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2 years ago

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Whatttuppppp I have nothing to say other than boo

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Marat540 [252]2 years ago

8 0

Whatsuppp how's life

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A sample of 4.50 g of methane occupies 12.7 dm3 at 310 K. (a) Calculate the work done when the gas expands isothermally against

valkas [14]

Answer:

(A) Work done will be 87.992 KJ

(B) Work done will be 167.4 KJ

Explanation:

We have given mass of methane m = 4.5 gram = 0.0045 kg

Volume occupies $V_1=12.7dm^3=12.7liters$

And volume is increased by $3.3dm^3$ so $V_2=12.7+3.3=16liters$

Temperature T = 310 K

Pressure is given as 200 Torr = 26664.5 Pa

(a) At constant pressure work done is given by

$W=P(V_2-V_1)=26664.5\times (16-12.7)=87992.85J=87.992kj$

(b) At reversible process work done is given by $W=nRTln\frac{V_2}{V_1}$

We have given mass = 4.5 gram

Molar mass of methane = 16

So number of moles $n=\frac{mass\ in\ gram}{mol;ar\ mass}=\frac{4.5}{16}=0.28125$

So work done $W=0.28125\times 8.314\times 310ln\frac{16}{12.7}=167.4J$

7 0

2 years ago

You are driving to the grocery store at 14 m/s. You are 115 m from an intersection when the traffic light turns red. Assume that

kogti [31]

Answer:

111.5 m

Explanation:

Given that You are driving to the grocery store at 14 m/s. You are 115 m from an intersection when the traffic light turns red. Assume that your reaction time is 0.50 s and that your car brakes with constant acceleration.

Use first equation of motion

V = U - at

Since the car is going to rest, V = 0 and a = negative

0 = 14 - a × 0.5

0.5a = 14

a = 14 /0.5

a = 28 m/s^2

Let us use second equation of motion

S = Ut - 1/2at^2

S = 14 × 0.5 - 0.5 × 28 × 0.5^2

S = 7 - 3.5

S = 3.5 m

115 - 3.5 = 111.5

Therefore, you are 111.5 metres from the intersection (in m) when you begin to apply the brakes.

6 0

2 years ago

Answer fast please !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

iogann1982 [59]

I believe it's the the third one. :)
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8 0

2 years ago

Read 2 more answers

Komok [63]

Answer:

A

Explanation:

6 0

2 years ago

Read 2 more answers

An electron passes through a point 2.83 cm 2.83 cm from a long straight wire as it moves at 35.5 % 35.5% of the speed of light p

igor_vitrenko [27]

Answer:

The magnitude of electron acceleration is $2.34 \times 10^{15}$ $\frac{m}{s^{2} }$

Explanation:

Given:

Distance from the wire to the field point $r = 2.83 \times 10^{-2}$ m

Speed of electron $v = 35.5 \%c$

Current $I = 17.7$ A

For finding the acceleration,

First find the magnetic field due to wire,

$B = \frac{\mu _{o}I }{2\pi r }$

Where $\mu_{o} = 4\pi \times 10^{-7}$

$B = \frac{4\pi \times 10^{-7} \times 17.7 }{2\pi (2.83 \times 10^{-2} ) }$

$B = 12.50 \times 10^{-5}$ T

The magnetic force exerted on the electron passing through straight wire,

$F = qvB$

$F = 1.6 \times 10^{-19} \times 0.355 \times 3 \times 10^{8} \times 12.50 \times 10^{-5}$

$F = 21.3 \times 10^{-16}$ N

From the newton's second law

$F = ma$

Where $m =$ mass of electron $= 9.1 \times 10^{-31}$ kg

So acceleration is given by,

$a = \frac{F}{m}$

$a = \frac{21.3 \times 10^{-16} }{9.1 \times 10^{-31} }$

$a = 2.34 \times 10^{15}$ $\frac{m}{s^{2} }$

Therefore, the magnitude of electron acceleration is $2.34 \times 10^{15}$ $\frac{m}{s^{2} }$

7 0

2 years ago