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3 years ago

11

Jill can use a force of 12.0 N to lift a single box up 5.00 m. How many of these boxes must she lift it in a minute to use 60.0

W of power?

1 answer:

Mama L [17]3 years ago

5 0

Answer:

60 boxes

Explanation:

The work done by lifting a single box is equal to the force applied (the weight of the box) times the displacement of the box:

$W_1 = Fd=(12.0 N)(5.00 m)=60 J$

Power is related to the work done by the equation:

$P=\frac{W}{t}$

where W is the work done and t is the time. In this problem, we are told that the power used is P=60.0 W, while the time taken is t = 1 min = 60 s, so the total work done must be

$W=Pt=(60.0 W)(60 s)=3600 J$

Therefore, the number of boxes that she has to lift in order to use this power is the total work divided by the work done in lifting each box:

$N=\frac{W}{W_1}=\frac{3,600 J}{60 J}=60$

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g The current through a 10 m long wire has a current density of 4 cross times 10 to the power of 6 space open parentheses bevell

olga55 [171]

Answer:

The value is $V = 2 V$

Explanation:

From the question we are told that

The length of the wire is $l = 10 \ m$

The current density is $J = 4*10^6 \ A/m^2$

The conductivity is $\sigma = 2*10^{7} \ S/m$

Generally conductivity is mathematically represented as

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Here R is the resistance which is mathematically represented as

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Here I is the current which is mathematically represented as

$I = J * A$

So

$R = \frac{V}{ J * A}$

And

$\sigma = \frac{l}{\frac{V}{ J * A} * A}$

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=> $V = \frac{l * J}{\sigma }$

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=> $V = 2 V$

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3 years ago

Air enters a nozzle steadily at 2.21 kg/m3 and 20 m/s and leaves at 0.762 kg/m3 and 150 m/s. If the inlet area of the nozzle is

saveliy_v [14]

Answer:

a) The mass flow rate through the nozzle is 0.27 kg/s.

b) The exit area of the nozzle is 23.6 cm².

Explanation:

a) The mass flow rate through the nozzle can be calculated with the following equation:

$\dot{m_{i}} = \rho_{i} v_{i}A_{i}$

Where:

$v_{i}$ : is the initial velocity = 20 m/s

$A_{i}$ : is the inlet area of the nozzle = 60 cm²

$\rho_{i}$ : is the density of entrance = 2.21 kg/m³

$\dot{m} = \rho_{i} v_{i}A_{i} = 2.21 \frac{kg}{m^{3}}*20 \frac{m}{s}*60 cm^{2}*\frac{1 m^{2}}{(100 cm)^{2}} = 0.27 kg/s$

Hence, the mass flow rate through the nozzle is 0.27 kg/s.

b) The exit area of the nozzle can be found with the Continuity equation:

$\rho_{i} v_{i}A_{i} = \rho_{f} v_{f}A_{f}$

$0.27 kg/s = 0.762 kg/m^{3}*150 m/s*A_{f}$

$A_{f} = \frac{0.27 kg/s}{0.762 kg/m^{3}*150 m/s} = 0.00236 m^{2}*\frac{(100 cm)^{2}}{1 m^{2}} = 23.6 cm^{2}$

Therefore, the exit area of the nozzle is 23.6 cm².

I hope it helps you!

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In which electric circuit would the voltmeter read 10 volts ?

ki77a [65]

Given that,

Voltage = 10 volt

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Put the value into the formula

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$\dfrac{1}{R}=\dfrac{1}{4}$

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Using ohm's law

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$I=\dfrac{V}{R}$

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Put the value into the formula

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Answer:

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