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3 years ago

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A red car with a mass of 3.0 kg traveling at 8 m/s collided with a blue car with a mass of 2.0 kg, which is at rest. The velocit

y of the blue car after the elastic collision is 9.6 m/s. What is the velocity of the red car?

1 answer:

anygoal [31]3 years ago

8 0

Answer:

460.8

Explanation:

just combine like terms

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The pressure of liquid varies as per<br>its depth

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Answer:

Yes.

Explanation:

The pressure varies as per the depth of the container

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Calculate the equivalent resistance for both circuits. Series circuit: 2 Ω and 4 Ω Parallel circuit: 2 Ω and 4 Ω Which circuit h

goldenfox [79]

Equivalent resistance is also known as the overall resistance.

For resistors in a series circuit, the total resistance is computed using the formula:

$R_{T} = R_{1}+ R_{2}+ R_{3}... R_{n}$

In other words, you just add up the resistance of each resistor in the series circuit. In your case you only have two resistors. You have 2Ω and 4Ω. So all you need to do is add that up.

$R_{T} = R_{1}+ R_{2}$
$R_{T} = 2 + 4=6$

The total resistance of the series circuit is 6Ω

In a parallel circuit you get the total resistance using the formula:
$\frac{1}{R_{T}} = \frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}...+\frac{1}{R_{n}}$

First you get the sum of all fractions and at the end take the reciprocal of the resulting fraction and divide. So let us take your problem into consideration where you have two resistors that have a resistance of 2Ω and 4Ω.

$\frac{1}{R_{T}} = \frac{1}{R_{1}}+\frac{1}{R_{2}}$
$\frac{1}{R_{T}} = \frac{1}{2}+\frac{1}{4}$
$\frac{1}{R_{T}} = \frac{2}{4}+\frac{1}{4}$
$\frac{1}{R_{T}} = \frac{3}{4}$

Get the reciprocal of the resulting fraction 3/4 and then divide. The reciprocal of 3/4 is 4/3.

4/3 = 1. 33Ω

So if you compare the equivalent resistance of the two circuits, the series circuit has a higher equivalent resistance.

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Estimate the speed of the water free surface and the time required to fill with water a cone-shaped container 1.5 m high and 1.5

Zolol [24]

Answer:

speed of water is 0.0007138m/s

Explanation:

From the law of conservation of mass

Rate of mass accumulation inside vessel = mass flow in - mass flow out

so, dm/dt = mass flow in - mass flow out

taking p as density

$d \frac{dQ}{dt} = pq_i_n$

where,

q(in) is the volume flow rate coming in

Q = is the volume of liquid inside tank at any time

But,

dQ = Adh

where ,

A = area of liquid surface at time t

h = height from bottom at time t

A = πr²

r is the radius of liquid surface

$r = (1.5/2) \div 1.5$ $h = \frac{h}{2}$

Hence,

$\pi( \frac{h}{2} )^2\frac{dh}{dt} =q_i_n$

$\frac{dh}{dt} = \frac{q_i_n}{\pi (\frac{h}{2})^2 } =\frac{4q_i_n}{\pi h^2}$

so, the speed of water surface at height h

$v = \frac{dh}{dt} =\frac{4q_i_n}{\pi h^2}$

where,

$q_i_n$ is 75.7 L/min = 0.0757m³/min

h = 1.5m

so,

$v = \frac{4 \times 0.0757}{\pi \times 1.5^2} \\\\v = 0.04283m/min$

v = 0.04283 /60

v = 0.0007138m/s

Hence, speed of water is 0.0007138m/s

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