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2 years ago

9

A person in car slides to the right when going through a very sharp left turn on the highway. Is that law of inertia or law of a

cceleration?

1 answer:

inessss [21]2 years ago

4 0

Answer:

Inertia :)

Explanation:

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A piston-cylinder device contains Helium gas initially at 150 kPa, 20 o C, and 0.5m 3 . The helium is now compressed in a polytr

Molodets [167]

Answer:

Explanation:

Given

$P_1=150 kPa$

$T_1=20^{\circ}C$

$V_1=0.5 m^3$

$T_2=140^{\circ}C$

$P_2=400 kPa$

R for Helium $R=2.076$

$c_v=3.115 kJ/kg-K$

mass of gas $m=\frac{P_1V_1}{RT_1}$

$m=\frac{150\times 0.5}{2.076\times 293}$

$m=0.123 kg$

Similarly $V_2$ can be found

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

$V_2=0.264 m^3$

Work done $W=\int_{V_1}^{V_2}PdV$

$W=\frac{P_2V_2-P_1V_1}{n-1}$

$W=\frac{mR(T_2_T_1)}{n-1}$

Since it is a polytropic Process

therefore $PV^n=c$

$P_1V_1^n=P_2V_2^n$

$(\frac{V_1}{V_2})^n=\frac{P_2}{P_1}$

$(\frac{0.5}{0.264})^n=\frac{400}{150}$

$n=\frac{\ln 2.66}{\ln 1.893}$

$n=1.533$

$W=\frac{0.123\times 2.076(140-20)}{1.533-1}$

$W=57.48 kJ$

From Energy balance

$E_{in}-E_{out}=\Delta E_{system}$

Neglecting kinetic and Potential Energy change

$Q_{in}+W_{in}=change\ in\ Internal\ Energy$

Change in Internal Energy $\Delta U=u_2-u_1$

$\Delta U=mc_v(T_2-T_1)$

$\Delta U=0.123\times 3.115(140-20)$

$\Delta U=45.977 kJ$

$Q_{in}+57.48=45.977$

$Q_{in}=-11.50 kJ$

i.e. Heat is being removed

3 0

3 years ago

Which of the following tools measures weight? A balance A scale A cyclometer A graduated cylinder

Kazeer [188]

I'm pretty sure its a scale.

8 0

3 years ago

"put a bowl of butter in the microwave and set a timer for twenty seconds. Observer the state and properties of the butter after

Kobotan [32]

Answer:

The physical change of the butter in the microwave is the butter melting

Explanation:

7 0

2 years ago

Read 2 more answers

How fast is the spaceship traveling towards the Sun? The radius of the orbit of Jupiter is 43.2 light-minutes, and that of the o

melisa1 [442]

Question:

A spaceship enters the solar system moving toward the Sun at a constant speed relative to the Sun. By its own clock, the time elapsed between the time it crosses the orbit of Jupiter and the time it crosses the orbit of Mars is 35.0 minutes

How fast is the spaceship traveling towards the Sun? The radius of the orbit of Jupiter is 43.2 light-minutes, and that of the orbit of Mars is 12.6 light-minutes.

Answer:

S = 5.508 × 10¹¹m

V = 2.62 × 10⁸ m/s

Explanation:

The radius of the orbit of Jupiter, Rj is 43.2 light-minutes

radius of the orbit of Mars, Rm is 12.6 light-minutes

Distance travelled S = (Rj - Rm)

= 43.2 - 12.6 = 30.6 light- minutes

= 30.6 × (3 ×10⁸m/s) × 60 s

= 5.508 × 10¹¹m

time = 35mins = (35 × 60 secs)

= 2100 secs

speed = distance/time

V = 5.508 × 10¹¹m / 2100 s

V = 2.62 × 10⁸ m/s

7 0

3 years ago

2. A certain object revolves at a rate of 30 rpm. Please determine the frequency and period of this

schepotkina [342]

Answer:

The time period of the motion is, T = 0.03 s

The frequency of the rotation is, f = 30 Hz

Explanation:

Given data,

The rotational speed of an object, ω = 30 rpm

ω = 188.5 rad/s

The time period of motion is,

T = 2π / ω

Substituting the given values in the above equation

= 2π / 188.5

T = 0.03 s

The time period of the motion is, T = 0.03 s

The frequency of rotation,

f = 1 /T

= 1 / 0.03

= 30 Hz

Hence, the frequency of the rotation is, f = 30 Hz

7 0

3 years ago