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3 years ago

14

Find the current passing through a circuit consisting of a battery and one resistor. The resistor has a resistance of 2 ohms and

the battery has a voltage of 12 volts.

2 answers:

ohaa [14]3 years ago

4 0

U=R*I=>I=U/R=12/2=6 A

andrew11 [14]3 years ago

4 0

Answer: 6 amps

i took the test and it was right :)

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What compound makes up carbohydrates?

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The monomer of glucose makes up all carbohydrates

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A light wave traveling through medium 1 strikes a boundary with medium 2 at at a 45 degree angle. the light then enters the seco

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2 years ago

What happens when a force is applied to an object in stable equilibrium

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the object is no longer in equilibrium .

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2 years ago

a car traveling at 24 m/s starts to decelerate steadily. It comes to a complete stop in 6 seconds what is its acceleration?

Irina18 [472]

We can solve for the acceleration by using a kinematic equation. First we should identify what we know so we can choose the correct equation.

We are given an original velocity of 24 m/s, a final velocity of 0 m/s, and a time of 6 s. We and looking for acceleration (a) in m/s^2.

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$v_f=v_i + at$

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3 years ago

Suppose that the hatch on the side of a Mars lander is built and tested on Earth so that the internal pressure just balances the

Kaylis [27]

Answer:

The value is $F_{net} = 4444 lb$

The force will be outward

Explanation:

From the question we are told that

The diameter of the disk is $d = 50.0 \ cm = \frac{50}{100} = 0.5 \ m$

The external pressure on Mars is $P = 650 \ N/m^2$

From the question we are told that

Internal pressure = External pressure

Generally the external Force on earth is

$F_E = P_{atm} * A$

Here $P_{atm}$ is the atmospheric pressure with value $P_{atm} = 1.013*10^{5}\ Pa$

So

$F_E = 1.013 *10^{5} * \pi * \frac{d^2}{4}$

=> $F_E = 1.013 *10^{5} *3.142 * \frac{0.50 ^2}{4}$

=> $F_E = 19893 \ N$

Generally the external Force on Mars is

$F= P * A$

$F = 650 * \pi * \frac{d^2}{4}$

=> $F = 650 *3.142 * \frac{0.5^2}{4}$

=> $F = 127.6 \ N$

Net force is mathematically represented as

$F_{net} = F_E -F$

=> $F_{net} = 19893 -127.6$

=> $F_{net} = 19765.6 \ N$ converting to pounds

$F_{net} = \frac{19765.6}{4.448}$

=> $F_{net} = 4444 lb$

Given that that the value is positive then the force will be outward

7 0

2 years ago