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1 year ago

What do tripped circuit breakers and blown fuses indicate.

Physics

1 answer:

vladimir2022 [97]1 year ago

7 0

Answer:

When a circuit breaker regularly trips or a fuse repeatedly blows, it is a sign that you are making excessive demands on the circuit and need to move some appliances and devices to other circuits. Or, it may indicate that your house has too few circuits and is in need of a service upgrade.

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To determine power, it is necessary to know the A. energy required and the time it takes. B. amount of work and the distance cov

Y_Kistochka [10]

The answer for this would be A. since power is Joules/seconds and energy is rated in Joules

5 0

2 years ago

Read 2 more answers

1.Do parallel universe exist? How?

Harlamova29_29 [7]

Answer:

Some physicists believe in a flatter version of multiple universes. ... But if the universe began at a finite point, as nearly every physicist agrees that it did, an alternate version of you likely doesn't exist, according to astrophysicist Ethan Siegel's 2015 Medium article.

7 0

1 year ago

A 4-L pressurecooker has an operating pressure of175 kPa. Initially, one-half of the volume is filled with liquid and the other

Soloha48 [4]

To solve this problem, it is necessary to apply the concepts related to the Energy balance and the mass balance that allow us to find in each state the data necessary to find the total Power of the system through heat exchange.

From the tables of properties of the Water it is possible to obtain at the given pressures the values of the specific volume, the specific energy and the specific enthalpy.

Given these pressures then we have to

$P_1 = 174kPA$

$\Rightarrow v_f = 0.001057m^3/kg\\\Rightarrow v_g = 1.0037m^3/kg\\\Rightarrow u_f = 486.82kJ/kg\\\Rightarrow u_g = 2524.5kJ/kg$

$P_2 = 175kPa \rightarrow$ Saturated vapor

$\Rightarrow v_2 = v_g = 1.0036 m^3/kg\\\Rightarrow v_2 = u_g = 2524.5kJ/kg$

$P_e = 175kPa \rightarrow$ Saturated vapor

$V = 4L$

Considering the process performed, the kinetic and potential energy can be neglected as well as the work involved by specific interactions in the system. Although it is an unstable process it can be treated as a uniform flow process. Considering these expressions we can perform a mass balance for which

$m_{in}-m_{out} = \Delta m_{system} \rightarrow m_e = m_1-m_2$

Similarly through the energy balance you can get that

$E_{in}-E_{out} = \Delta E_{system}$

$Q_{in} -m_eh_e = m_2u_2-m_1u_1 \Rightarrow$ Since $W=ke=pe=0$

The initial mass, initial internal energy, and final mass in the tank are

$m_1 = mf_+m_g = \frac{V_f}{v_f}+\frac{V_g}{v_g}$

$m_1 = \frac{0.002m^3}{0.001057m^3/kg}+\frac{0.002m^3}{1.0036m^3/kg}$

$m_1 = 1.893+0.002= 1.895Kg$

At the same time the internal energy can be defined from the mass in state 1 as,

$U_1 = m_1 u_1$

$U_1 = m_fu_f+m_gu_g$

$U_1 = 1.893*486.82+0.002*2524.5$

$U_1 = 926.6kJ$

The calculation of mass in state 2 can be defined as

$m_2 = \frac{V}{v_2} = \frac{0.004m^3}{1.0037m^3/kg}$

$m_2 = 0.004Kg$

Then from the mass and energy balances,

$m_e = m_1-m_2\\m_e = 1.895-0.004\\m_e = 1.891Kg$

In this way the calculation of the heat of entry would be subject to

$Q_{in} = m_eh_e+m_2u_2-m_1u_1$

$Q_{in} = 1.891*2700.2+0.004*2524.5-926.6$

$Q_{in} = 4188kJ$

Therefore the Power would be given as

$\dot{Q} = \frac{Q}{\Delta t} = \frac{4188kJ}{3600s} = 1.163kW$

Therefore the highest rate of heat transfer allowed is 1.163kW

3 0

2 years ago

A 0.12-mu or micro FF capacitor, initially uncharged, is connected in series with a 10-kohm resistor and a 12-V battery of negli

miv72 [106K]

Answer:

The time is 2.8 ms.

Explanation:

Given that,

Capacitor = 0.12 μF

Resistance = 10 kohm

Voltage = 12 V

Charge Q = 0.9 Q₀

We need to calculate the time constant

Using formula of time constant

$T=RC$

Put the value into the formula

$T=10\times10^{3}\times0.12\times10^{-6}$

$T=0.0012\ sec$

We need to calculate the time

Using formula of time

$Q=Q_{0}(1-e^{\frac{-t}{T}})$

Put the value into the formula

$0.9Q_{0}=Q_{0}(1-e^{\frac{-t}{0.0012}})$

$ln (0.1)=\dfrac{-t}{0.0012}$

$t=0.00276\ sec$

$t=2.8\ ms$

Hence, The time is 2.8 ms.

6 0

2 years ago

A 520kg rocket sled at rest is propelled along the ice by an engine developing a constant thrust of 12 000 N over a distance of

Maurinko [17]

<u>Answer:</u>

a) 480 000 J

b) 43 m/s

<u>Explanation:</u>

To calculate the work done by the engine, we can use the following formula:

$\boxed{\mathrm{Work \space\ done = Force \times Distance}}$ .

⇒ $12000 \times 40$

⇒ $480000 \space\ \mathrm J$

We know that the work done by the propeller was converted into motion of the sled, which means the propeller provided the sled with kinetic energy.

∴ work done by propeller = kinetic energy gained by sled

⇒ $480000 = \frac{1}{2} mv^2$

⇒ $480000 = \frac{1}{2} \times 520 \times v^2$

⇒ $480000 = 260v^2$

⇒ $v^2 = \frac{480000}{260}$

⇒ $v =\bf 43 \space\ m/s$

This means that its velocity after 40 m was 43 m/s.

4 0

1 year ago