Chiếu tia sáng lên một gương phẳng biết góc tới bằng 15⁰ hãy tính góc phản xạ bằng bao nhiêu

The type of graph used to show how a part of something relates to the whole is which of the following?

Slav-nsk [51]

Answer:

Circle or pie graph is used to show how a part of something relates to the whole.

Explanation:

Pie graphs are easy to read and can present a very clear picture of the relationships.

4 0

3 years ago

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Describe what is meant by “a constant change of direction”. Identify whether the examples provided

MaRussiya [10]

A because I heard it is

3 0

3 years ago

An installation consists of a 30-kVA, 3-phase transformer, a 480-volt primary, and a 240-volt secondary. Calculate the largest s

Nikitich [7]

Answer: 45 A

Explanation:

Primary only protection 3-phase

I = 3 phase kVA / ( 1.723 * V)

I = 30000 / ( 1.732 * 480 ) = 36.085 A

Table 450.3(B)

Currents of 9A or more column

primary only protection = 125%

Max OCPD pri = 125% of I = 1.25 * 36.085 = 45.11 A

Table 450.3(B) Note 1 does not apply, use next smaller Table 240.6(A)

Next smaller = 45 A

5 0

1 year ago

une lampe soumise à une tension continue de 12V , consomme unepuissance de 1.8 W. L'intensité du courant traversant la lampe est

Eduardwww [97]

Answer:

I = 0.15 A

Explanation:

The question is "a lamp subjected to a continuous voltage of 12V , consumes a power of 1.8 W. The intensity of the current passing through the lamp is what"

Given that,

Power of the lamp, P = 1.8 W

Voltage, V = 12 V

We need to find the current through the lamp. The power of the lamp is given by :

$P=V\times I\\\\I=\dfrac{P}{V}\\\\I=\dfrac{1.8}{12}\\\\I=0.15\ A$

So, the current through the lamp is 0.15 A.

6 0

3 years ago

Consult Multiple-Concept Example 15 to review the concepts on which this problem depends. Water flowing out of a horizontal pipe

kiruha [24]

Answer:

The pressure of the water in the pipe is 129554 Pa.

Explanation:

<em>There are wrongly written values on the proposal, the atmospheric pressure must be 101105 Pa, and the density of water 1001.03 kg/m3, those values are the ones that make sense with the known ones.</em>

We start usign the continuity equation, and always considering point 1 a point inside the pipe and point 2 a point in the nozzle:

$A_1v_1=A_2v_2$

We want $v_2$ , and take into account that the areas are circular:

$v_2=\frac{A_1v_1}{A_2}=\frac{\pi r_1^2 v_1}{\pi r_2^2}=\frac{r_1^2 v_1}{r_2^2}$

Substituting values we have (we don't need to convert the cm because they cancel out between them anyway):

$v_2=\frac{r_1^2 v_1}{r_2^2}=\frac{(1.8cm)^2 (0.56m/s)}{(0.49cm)^2}=7.56m/s$

For determining the absolute pressure of the water in the pipe we use the Bernoulli equation:

$P_1+\frac{\rho v_1^2}{2}+\rho gh_1=P_2+\frac{\rho v_2^2}{2}+\rho gh_2$

Since the tube is horizontal $h_1=h_2$ and those terms cancel out, so the pressure of the water in the pipe will be:

$P_1=P_2+\frac{\rho v_2^2}{2}-\frac{\rho v_1^2}{2}=P_2+\frac{\rho (v_2^2-v_1^2)}{2}$

And substituting for the values we have, considering the pressure in the nozzle is the atmosphere pressure since it is exposed to it we obtain:

$P_1=101105 Pa+\frac{1001.03Kg/m^3 ((7.56m/s)^2-(0.56m/2)^2)}{2}=129554 Pa$

3 0

3 years ago