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OlgaM077 [116]
3 years ago
15

A 51-kg woman runs up a vertical flight of stairs in 5.0 s. Her net upward displacement is 5.0 m. Approximately, what average po

wer did the woman exert while she was running?
Physics
1 answer:
Natalka [10]3 years ago
7 0

Answer:

510

Explanation:

Power= Rate of doing work i.e work done\time

=force*displacement\time

=mass*g*d\t

=51*10*5/5

=510watt

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The electric potential at the dot in the figure is 3160 V. What is charge q?
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Hi there!

Recall the equation for electric potential of a point charge:

V = \frac{kQ}{r}

V = Electric potential (V)
k = Coulomb's Constant(Nm²/C²)

Q = Charge (C)
r = distance (m)

We can begin by solving for the given electric potentials. Remember, charge must be accounted for. Electric potential is also a SCALAR quantity.

Upper right charge's potential:

V = \frac{(8.99*10^9)(-5 * 10^{-9})}{0.04} = -1123.75 V

Lower left charge's potential:

V = \frac{(8.99*10^9)(5*10^{-9})}{0.02} = 2247.5 V

Add the two, and subtract from the total EP at the point:

3160 + 1123.75 - 2247.5 = 2036.25


The remaining charge must have a potential of 2036.25 V, so:

2036.25 = \frac{(8.99*10^9)(Q)}{\sqrt{0.02^2 + 0.04^2}}\\\\2036.25 = \frac{(8.99*10^9)Q}{0.0447} \\\\Q = 0.000000010127 = \boxed{10.13nC}


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A small water pump is used in an irrigation system. The pump takes water in from a river at 10oC, 100 kPa at a rate of 5 kg/s. T
sergij07 [2.7K]

Answer:

0.98kW

Explanation:

The conservation of energy is given by the following equation,

\Delta U = Q-W

\dot{m}(h_1+\frac{1}{2}V_1^2+gz_1)-\dot{W} = \dot{m}(h_2+\frac{1}{2}V_2^2+gz_)

Where

\dot{m} = Mass flow

h_1 =Specific Enthalpy (IN)

h_2 = Specific Enthalpy (OUT)

g = Gravity

z_{1,2} = Heigth state (In, OUT)

V_{1,2} =Velocity (In, Out)

Our values are given by,

T_i = 10\°C

P_1 = 100kPa

\dot{m} = 5kg/s

z_2 = 20m

For this problem we know that as pressure, temperature as velocity remains constant, then

h_1 = h_2

V_1 = V_2

Then we have that our equation now is,

\dot{m}(gz_1) = \dot{m}(gz_2)+\dot{W}

\dot{W} = \frac{(5)(9.81)(0-20)}{1000}

\dot{W} = -0.98kW

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