PHYSICS COLLEGE

If anyone can help with this, before 8 am eastern time i’ll give you 20$. Thank you

Answers

Answer 1

Answer:

Answer:

1. **Position vs. Time Plot and Acceleration**:

To create a position vs. time plot, we can use the data provided. I'll first list the data in a table format:

| t (s) | x (m) |

|-------|-------|

| 0 | 0 |

| 0.05 | 1 |

| 1 | -12 |

| 3 | -40 |

| 5 | -105 |

| 8 | -175 |

| 10 | 15 |

| 22 | -2060 |

| 33 | ??? |

However, there seems to be a discrepancy in the data at t = 10. The position is given as both 15 and -410 -900. Please clarify this point, and I'll calculate the acceleration once we have the correct data.

2. **Estimate Number of Breaths in One Week**:

To estimate the number of breaths a person takes in one week, we can make some assumptions:

- An average person takes about 12-20 breaths per minute at rest.

- Let's assume 15 breaths per minute on average.

- In an hour, that's 15 breaths/minute * 60 minutes/hour = 900 breaths.

- In a day, it's 900 breaths/hour * 24 hours/day = 21,600 breaths.

- In a week (7 days), it's 21,600 breaths/day * 7 days/week = 151,200 breaths in one week.

3. **Narrative of Motion and Calculations**:

I'll need more information about the graph you mentioned for part 3. Could you describe the graph or provide the relevant data or equations related to it? This will allow me to answer parts a, b, c, and d accurately.

Related Questions

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Acetone, a component of some types of fingernail polish, has a boiling point of 56°C. What is its boiling point in units of kelvin? Report your answer to the correct number of significant figures.

Calculate the final temperature of a mixture of 0.350 kg of ice initially at 218°C and 237 g of water initially at 100.0°C.

Answers

Answer:

115 ⁰C

Explanation:

Step 1: The heat needed to melt the solid at its melting point will come from the warmer water sample. This implies

$q_(1) +q_(2) =-q_(3)$ -----eqution 1

where,

$q_(1)$ is the heat absorbed by the solid at 0⁰C

$q_(2)$ is the heat absorbed by the liquid at 0⁰C

$q_(3)$ the heat lost by the warmer water sample

Important equations to be used in solving this problem

$q=m *c*\delta {T}$ , where -----equation 2

q is heat absorbed/lost

m is mass of the sample

c is specific heat of water, = 4.18 J/0⁰C

$\delta {T}$ is change in temperature

Again,

$q=n*\delta {_f_u_s}$ -------equation 3

where,

q is heat absorbed

n is the number of moles of water

tex]\delta {_f_u_s}[/tex] is the molar heat of fusion of water, = 6.01 kJ/mol

Step 2: calculate how many moles of water you have in the 100.0-g sample

$=237g *(1 mole H_(2) O)/(18g) = 13.167 moles of H_(2)O$

Step 3: calculate how much heat is needed to allow the sample to go from solid at 218⁰C to liquid at 0⁰C

$q_(1) = 13.167 moles *6.01(KJ)/(mole) = 79.13KJ$

This means that equation (1) becomes

79.13 KJ + $q_(2) = -q_(3)$

Step 4: calculate the final temperature of the water

$79.13KJ+M_(sample) *C*\delta {T_(sample)} =-M_(water) *C*\delta {T_(water)$

Substitute in the values; we will have,

$79.13KJ + 237*4.18(J)/(g^(o)C)*(T_(f)-218}) = -350*4.18(J)/(g^(o)C)*(T_(f)-100})$

79.13 kJ + 990.66J* $(T_(f)-218})$ = -1463J* $(T_(f)-100})$

Convert the joules to kilo-joules to get

79.13 kJ + 0.99066KJ* $(T_(f)-218})$ = -1.463KJ* $(T_(f)-100})$

$79.13 + 0.99066T_(f) -215.96388= -1.463T_(f)+146.3$

collect like terms,

2.45366 $T_(f)$ = 283.133

∴ $T_(f) =$ = 115.4 ⁰C

Approximately the final temperature of the mixture is 115 ⁰C

In a two-slit experiment, the slit separation is 3.00 × 10-5 m. The interference pattern is recorded on a flat screen-like detector that is 2.00 m away from the slits. If the seventh bright fringe on the detector is 10.0 cm away from the central fringe, what is the wavelength of the light passing through the slits? (The central bright fringe is zeroth one).

