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2 years ago

What are the variables of Boyle's law?

Physics

1 answer:

telo118 [61]2 years ago

4 0

Pressure and number of molecules of gas,volume and number of molecules of gas.

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A transformer with X turns in primary coil and Y turns in secondary coil is used to change the magnitude of voltage to 240 V. Ca

Vinil7 [7]

The input voltage is 120 V and the transformer is a step up transformer due to increase in the voltage induced in the secondary coil.

<h3>Input voltage </h3>

The input voltage of the transformer is the voltage of the primary coil and it is calculated as follows;

Ns/Np = Es/Ep

where;

Ns is the number of turn in the secondary coil
Np is the number of turn in the primary coil
Es is the secondary voltage
Ep is the primary voltage

2X/X = 240/Ep

2 = 240/Ep

Ep = 240/2

Ep = 120 V

Thus, the transformer is a step up transformer due to increase in the voltage induced in the secondary coil.

Learn more about transformer here: brainly.com/question/25886292

#SPJ1

3 0

2 years ago

In this diagram,the distance known as the amplitude is shown by choice

max2010maxim [7]

Answer:

Explanation:

the amplitude is the distance from the resting point to the crest/trough.

5 0

2 years ago

Read 2 more answers

A 6 kg cart starting from rest rolls down a hill and at the bottom has a speed of 10 m/s. What is the height of the hill?

Arisa [49]

Answer:

h = 5.09 m

Explanation:

Applying the Law of conservation of energy to this situation, we can write:

$Kinetic\ Energy\ Gained\ by\ the\ Cart = Potential\ Energy\ Lost\ by\ the\ Cart\\\frac{1}{2}mv^2 = mgh\\\\h = \frac{v^2}{2g}$

where,

h = height of the hill = ?

v = speed of cart at the end = 10 m/s

g = acceleration due to gravity = 9.81 m/s²

Therefore,

$h = \frac{(10\ m/s)^2}{(2)(9.81\ m/s^2)}\\\\$

4 0

2 years ago

A bowling ball of mass 5.8 kg moves in a straight line at 4.34 m/s How fast must a Ping-Pong ball of mass 2.214 g move in a stra

lilavasa [31]

Answer: 11369.46 m/s

Explanation:

We have the following data:

$m_{1}=5.8 kg$ is the mass of the bowling ball

$V_{1}=4.34 m/s$ is the velocity of the bowling ball

$m_{2}=2.214 g \frac{1 kg}{1000 g}=0.002214 kg$ is the mass of the ping-pong ball

$V_{2}$ is the velocity of the ping-pong ball

Now, the momentum $p_{1}$ of the bowling ball is:

$p_{1}=m_{1}V_{1}$ (1)

$p_{1}=(5.8 kg)(4.34 m/s)$

$p_{1}=25.172 kg m/s$ (2)

And the momentum $p_{2}$ of the ping-pong ball is:

$p_{2}=m_{2}V_{2}$ (3)

If the momentum of the bowling ball is equal to the momentum of the ping-pong ball:

$p_{1}=p_{2}$ (4)

$m_{1}V_{1}=m_{2}V_{2}$ (5)

Isolating $V_{2}$ :

$V_{2}=\frac{m_{1}V_{1}}{m_{2}}$ (6)

$V_{2}=\frac{25.172 kg m/s}{0.002214 kg}$ (7)

Finally:

$V_{2}=11369.46 m/s$

6 0

2 years ago

My name is scooter hey fat boy you waa fight

zavuch27 [327]

Answer:

HEHE

YES

Explanation:

6 0

2 years ago

Read 2 more answers