PHYSICS COLLEGE

The position of a particle as a function of time is given by x = (2.0 m/s)t + (-3.0 m/s2)t2. (a) plot x-versus-t for time from t = 0 to t = 1.0 s. (do this on paper. your instructor may ask you to turn in this plot.) this answer has not been graded yet. (b) find the average velocity of the particle from t = 0.45 s to t = 0.55 s. m/s (c) find the average velocity from t = 0.49 s to t = 0.51 s.

Answers

Answer 1

Answer:

Part a)

Equation of position with time is given as

$x = (2.0 m/s)t + (-3.0 m/s2)t^2$

since this equation is a quadratic equation

so it will be a parabolic graph between t = 0 to t = 1

part b)

at t = 0.45 s

$x = 2* 0.45 - 3 * 0.45^2$

$x_1 = 0.2925 m$

at t = 0.55 s

$x = 2* 0.55 - 3*0.55^2$

$x_2 = 0.1925$

now the displacement is given as

$d = x_2 - x_1$

$d = 0.1925 - 0.2925 = -0.1 m$

so the average velocity is given by

$v = (d)/(t)$

$v = \frac{-0.1}{0.1) = -1 m/s$

part c)

at t = 0.49 s

$x = 2* 0.49 - 3 * 0.49^2$

$x_1 = 0.2597 m$

at t = 0.51 s

$x = 2* 0.51 - 3*0.51^2$

$x_2 = 0.2397 m$

now the displacement is given as

$d = x_2 - x_1$

$d = 0.2397 - 0.2597 = -0.02 m$

so the average velocity is given by

$v = (d)/(t)$

$v = \frac{-0.02}{0.02) = -1 m/s$

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A piston-cylinder device initially contains 1.4 kg saturated liquid water at 200oC. Now heat is transferred to the water until the volume quadruples and the cylinder contains saturated vapor only. Determine (a) the volume of the cylinder, (b) the final temperature.

Answers

Answer:

Explanation:

Given

mass of saturated liquid water $m=1.4\ kg$

at $200^(\circ)$ specific volume is $\nu =0.001157\ m^3\kg$ (From Table A-4,Saturated water Temperature table)

$V_1=m\nu _1$

$V_1=1.4* 0.001157$

$V_1=1.6198* 10^(-3)\ m^3$

Final Volume $V_2=4V_1$

$V_2=4* (1.6198* 10^(-3))$

$V_2=6.4792* 10^(-3)\ m^3$

Specific volume at this stage

$\nu _2=(V_2)/(m)$

$\nu _2=(6.4792* 10^(-3))/(1.4)$

$\nu _2=0.004628\ m^3/kg$

Now we see the value and find the temperature it corresponds to specific volume at vapor stage in the table.

$T_2=T_1^(*)+(T_2^(*)-T_1^(*))/(\alpha _2^(*)-\alpha _1^(*))* (\alpha _2-\alpha _1^(*))$

$T_2=370^(\circ)+(373.95-370)/(0.003106-0.004953)* (0.004628-0.004953)$

$T_2=370.7^(\circ) C$

Final answer:

The problem in the question is solved using the principles of thermodynamics. The volume of the device after the heat transfer is 6311.2 cm³. The final temperature inside the cylinder, when the water reached the state of saturated vapor, is approximately 240°C.

Explanation:

The subject question is a thermodynamics problem; more specifically dealing with changes of state, volume, and temperature in a system under certain conditions.

For solving part (a), one would first need to find the specific volume (v) at the initial state, which is saturated liquid at 200°C. Looking up in the property tables, we see that v = 1.127 cm³/g for saturated water at 200°C. Then, the initial volume (V) is mass times specific volume, so V = 1.4 kg x 1.127cm³/g x 1000g/kg = 1577.8 cm³. Because volume quadrupled, the final volume is 4 x 1577.8 cm³ = 6311.2 cm³.

