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3 years ago

How much power is generated by a motor that can do 3,000j of work in 75s

Physics

1 answer:

puteri [66]3 years ago

6 0

Answer:

40watts

Explanation:

Given parameters:

Work done = 3000J

Time = 75s

Unknown:

Power generated = ?

Solution:

Power is the rate at which work is done;

Power = $\frac{work done}{time}$

Now, insert the parameters and solve;

Power = $\frac{3000}{75}$ = 40watts

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ICE Princess25 [194]

Answer:

1,323 days left

Explanation:

147 x 10 = 1,470

1470 - 147 = 1,323

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3 years ago

The field between two charged parallel plates is kept constant. If the two plates are brought closer together, the potential dif

Komok [63]

Since the electric field between the plates is constant, If the two plates are brought closer together, the potential difference between the two plates decreases

The relation between potential difference and the electric field is given by ΔV = E.d

Since the electric field is maintained constant, the potential difference is directly inversely proportional to the distance between the plates.

The potential difference between the plates will therefore likewise decrease if the distance between the plates is reduced, we will state in this case.

The energy required to move a unit charge, or one coulomb, from one point to the other in a circuit is measured as the potential difference between the two points. Potential difference is measured in volts or joules per coulomb.

Refer to more about the potential difference here

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2 years ago

Why doesn’t a ball roll on forever after being kicked at a soccer game?

Dimas [21]

Answer:

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3 years ago

Read 2 more answers

Air enters a turbine operating at steady state at 8 bar, 1600 K and expands to 0.8 bar. The turbine is well insulated, and kinet

kobusy [5.1K]

Answer:

the maximum theoretical work that could be developed by the turbine is 775.140kJ/kg

Explanation:

To solve this problem it is necessary to apply the concepts related to the adiabatic process that relate the temperature and pressure variables

Mathematically this can be determined as

$\frac{T_2}{T_1} = (\frac{P_2}{P_1})^{(\frac{\gamma-1}{\gamma})}$

Where

Temperature at inlet of turbine

Temperature at exit of turbine

Pressure at exit of turbine

The steady flow Energy equation for an open system is given as follows:

$m_i = m_0 = mm(h_i+\frac{V_i^2}{2}+gZ_i)+Q = m(h_0+\frac{V_0^2}{2}+gZ_0)+W$

Where,

m = mass

m(i) = mass at inlet

m(o)= Mass at outlet

h(i)= Enthalpy at inlet

h(o)= Enthalpy at outlet

W = Work done

Q = Heat transferred

v(i) = Velocity at inlet

v(o)= Velocity at outlet

Z(i)= Height at inlet

Z(o)= Height at outlet

For the insulated system with neglecting kinetic and potential energy effects

$h_i = h_0 + WW = h_i -h_0$

Using the relation T-P we can find the final temperature:

$\frac{T_2}{T_1} = (\frac{P_2}{P_1})^{(\frac{\gamma-1}{\gamma})}\\$

$\frac{T_2}{1600K} = (\frac{0.8bar}{8nar})^{(\frac{1.4-1}{1.4})}\\ = 828.716K$

From this point we can find the work done using the value of the specific heat of the air that is 1,005kJ / kgK

$W = h_i -h_0W = C_p (T_1-T_2)W = 1.005(1600 - 828.716)W = 775.140kJ/Kg$

the maximum theoretical work that could be developed by the turbine is 775.140kJ/kg

4 0

3 years ago

The objective lens and the eyepiece of a telescope are spaced 85 cm apart. If the eyepiece is123 D what is the total magnificati

dedylja [7]

Answer:

The magnification would be "103.55". A further explanation is given below.

Explanation:

The given values are:

Distance between lens and eyepiece,

L = 85 cm

Eyepiece is,

= 123 D

Now,

The refractive power of eye piece will be:

⇒ $\frac{1}{f_e}=123D$

$f_e=\frac{1}{123D}$

$f_e=0.813 \ cm$

The length of the telescope will be:

⇒ $L=f_0+f_e$

⇒ $f_0=L-f_e$

On substituting the values, we get

⇒ $=85-0.813$

⇒ $=84.187 \ cm$

Now,

The magnification of the telescope will be:

⇒ $M=\frac{f_0}{f_e}$

⇒ $=\frac{84.187}{0.813}$

⇒ $=103.55$

5 0

3 years ago