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3 years ago

12

How do I do this question?

1 answer:

Mkey [24]3 years ago

5 0

a. 27

after that all you need to do is use the work formula as you will be able to acheive results

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Materials that are soft and porous will absorb energy causing a decrease in amplitude and energy of the sound. This is called __

Vedmedyk [2.9K]

Damping of the sound

5 0

2 years ago

Convert the number from scientific into standard notation: 5.9 x 10-2

guapka [62]

Move the decimal point to:
Left : (if the exponent of ten is a negative number -) ... OUR CASE HERE (-2)
or to
Right : (if the exponent is positive +).

You should move the point as many times as the exponent indicates.
Do not write the power of ten anymore.

So, standard form is:
Two points to the left {Exponent of Ten is Negative (-2)}
0.059 ... (without the 10)

6 0

2 years ago

An object of mass, m1 with a velocity, v1 collides with another object at rest (v2 = 0) with a mass, m2. After the collision, m1

goblinko [34]

Answer:

$v"_{1} = v_{1} tan$ Θ

$v^{"} _{2} = \frac{m_{1}v_{1}}{m_{2}cos}$ Θ

Θ = $tan^{-1}(\frac{v^{"} _{1} }{v_{1} } )$

Explanation:

Applying the law of conservation of momentum, we have:

Δ $p_{x = 0}$

$p_{x} = p"_{x}$

$m_{1}v_{1} = m_{2}v"_{2} cos$ Θ (Equation 1)

Δ $p_{y} = 0$

$p_{y} = p"_{y}$

$0 = m_{1} v"_{1} - m_{2} v"_{2} sin$ Θ (Equation 2)

From Equation 1:

$v"_{2} = \frac{m_{1}v_{1}}{m_{2}cos}$ Θ

From Equation 2:

$m_{2} v"_{2}$ sinΘ = $m_{1} v_{1}$

$v"_{1} = \frac{m_{2} v"_{2}sinΘ}{m_{1} }$

Replacing Equation 3 in Equation 4:

$v"_{1}=\frac{m_{2}\frac{m_{1}v_{1}}{m_{2}cosΘ}sinΘ}{m_{1}}$

$v"_{1}=v_{1}\frac{sinΘ}{cosΘ}$

$v"_{1}=v_{1}tan$ Θ (Equation 5)

And we found Θ from the Equation 5:

tanΘ= $\frac{v"_{1}}{v_{1}}$

Θ= $tan^{-1}(\frac{v"_{1}}{v_{1}})$

7 0

3 years ago

When water is boiled under a pressure of 2.00atm, the heat of vaporization is 2.20×106J/kg and the boiling point is 120∘C. At th

Vikki [24]

Answer:

2033219.05 J

Explanation:

V = Volume

P = Pressure = 2 atm

m = Mass of water = 1 kg

$L_v$ = Heat of vaporization = $2.2\times 10^6\ J/kg$

Work done in an isobaric system is given by

$W=-P\Delta V\\\Rightarrow W=-2\times 101325\times (1\times 10^{-3}-0.824)\\\Rightarrow W=166780.95\ J$

Work done is 166780.95 J

Change in internal energy is given by

$\Delta U=Q-W$

Heat is given by

$Q=mL_v\\\Rightarrow Q=1\times 2.2\times 10^6\\\Rightarrow Q=2.2\times 10^6\ J$

$\Delta U=2.2\times 10^6-166780.95\\\Rightarrow \Delta U=2033219.05\ J$

The increase in internal energy of the water is 2033219.05 J

6 0

3 years ago

A 12 volt battery in a motor vehicle is capable of supplying the starter motor with 150 A. It is noticed that the terminal volta

qwelly [4]

Answer: $0.0133 \Omega$

Explanation:

Given

Voltage=12 V

Current(I)=150 A

$V_{terminal}=10 V$

$r_{internal}=\frac{\Delta V}{I}$

$r_{internal}=\frac{12-10}{150}$

$r_{internal}=\frac{2}{150}=0.0133 \Omega$

4 0

3 years ago