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Amiraneli [1.4K]
1 year ago
13

A rightward force of 12.0 N is applied to a 2.0-kg object to accelerate it across a horizontal

Physics
1 answer:
ad-work [718]1 year ago
8 0

Answer:

below

Explanation:

Net accelerating force becomes  12-8 = 4 N

F = ma

4 = 2 * a

a = 2 m/s^2

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Two bumper cars colide into each other and each one bounces bach. wich of Newton's laws is it ?
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One mole of an ideal gas expands isothermally from 0.01 m3 to 0.05 m3. The final pressure of gas is 1.2 atm. [15 Marks] determin
erik [133]

Answer:

Initial pressure = 6 atm. Work = 0.144 J

Explanation:

You need to know the equation P1*V1=P2*V2, where P1 is the initial pressure, V1 is the initial volume, and P2 and V2 are the final pressure and volume respectively. So you can rearrange the terms and find that (1.2*0.05)/(0.01) = initial pressure = 6 atm. The work done by the system can be obtained calculating the are under the curve, so it is 0.144J

4 0
3 years ago
A monatomic gas is adiabatically compressed to 0.250 of its initial volume. Do each of the following quantities change?
Len [333]

Answer:

Given that

V2/V1= 0.25

And we know that in adiabatic process

TV^န-1= constant

So

T1/T2=( V1 /V2)^ န-1

So = ( 1/0.25)^ 0.66= 2.5

Also PV^န= constant

So P1/P2= (V2/V1)^န

= (1/0.25)^1.66 = 9.98

A. RMS speed is

Vrms= √ 3RT/M

But this is also

Vrms 2/Vrms1= (√T2/T1)

Vrms2=√2.5= 1.6vrms1

B.

Lambda=V/4π√2πr²N

So

Lambda 2/lambda 1= V2/V1 = 0.25

So the mean free path can be inferred to be 0.25 times the first mean free path

C. Using

Eth= 3/2KT

So Eth2/Eth1= T2/T1

So

Eth2= 2.5Eth1

D.

Using CV= 3/2R

Cvf= Cvi

So molar specific heat constant does not change

5 0
3 years ago
Read 2 more answers
A 234.0 g piece of lead is heated to 86.0oC and then dropped into a calorimeter containing 611.0 g of water that initally is at
Vaselesa [24]

Answer:24.70 ^{\circ}C

Explanation:

Given

mass of lead piece m_l=234 gm\approx 0.234 kg

mass of water in calorimeter m_w=611 gm\approx 0.611 kg

Initial temperature of water T_w=24^{\circ}C

Initial temperature of lead piece T_l=24^{\circ}C

we know heat capacity of lead and water are 125.604 J/kg-k and 4.184 kJ/kg-k respectively

Let us take T ^{\circ}C be the final temperature of the system

Conserving energy

heat lost by lead=heat gained by water

m_lc_l(T_l-T)=m_wc_w(T-T_w)

0.234\times 125.604(86-T)=0.611\times 4.184\times 1000(T-24)

86-T=\frac{0.611\times 4.184\times 1000}{29.391}(T-24)

86-T=86.97T-2087.49

T=\frac{2173.491}{87.97}=24.70^{\circ}C

3 0
3 years ago
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