What do many geologist think is the driving force of plate tectonics? explain?

1 answer:

7 0

The leading theory of the driving force behind tectonic plate motion envisaged large scale convection current in the upper mantle, which can be transmitted throug the asthenosphere.

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A student sees a newspaper ad for an apartment that has 1730 square feet (ft2) of floor space. How many square meters of area ar

SVETLANKA909090 [29]

The correct answer for this question is 160.89 m².

The square footage of a newspaper ad for an apartment is = 1730ft²

We know that one square foot equals 0.093 square meters.

We can determine the area in square meters using the above conversion.

Total square meters = 0.093×1730ft² (or) 1730÷10.7

= 160.89 m²

To know more about conversion calculations, click here:

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6 0

2 years ago

A pulley system has an efficiency of 87.5 percent.

Gelneren [198K]

Answer:

work done on desk = m g h = 105 * 9.81 * 2.46 = 2534 Joules

Explanation:

so work in = 2534 / 0.875 = 2896 Joules

that is your 648 times distance your hand moved holding the rope

2896 = 648 * x

4.47 meters

I think maybe a typo, 648 N is too big, maybe 64.8 ? Any block and tackle system does better than that.

7 0

2 years ago

You have 10 ohm and a 100 ohm resistor in parallel. You place this equivalent resistance in series with an LED, which is rated t

Nataly [62]

Answer:

Approximately $\rm 2.0\; V$ .

Approximately $\rm 30 \; mA$ . (assumption: the LED here is an Ohmic resistor.)

Explanation:

The two resistors here $R_1= 10\; \Omega$ and $R_2= 100\; \Omega$ are connected in parallel. Their effective resistance would be equal to

$\displaystyle \frac{1}{\dfrac{1}{R_1} + \dfrac{1}{R_2}} = \frac{1}{\dfrac{1}{10} + \dfrac{1}{100}} = \frac{10}{11} \; \Omega$ .

The current in a serial circuit is supposed to be the same everywhere. In this case, the current through the LED should be $20\; \rm mA = 0.020\; \rm A$ . That should also be the current through the effective $\displaystyle \rm \frac{10}{11} \; \Omega$ resistor. Make sure all values are in standard units. The voltage drop across that resistor would be

$V = I \cdot R = 0.020 \times \dfrac{10}{11} \approx 0.182\; \rm V$ .

The voltage drop across the entire circuit would equal to

the voltage drop across the resistors, plus
the voltage drop across the LED.

In this case, that value would be equal to $1.83 + 0.182 \approx 2.0\; \rm V$ . That's the voltage that needs to be supplied to the circuit to achieve a current of $20\; \rm mA$ through the LED.