =  720  K

Heat added (Q) can be calculated as;  

Q  =  n C p Δ T  = nC×(T2 - T1)

Q =  0.3 × 29.1 × ( 720  −  360 )

Q =  3142.8   J .

(c) Internal-energy change of the gas during the initial expansion:

From first law of thermodynamics ;

Q  =  Δ U + W

where ,

Q is the heat added or extracted,

Δ U  is the change in internal energy,

W is the work done on or by the system.

Put the previously calculated values of Q and W in the above formula to calculate  Δ U  as;

Δ U  =  Q  −  W

ΔU =  3142.8  −  900

ΔU =  2242.8   J.

(d) The work done during the final cooling:

The final cooling is a constant volume or isochoric process. There is no change in volume and thus the work done is zero.

(e) Heat added during the final cooling:

The final process is a isochoric process and for this, the first law equation becomes ,

Q  =  Δ U  

The molar specific heat at constant volume is given as;

C v  =  5 /2 ×R

Cv =  5 /2 × 8.314

Cv =  20.785  J / m o l   K

The change in internal energy and thus the heat added can be calculated as;  

Q  = Δ U  =  n C v Δ T

Q =  0.3 × 20.785 × ( 720 - 360 )

Q =   2244.78   J.

(f) Internal-energy change during the final cooling:

Internal-energy change during the final cooling  is equal to the heat added during the final cooling Q  =  Δ U  .

(g) The internal-energy change during the isothermal compression:

For isothermal compression,

Δ U  =  n C v Δ T

As their is no change in temperature for isothermal compression,  

Δ T = 0 ,  then,

Δ U  =  0.

8 0
3 years ago
The device shows the relative humidity at 22 degrees celsius. What's the water vapor density if the maximum water vapor in air a
sweet [91]

Answer:

11.6 g/m³

Explanation:

The complete question is,

The device shows the relative humidity at 22 degrees celsius. What's the water vapor density if the maximum water vapor in air at this temperature is 20 gram/cubic meter? a device showing that at 22 degrees celsius the relative humidity is 58%.

Solution:

saturation of water in air is showed by the relative humidity. As it is 58% and not 100%. Lets scale density that we have, if humidity is 100%.

density is 20 gram/m³ when the humidity is 100%

When humidity is 58%

0.58 x 20 = 11.6 g/m³

Therefore, density is 11.6g/m³

6 0
3 years ago
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