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3 years ago

11

a circuit is connected to 8 volt battery and it has 2 amperes of current flowing through the wires.What is the resistance of the

circuit

1 answer:

guajiro [1.7K]3 years ago

4 0

Answer:

4 ohms

Explanation:

Current = Voltage/resistance

2 = 8/R

2R = 8

R = 4

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Determine the thrust produced if 1.5 x 10^3 kg of gas exits the combustion chamber each second, with a speed of 4.00 x 10^3 m/s.

ozzi

Answer:

The thrust is $6\times 10^6\ N$

Explanation:

Given that,

Mass of gas, $m=1.5\times 10^3\ kg$

The rate at which the gas is expelling, $\dfrac{dv}{dt}=4\times 10^{3}\ m/s$

We need to find the thrust produced by the gas.

We know that force is equal to the rate of change of momentum. So,

$F=\dfrac{p}{t}$

Also, p = mv

$F=\dfrac{mv}{t}$

So,

$F=1.5\times 10^3\times 4\times 10^3\\\\F=6\times 10^6\ N$

So, the thrust is $6\times 10^6\ N$

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3 years ago

What are the opposite ends of a magnet called? A. north and south terminals B. north and south poles C. magnetic fields D. magne

Bezzdna [24]

I think the answer is B.

Hope this helps.

6 0

2 years ago

Read 2 more answers

Which one of the following is not a fundamental physical quantity??! A. Temperature b. Current c. Area d. Mass

Hatshy [7]

Answer:

Out of this, Area is not a fundamental physical quantity.

3 0

2 years ago

If a flea can jump straight up to a height of 0.410 m , what is its initial speed as it leaves the ground?

aivan3 [116]

Initial velocity = $v_0$

acceleration in the downward direction = -9.8 $\frac {m}{s^2}$

Final velocity at the highest point = 0

Maximum height reached = 0.410 m

Now, Using third equation of motion:

$\(v^2 = {v_0}^{2} + 2aH$

$\(0^2 = {v_0}^{2} - 2 \times 9.8 \times 0.410$

${v_0}^{2} = 2 \times 9.8 \times 0.410$

$v_0 = 2.834 \frac {m}{s}$

Speed with which the flea jumps = $2.834 \frac {m}{s}$

4 0

3 years ago

What is approximate resistance between P and Q?<br> a) 0.5<br> b)0.8<br> c)2.0<br> d)2.2<br> e)3.6

Rufina [12.5K]

Answer:

0.5 Ohms

Explanation:

We note that the node Q is also between the resistors of 1ohm and 2ohms.

We note that the node P is also between the resistors of 2ohms and 3ohms.

Thus, all these resistors are in parallel, beween nodes P and Q

1/Re=1/R1+1/R2+1/R3

1/Re=1/1+1/2+1/3=(6+3+2)/6=11/6 [ohm^(-1)]

Re=6/11=0.54ohms

Rounding to the tenth: Re=0.5 ohms

8 0

2 years ago