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3 years ago

Thiết bị nào sau đây không phải là nguồn điện

Physics

1 answer:

kumpel [21]3 years ago

7 0

Thiết bị không phải nguồn điện:

Đáp án D

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Water at 20oC flows through a long elliptical duct 30 cm wide and 22 cm high. What average velocity, in m/s, would cause the wei

mars1129 [50]

Explanation:

The given data is as follows.

Fluid is water so, density $\rho = 1000 kg/m^{3}$

Weight flow rate = 500 lbf/s = 2224.11 N/sec

Cross-sectional area (A) = $\pi \times \frac{30}{2} \times \frac{22}{2}$

= 0.05184 $m^{2}$

Hence, weight flow rate will be given as follows.

w = $\rho \times g \times A \times V$

2224.11 N/sec = $\rho \times g \times A \times V$

V = $\frac{2224.11}{1000 \times 9.81 \times 0.05184}$ m/s

= 4.373 m/s

Thus, we can conclude that average velocity in the given case is 4.373 m/s.

8 0

3 years ago

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Vector A⃗ points in the positive y direction and has a magnitude of 12 m. Vector B⃗ has a magnitude of 33 m and points in the ne

gavmur [86]

vector A has magnitude 12 m and direction +y

so we can say

$\vec A = 12 \hat j$

vector B has magnitude 33 m and direction - x

$\vec B = -33 \hat i$

Now the resultant of vector A and B is given as

$\vec A + \vec B = 12 \hat j - 33 \hat i$

now for direction of the two vectors resultant will be given as

$\theta = tan^{-1}\frac{12}{-33}$

$\theta = 160 degree$

so it is inclined at 160 degree counterclockwise from + x axis

magnitude of A and B will be

$R = \sqrt{A^2 + B^2}$

$R = \sqrt{12^2 + 33^2} = 35.11 m$

so magnitude will be 35.11 m

6 0

3 years ago

Scientists have designed solar cells to trap solar energy and convert it to _____ energy.

lapo4ka [179]

Scientists have designed solar cells to trap solar energy and convert it to D electrical energy. Its D because converting something has to do with recharging something and electronics have to charge. I hope I helped.

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3 years ago

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This force is involves the attraction between charged particles

Nikolay [14]

That's the 'electrostatic' force.

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3 years ago

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Find the potential inside and outside a uniformly charged solid sphere whose radius is R and whose total charge is q. Use infini

amid [387]

Answer:

Recall that the electric field outside a uniformly charged solid sphere is exactly the same as if the charge were all at a point in the centre of the sphere:

$E_{outside} =\frac{1}{4\pi(e_{0})}\frac{Q}{r^{2} } r^{'}$

lnside the sphere, the electric field also acts like a point charge, but only for the proportion of the charge further inside than the point r:

$E_{inside} =\frac{1}{4\pi(e_{0})}\frac{Q}{R^{2} } \frac{r}{R} r^{'}$

To find the potential, we integrate the electric field on a path from infinity (where of course, we take the direct path so that we can write the it as a 1 D integral):

$V(r>R)=\int\limits^r_\infty {\frac{1}{4\pi(e_{0)} }\frac{Q}{r^2} } \, dr=\frac{q}{4\pi(e_{0)} } \frac{1}{r} \\V(r$

= $\frac{q}{4\pi e_{0} } [\frac{1}{R} -\frac{r^{2}-R^{2} }{2R^{3} } ]$

∴NOTE: Graph is attached

8 0

4 years ago