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qaws [65]
4 years ago
11

1. Find the current ix in the circuits in Fig. E2.6.

Physics
1 answer:
artcher [175]4 years ago
6 0

It's hard to exaggerate how much easier that task would be,
if only I could see Fig. E2.6.

I can't imagine what reason you fantasized for why that figure
is printed alongside the problem, or whether the question ever
occurred to you.

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The universe could be considered an isolated system because
Stolb23 [73]

Answer:

A(many people think that no energy or matter exists outside the universe)

Explanation:

4 0
3 years ago
Water (density = 1x10^3 kg/m^3) flows at 15.5 m/s through a pipe with radius 0.040 m. The pipe goes up to the second floor of th
RUDIKE [14]

Answer:

The speed of the water flow in the pipe on the second floor is approximately 13.1 meters per second.

Explanation:

By assuming that fluid is incompressible and there are no heat and work interaction through the line of current corresponding to the pipe, we can calculate the speed of the water floor in the pipe on the second floor by Bernoulli's Principle, whose model is:

P_{1} + \frac{\rho\cdot v_{1}^{2}}{2}+\rho\cdot g\cdot z_{1} = P_{2} + \frac{\rho\cdot v_{2}^{2}}{2}+\rho\cdot g\cdot z_{2} (1)

Where:

P_{1}, P_{2} - Pressures of the water on the first and second floors, measured in pascals.

\rho - Density of water, measured in kilograms per cubic meter.

v_{1}, v_{2} - Speed of the water on the first and second floors, measured in meters per second.

z_{1}, z_{2} - Heights of the water on the first and second floors, measured in meters.

Now we clear the final speed of the water flow:

\frac{\rho\cdot v_{2}^{2}}{2} = P_{1}-P_{2}+\rho \cdot \left[\frac{v_{1}^{2}}{2}+g\cdot (z_{1}-z_{2}) \right]

\rho\cdot v_{2}^{2} = 2\cdot (P_{1}-P_{2})+\rho\cdot [v_{1}^{2}+2\cdot g\cdot (z_{1}-z_{2})]

v_{2}^{2}= \frac{2\cdot (P_{1}-P_{2})}{\rho}+v_{1}^{2}+2\cdot g\cdot (z_{1}-z_{2})

v_{2} = \sqrt{\frac{2\cdot (P_{1}-P_{2})}{\rho}+v_{1}^{2}+2\cdot g\cdot (z_{1}-z_{2}) } (2)

If we know that P_{1}-P_{2} = 0\,Pa, \rho=1000\,\frac{kg}{m^{3}}, v_{1} = 15.5\,\frac{m}{s}, g = 9.807\,\frac{m}{s^{2}} and z_{1}-z_{2} = -3.5\,m, then the speed of the water flow in the pipe on the second floor is:

v_{2}=\sqrt{\left(15.5\,\frac{m}{s} \right)^{2}+2\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (-3.5\,m)}

v_{2} \approx 13.100\,\frac{m}{s}

The speed of the water flow in the pipe on the second floor is approximately 13.1 meters per second.

4 0
3 years ago
A weather balloon is inflated to a volume of 26.8 LL at a pressure of 744 mmHgmmHg and a temperature of 31.2 ∘C∘C. The balloon r
elena-s [515]

Answer:

43.96 L

Explanation:

We are given that

V_1=26.8 L

P_1=744mm Hg

T_1=31.2^{\circ} C=31.2+273=304.2K

P_2=385mmHg

T_2=-14.8^{\circ}=273-14.8=258.2K

We know that

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

V_2=\frac{P_1V_1T_2}{P_2T_1}

Substitute the values

V_2=\frac{744\times 26.8\times 258.2}{385\times 304.2}

V_2=43.96 L

Hence, the volume of balloon at -14.8 degree Celsius=43.96 L

5 0
3 years ago
Force F=2.0N i - 3.0N k acts on a pebble with position vectorr=0.50m j - 2.0m k relative to the origin. In unit vector notation,
klasskru [66]

Answer with Explanation:

We are given that

Force acts on a pebble=2\hat{i}-3\hat{k} N

Position vector=r=0.5\hat{j}-2\hat{k} m

a.We have to find the resulting torque on the pebble about origin.

Torque=r\times F

Substitute the values then we get

Torque= (0.5j-2k)\times (2i-3k)

Torque=-k-1.5i-4jN-m

By using i\times j=k,j\times k=i,k\times i=j,j\times i=-k,k\times j=-i,i\times k=-j,i\times i=j\times j=k\times k=0

b.r=2i-3k

r-r_1=(0.5j-2k)-(2i-3k)=-2i+0.5j+k

Torque about point (2,0,-3)

\tau=(-2i+0.5j+k)\times (2i-3k)

\tau=-6j-k-1.5i+2j=-1.5i-4j-kN-m

6 0
3 years ago
Where is the sun in the life cycle? HELP ITS A FINAL EXAM
scZoUnD [109]

Answer:

At the very beginning it is what states the whole life cycle

8 0
3 years ago
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