Answers

The wavelength of the light passing through the slit is 214 nm.

What is the wavelength?

The wavelength is the distance between identical points in the adjacent cycles of a waveform.

Given that the separation between two slits d is 3.00 × 10^-5 m and the distance from the slit to screen r is 2 m. The distance from the central spot to fringe s is 10.0 m and the bright bands of the spectrum m are 7 for the seventh bright fringe.

The wavelength of the light passing through the slit is calculated as given below.

$\lambda = \frac {sd}{mr}$

$\lambda = \frac {10* 10^(-2) * 3.00 * 10^(-5)}{7 * 2}$

$\lambda = 2.14 * 10^(-7)\;\rm m$

$\lambda = 214 \;\rm nm$

Hence we can conclude that the wavelength of the light passing through the slit is 214 nm.

To know more about the wavelength, follow the link given below.

brainly.com/question/7143261.

To solve this problem, the concepts related to destructive and constructive Interference of light spot and dark spot are necessary.

By definition in the principle of superposition, light interference is defined as

$Y = (m\lambda R)/(d)$

Where,

d = Separation of the two slits

$\lambda = Wavelength$

R = Distance from slit to screen

m= Any integer, which represents the repetition of the spectrum. The order of m equal to 1,2,3,4,5 represent bright bands and the order of m equal to 1.5,2.5,3.5 represent the dark bands.

Y = Distance from central spot to fringe.

Re-arrange the equation to find \lambda we have that

$\lambda = (Yd)/(mR)$

Our values are gives according the problem as,

$Y = 10*10^(-2)m$

$d = 3*10^(-5)$

m = 7 (The seventh bright fringe)

R = 2m

$\lambda = ((10*10^(-2))(3*10^(-5)))/(7*2)$

$\lambda = 214nm$

Therefore the wavelength of the light passing through the slits is 214nm

Patricia places 1500g of fruit on a scale compressing it by 0.02m. What is the force constant of the spring on the scale. Hint: you need 2 equations to solve this.

Answers

Answer:

The force constant of the spring is 735 N/m.

Explanation:

It is given that,

Mass of fruit, m = 1500 g = 1.5 kg

Compression in the scale, x = 0.02 m

We need to find the force constant of the spring on the scale. The force acting on the scale is given by using Hooke's law. So,

$F=-kx$

Also, F = mg

$mg=kx$

k is force constant

$k=(mg)/(x)\n\nk=(1.5* 9.8)/(0.02)\n\nk=735\ N/m$

So, the force constant of the spring is 735 N/m.

If Jim could drive a Jetson's flying car at a constant speed of 490 km/hr across oceans and space, approximately how long (in millions of years, in 106 years) would he take to drive to a nearby star that is 4.5 light-years away? Use 9.461 × 1012 km/light-year and 8766 hours per year (365.25 days). unanswered

Answers

Answer:

109.5 million years

Explanation:

The question asked us to find the time.

Remember that

Rate of velocity = distance / time, and this,

time taken = distance/rate

Due to the confusing nature of the units, we would have to be converting them to a more uniform one.

1 km is equal to 9.461*10^12 km/light-year, that's if we try to convert km to light year.

Since the speed is in km, the distance has to be in km also, and therefore, we convert ly to km:

4.5 light-years = 9.461*10^12 km/light-year) = 42.57*10^13 km

We that this value as our distance, in km.

