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Vanyuwa [196]
3 years ago
8

According to the picture, which is the least dense?

Physics
2 answers:
STatiana [176]3 years ago
7 0

Answer:

a

Explanation:

alexdok [17]3 years ago
4 0

Answer:

a the chess peice

Explanation:

my head

You might be interested in
Help with TEST pls!!
Westkost [7]

Answer:

12. D.

13. C.

14. H.

15. F.

16. G.

17. A.

18. D.

19. B.

20. C.

Hope this helps :)

Explanation:

8 0
3 years ago
A 67 Vrms source is placed across a load that consists of a 12 ohm resistor in series with an capacitor whose reactance is 5 ohm
SVETLANKA909090 [29]

a) The average true power is 318.3 W

b) The reactive power is 132.6 W

c) The apparent power is 344.8 W

d) The power factor is 0.92

Explanation:

a)

For a circuit made of a resistor and a capacitor, the average (true) power is given by the resistive part of the circuit only.

Therefore, the average true power is given by:

P=I^2R

where

I is the current

R is the resistance

In this problem, we have

V = 67 V (rms voltage)

R=12 \Omega is the resistance of the load

X=5\Omega is the reactance of the circuit

First we have to find the impedance of the circuit:

Z=\sqrt{R^2+X^2}=\sqrt{12^2+5^2}=13 \Omega

Then we can find the current in the circuit by using Ohm's law:

I=\frac{V}{Z}=\frac{67}{13}=5.15 A

Therefore, the average true power is

P=I^2R=(5.15)^2(12)=318.3 W

b)

The reactive power of a circuit consisting of a resistor and a capacitor is the power given by the capacitive part of the circuit.

Therefore, it is given by

Q=I^2X

where

I is the current

X is the reactance of the circuit

In this circuit, we have

I=5.15 A (current)

X=5 \Omega (reactance)

Therefore, the reactive power is

Q=(5.15)^2(5)=132.6W

c)

In a circuit with a resistor and a capacitor, the apparent power is given by both the resistive and capacitive part of the circuit.

Therefore, it is given by

S=I^2Z

where

I is the current

Z is the impedance of the circuit

Here we have

I = 5.15 A (current)

Z=13 \Omega (impedance)

Therefore, the apparent power is

S=I^2 Z=(5.15)^2(13)=344.8 W

d)

For a circuit with a resistor and a capacitor, the power factor is the ratio between the true power and the apparent power. Mathematically:

PF=\frac{P}{S}

where

P is the true power

S is the apparent power

For this circuit, we have

P = 318.3 W (true power)

S = 344.8 W (apparent power)

So, the power factor is

PF=\frac{318.3}{344.8}=0.92

Learn more about power and circuits:

brainly.com/question/4438943

brainly.com/question/10597501

brainly.com/question/12246020

#LearnwithBrainly

3 0
3 years ago
A proton is confined within an atomic nucleus of diameter 3.60 fm. part a estimate the smallest range of speeds you might find f
Cerrena [4.2K]
The answer for this problem would be:
Assuming non-relativistic momentum, then you have: 
ΔxΔp = mΔxΔv = h / (4) 
Δv = h / (4πmΔx) 
m ~ 1.67e-27 h ~ 6.62e-34,Δx = 4e-15 --> 
Δv ~ 6.62e-34 / (4π * 1.67e-27 * 4e-15) ~ 7,886,270 m/s ~ 7.89e6 m/s 
That's about 1% of the speed of light, the assumption that it's non-relativistic.
3 0
3 years ago
In the diagram, q1 = +1.39*10^-9 C and q2 = +3.22*10^-9 C. The electric field at point P is zero. What is the distance from P to
Sphinxa [80]

Answer:

0.388

Explanation:

6 0
4 years ago
An engine extracts 441.3kJ of heat from the burning of fuel each cycle, but rejects 259.8 kJ of heat (exhaust, friction,etc) dur
Lilit [14]

Answer:

  • The thermal efficiency is 0.4113.

Explanation:

We know that the thermal efficiency is the ratio of work done by the engine over the heat taken

\eta = \frac{W_{made}}{Q_{in}}

Now, how much work the engine do in a cycle?

We know that the work done in a cycle must be equal to the heat taken minus the heat rejected

W{made} = Q_{in} - Q_{rejected}

So, the thermal efficiency will be:

\eta = \frac{Q_{in} - Q_{rejected}}{Q_{in}}

\eta = \frac{Q_{in}}{Q_{in}} - \frac{Q_{rejected}}{Q_{in}}

\eta = 1 - \frac{Q_{rejected}}{Q_{in}}

Putting the values of the problem

\eta = 1 - \frac{259.8 kJ }{441.3kJ}

\eta = 0.4113

7 0
4 years ago
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