For part (b), at the final state, the water is a saturated vapor. The specific volume at the final state is the final volume divided by the mass, which equals to 6311.2 cm³ / 1.4kg / 1000g/kg = 4.507 cm³/g. Look this value up in the property table to find the corresponding temperature. We get a final temperature of about 240°C.

Learn more about Thermodynamics here:

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The animal that is hunted and consumed is considered the

Answers

Predator? like they hunt their prey

Answer:

prey

Explanation:

A 2100 g block is pushed by an external force against a spring (with a 22 N/cm spring constant) until the spring is compressed by 11 cm from its uncompressed length. The compressed spring and block rests at the bottom of an incline of 28◦ with the spring lying along the surface of the ramp.After all the external forces are removed (so the compressed spring releases the mass) how far D along the plane will the block move before coming to a stop? Answer in units of m.

Answers

Answer:

6.5e-4 m

Explanation:

We need to solve this question using law of conservation of energy

Energy at the bottom of the incline= energy at the point where the block will stop

Therefore, Energy at the bottom of the incline consists of the potential energy stored in spring and gravitational potential energy= $(1)/(2) kx^(2) +PE1$

Energy at the point where the block will stop consists of only gravitational potential energy= $PE2$

Hence from Energy at the bottom of the incline= energy at the point where the block will stop

⇒ $(1)/(2) kx^(2) +PE1=PE2$

⇒ $PE2-PE1=(1)/(2) kx^(2)$

Also $PE2-PE2=mgh$

where $m$ is the mass of block

$g$ is acceleration due to gravity=9.8 m/s

$h$ is the difference in height between two positions

⇒ $mgh=(1)/(2) kx^(2)$

Given m=2100kg

k=22N/cm=2200N/m

x=11cm=0.11 m

∴ $2100*9.8*h=(1)/(2)*2200*0.11^(2)$

⇒ $20580*h=13.31$

⇒ $h=(13.31)/(20580)$

⇒h=0.0006467m= $6.5e-4$

In the circuit shown in the figure above, suppose that the value of R1 is 100 k ohms and the value of R2 is 470 k ohms. At which of the following locations in the circuit would you measure the highest voltage with your meter? A. Between points A and B
B. Between points A and C
C. Between points B and E
D. Between points B and C

Answers

Answer:

Explanation:

voltage between A and C is equal battery's voltage.

Which of the following is a good example of a contact force?ОА.
Earth revolving around the Sun
OB.
a bridge suspended by cables
OC.
a ball falling downward a few seconds after being thrown upward
OD. electrically charged hairs on your head repelling each other and standing up

Answers

Answer:

A bridge suspended by cables

Explanation:

Both objects represent a contact force (in this case, normal force) acting on each other. The force occurs since both objects are in direct physical contact.

An engineer is designing a small toy car that a spring will launch from rest along a racetrack. She wants to maximize the kinetic energy of the toy car when it launches from the end of a compressed spring onto the track, but she can make only a slight adjustment to the initial conditions of the car. The speed of the car just as it moves away from the spring onto the track is called the launch speed. Which of the following modifications to the car design would have the greatest effect on increasing the kinetic energy of the car? Explain your reasoning.Decrease the mass of the car slightly.
Increase the mass of the car slightly.
Decrease the launch speed of the car slightly.
Increase the launch speed of the car slightly.

Answers

Answer:

we see that to increase the energy of the expensive we must increase the launch speed, since it increases quadratically

Explanation:

Kinetic energy is

K = ½ m v²

the speed of the expensive we can find it r

v² = v₀² + 2 a x

we can find acceleration with Newton's second law

F = m a

a = F / m

F= cte

substitute in the velocity equation

v² = v₀² + 2 F/m x

let's substitute in the kinetic energize equation

K = ½ m (v₀² + 2 F/m x)

K = ½ m v₀² + f x

we see that the kinetic energy depends on two tomines

in January in these systems the force for launching is constant, which is why decreasing the mass increases the speed of the vehicle and therefore increases the kinetic energy

As the launch speed increases the initial energy increases quadratically

we see that to increase the energy of the expensive we must increase the launch speed, since it increases quadratically

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