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3 years ago

Do you think energy can be “used up?” Why or why not?

Physics

1 answer:

My name is Ann [436]3 years ago

8 0

I dint think energy can be used up because energy cant be created nor destroyed

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This parallel circuit has three resistors R1-120 ohm; R2-45 ohm: R3=360

Nina [5.8K]

Answer:

the answer is 30 Ohms. Not found in the options provided

5 0

3 years ago

Wire A carries 4 A into a junction, wire B carries 5 A into the same junction, and another wire is connected to the junction. Wh

nydimaria [60]

Answer:9A

Explanation:

let the last wire be wire C

According to Kirchhoff's rule

the sum of all currents entering a junction must be equal to the sum of all currents leaving a junction

Ic=Ia+Ib

Ic= 4+5

Ic=9A

7 0

3 years ago

Given the displacement vector D = (4î − 8ĵ) m, find the displacement vector R (in m) so that D + R = −3Dĵ. (Express your answer

Elena-2011 [213]

Answer:

--------------------------------

Explanation:

4 0

3 years ago

(8 points) Air is used as the working fluid in a simple ideal Brayton cycle that has a pressure ratio of 12, a compressor inlet

trapecia [35]

Answer:

a ) $\dot m = 351.49 kg/s$

b) $\dot m_{actual} = 1046.15 kg/s$

Explanation:

given data:

pressure ration rp = 12

inlet temperature = 300 K

TURBINE inlet temperature = 1000 K

AT the end of isentropic process (compression) temperature is

$\frac{T_2'}{T_1} = rp ^{\frac{\gamma -1}{\gamma}}$

$\frac{T_2'}{300} = 12^{\frac{1.4 -1}{1.4}}$

$T_2' = 610.181 K$

AT the end of isentropic process (expansion) temperature is

$\frac{T_3}{T_4'} = rp ^{\frac{\gamma -1}{\gamma}}$

$\frac{1000'}{T_4'} = 12^{\frac{1.4 -1}{1.4}}$

$T_4' = 491.66 K$

isentropic work is given as

$w(compressor) = CP (T_2' -T_1)$

w = 1.005(610.18 - 300)

w = 311.73 kJ/kg

w(turbine) = 1.005( 1000 - 491.66)

w(turbine) = 510.88 kJ/kg

a) mass flow rate for isentropic process is given as

$\dot m = \frac{70000}{510.88 - 311.73}$

$\dot m = 351.49 kg/s$

b) actual mass flow rate uis given as

$\dot m_{actual} = \frac{70000}{51.088\times 0.85 - \frac{311.73}{0.85}}$

$\dot m_{actual} = 1046.15 kg/s$

6 0

3 years ago

Can power be negative in physics? I am doing a lab in which I run down the stairs as quickly as possible. They are 5.4m tall and

RUDIKE [14]

Answer:

Yes.

Explanation:

A negative power would just represent a loss of power. So in your case it lost -1252.16 W

8 0

3 years ago