Also,

Time = distance/speed

Time = 45.57*10^13 km / 490 km/hr = 9.3*10^11 hr

Now the next step is to convert hours to years, using the conversion factor 8766 hr/yr.

time (in years) = 9.6*10^11 hr / 8766 hr/yr) = 10.95*10^7 years

the final step is to divide the time in years by 10^6 years/million years, which gives the final answer as the trip takes 109.5 million years.

An insulated rigid tank contains 3 kg of H2O in the form of a saturated mixture of liquid and vapor at a pressure of 150 kPa and a quality of 0.25. An electric heater inside the tank is turned on to heat this H2O until the pressure increases to 200 kPa. Please determine the change in total entropy of water during this process. Hint: See if you can find the electrical work consumed during this process.

Answers

Answer:

change in entropy is 1.44 kJ/ K

Explanation:

from steam tables

At 150 kPa

specific volume

Vf = 0.001053 m^3/kg

vg = 1.1594 m^3/kg

specific entropy values are

Sf = 1.4337 kJ/kg K

Sfg = 5.789 kJ/kg

initial specific volume is calculated as

$v_1 = vf + x(vg - vf)$

$= 0.001053 + 0.25(1.1594 - 0.001053)$

$v_1 = 0.20964 m^3/kg$

$s_1 = Sf + x(Sfg)$

$= 1.4337 + 0.25 * 5.7894 = 2.88 kJ/kg K$

FROM STEAM Table

at 200 kPa

specific volume

Vf = 0.001061 m^3/kg

vg = 0.88578 m^3/kg

specific entropy values are

Sf = 1.5302 kJ/kg K

Sfg = 5.5698 kJ/kg

constant volume so $v_1 - v_2 = 0.29064 m^3/kg$

$v_2 = v_1 = vf + x(vg - vf)$

$=0.29064 = x_2(0.88578 - 0.001061)$

$x_2 = 0.327$

$s_2 = 1.5302 + 0.32 * 5.5968 = 3.36035 kJ/kg K$

Change in entropy $\Delta s = m(s_2 - s_1)$

=3( 3.36035 - 2.88) = 1.44 kJ/kg

A 0.454-kg block is attached to a horizontal spring that is at its equilibrium length, and whose force constant is 21.0 N/m. The block rests on a frictionless surface. A 5.30×10?2-kg wad of putty is thrown horizontally at the block, hitting it with a speed of 8.97 m/s and sticking.Part AHow far does the putty-block system compress the spring?

Answers

The distance the putty-block system compress the spring is 0.15 meter.

Given the following data:

Mass = 0.454 kg

Spring constant = 21.0 N/m.

Mass of putty = $5.30* 10^(-2)\;kg$

Speed = 8.97 m/s

To determine how far (distance) the putty-block system compress the spring:

First of all, we would solver for the initialmomentum of the putty.

$P_p = mass * velocity\n\nP_p = 5.30* 10^(-2)* 8.97\n\nP_p = 47.54 * 10^(-2) \;kgm/s$

Next, we would apply the law of conservation of momentum to find the final velocity of the putty-block system:

$P_p = (M_b + M_p)V\n\n47.54* 10^(-2) = (0.454 + 5.30* 10^(-2))V\n\n47.54* 10^(-2) = 0.507V\n\nV = (0.4754)/(0.507)$

Velocity, V = 0.94 m/s

To find the compression distance, we would apply the law of conservation of energy:

$U_E = K_E\n\n(1)/(2) kx^2 = (1)/(2) mv^2\n\nkx^2 =M_(bp)v^2\n\nx^2 = (M_(bp)v^2)/(k) \n\nx^2 = ((0.454 + 5.30* 10^(-2)) * 0.94^2)/(21)\n\nx^2 = ((0.507 * 0.8836))/(21)\n\nx^2 = ((0.4480))/(21)\n\nx=√(0.0213)$