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3 years ago

Which of these statements is true?

Physics

2 answers:

balu736 [363]3 years ago

8 0

Imma go with A.

Hope this helps:)

GaryK [48]3 years ago

7 0

It is true that <span>being tired makes it harder to have a good attitude. </span>

You might be interested in

20 quantities and classified them in vector and scaler quantities

ss7ja [257]

Answer:

A scalar quantity is defined as the physical quantity that has only magnitude, for example, mass and electric charge. On the other hand, a vector quantity is defined as the physical quantity that has both magnitude as well as direction like force and weight.

5 0

2 years ago

Argon gas enters steadily an adiabatic turbine at 900 kPa and 450C with a velocity of 80 m/s and leaves at 150 kPa with a veloc

Crazy boy [7]

Answer:

Temperature at the exit = $267.3 C$

Explanation:

For the steady energy flow through a control volume, the power output is given as

$W_{out}= -m_{f}(h_{2}-h_{1} + \frac{v_{2}^{2}}{2} - \frac{v_{1}^{2}}{2})$

Inlet area of the turbine = $60cm^{2}= 0.006m^{2}$

To find the mass flow rate, we can apply the ideal gas laws to estimate the specific volume, from there we can get the mass flow rate.

Assuming Argon behaves as an Ideal gas, we have the specific volume $v_{1}$

$v_{1}=\frac{RT_{1}}{P_{1}}=\frac{0.2081\times723}{900}=0.1672m^{3}/kg$

$m_{f}=\frac{1}{v_{1}}\times A_{1}V_{1} = \frac{1}{0.1672}\times(0.006)(80)=2.871kg/sec$

for Ideal gasses, the enthalpy change can be calculated using the formula

$h_{2}-h_{1}=C_{p}(T_{2}-T_{1})$

hence we have

$W_{out}= -m_{f}((C_{p}(T_{2}-T_{1}) + \frac{v_{2}^{2}}{2} - \frac{v_{1}^{2}}{2})$

$250= -2.871((0.5203(T_{2}-450) + \frac{150^{2}}{2\times 1000} - \frac{80^{2}}{2\times 1000})$

<em>Note: to convert the Kinetic energy term to kilojoules, it was multiplied by 1000</em>

evaluating the above equation, we have $T_{2}=267.3C$

Hence, the temperature at the exit = $267.3 C$

5 0

3 years ago

Three charges, q1 = +2.06 x 10-9 C, q2 = -3.27 x 10-9 C, and q3 = +1.05 x 10-9 C, are located on the x-axis at x1 = 0, x2 = 10.0

lisov135 [29]

Answer:

The resultant force on charge 3 is Fr= -2,11665 * 10^(-7)

Explanation:

Step 1: First place the three charges along a horizontal axis. The first positive charge will be at point x=0, the second negative charge at point x=10 and the third positive charge at point x=20. Everything is indicated in the attached graph.

Step 2: I must calculate the magnitude of the forces acting on the third charge.

F13: Force exerted by charge 1 on charge 3.

F23: Force exerted by charge 2 on charge 3.

K: Constant of Coulomb's law.

d13: distance from charge 1 to charge 3.

d23: distance from charge 2 to charge 3

Fr: Resulting force.

q1=+2.06 x 10-9 C

q2= -3.27 x 10-9 C

q3= +1.05 x 10-9 C

K=9-10^9 N-m^2/C^2

d13= 0,20 m