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2 years ago

2. How long must a 400 W electrical engine work in order to produce 300 kJ of work?

Physics

1 answer:

yanalaym [24]2 years ago

7 0

Answer:

Explanation:

400 W = 400 J/s

300000 J / 400 J/s = 750 s or 12.5 minutes

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A student took a calibrated 250.0 gram mass, weighed it on a laboratory balance, and found it read 266.5 g. What was the student

Kruka [31]

Answer:

B. 6.6%

Explanation:

The percentage error of a measurement can be calculated using the formula;

Percent error = (experimental value - accepted value / accepted value) × 100

In this question, the calibrated 250.0 gram mass is the accepted value while the weighed mass of 266.5 g is the experimental or measured value.

Hence, the percentage error can be calculated thus;

Percent error = (266.5-250.0/250.0) × 100

Percent error = 16.5/250 × 100

Percent error = 0.066 × 100

Percent error = 6.6%

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3 years ago

How does Mr. Anderson define solubility in the video?

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Answer:

7.3

Explanation:

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2 years ago

Gravity on Earth is 9.8 m/s^2, and gravity on Jupiter is 23.1 m/s^2. So, if the mass of a rock is 70 kilograms, it's weight on E

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~686newtons on earth and
~1617 newtons on jupiter
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2 years ago

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The resistivity of gold is at room temperature. A gold wire that is 0.9 mm in diameter and 14 cm long carries a current of 940 m

IgorC [24]

Answer:

0.0360531138247 V/m

Explanation:

$\rho$ = Resistivity of gold = $2.44\times 10^{-8}\ \Omega .m$ (General value)

I = Current = 940 mA

d = Diameter = 0.9 mm

A = Area = $\dfrac{\pi}{4}d^2$

E = Electric field

Resistivity is given by

$\rho=\dfrac{EA}{I}\\\Rightarrow E=\dfrac{\rho I}{A}\\\Rightarrow E=\dfrac{2.44\times 10^{-8}\times 940\times 10^{-3}}{\dfrac{\pi}{4}(0.9\times 10^{-3})^2}\\\Rightarrow E=0.0360531138247\ V/m$

The electric field in the wire is 0.0360531138247 V/m

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3 years ago

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A 500-N weight sits on the small piston of a hydraulic machine. The small piston has area 2.0 cm2. If the large piston has area

weqwewe [10]

Answer:

W₂= 10000 N

Explanation:

Pascal´s Principle can be applied in the hydraulic press:

If we apply a small force (F1) on a small area piston A1, then, a pressure (P) is generated that is transmitted equally to all the particles of the liquid until it reaches a larger area piston and therefore a force (F2) can be exerted that is proportional to the area (A2) of the piston:

Pressure is defined as the force (F) applied per unit area (A)

P=F/A (N/m²)

P1=P2

$\frac{F_{1} }{A_{1} } = \frac{F_{2} }{A_{2} }$

$F_{2} = \frac{F_{1}*A_{2} }{A_{1}}$ Equation (1)

Data

W₁ = weight sits on the small piston

F₁ = W₁= 500 N

A₁ = 2.0 cm²

A₂ = 40 cm²

Calculation of the weight (W₂) can the large piston support

We replace data in the equation (1)

$F_{2} = \frac{(500)*(40) }{2}$

F₂ = 10000 N

W₂= F₂= 10000 N

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3 years ago