5.Which of the following does not affect rate of evaporation?
O Wind speed
O Surface area
O Temperature
O Insoluble heavy impurities

Answers

Answer 1
Insoluble heavy impurities
Answer 2

Answer:

D

Explanation:

Insoluble impurities would not change the constituent of the substance. Soluble would for example salt water takes longer time for the water to become vapour when subjected to the same temperature that normal water.

Wind would affect, the more windy the tendency for particles of the liquid to be moved into the atmosphere.

With an increase in surface area, the evaporation rate increase . Take a clue from water placed on the ground and exposed to the atmosphere and that same quantity of water is placed in a cup. That on the floor would evaporate faster.

Similarly the higher the temperature a substance is subjected to the easier is it's rate of evaporation. Take for instance water in a cup placed in the sun and that same placed in a room with mild temperatures than that of the sun.With time that in the sun decreases in volume faster than that in the room.


Related Questions

A 1100 kg car pushes a 1800 kg truck that has a dead battery. When the driver steps on the accelerator, the drive wheels of the car push against the ground with a force of 4500 N.A) What is the magnitude of the force of the car on the truck?B) What is the magnitude of the force of the truck on the car?

Answers

Answer:The answer is 3000 N.

Force (F) is the multiplication of mass (m) and acceleration (a).

F = m · a

It is given:

mc = 1000 kg

mt = 2000 kg

total force: F = 4500 N 

total mass: m = mc + mt

Let's calculate acceleration which is common:

a = F/m = F/(mc + mt) = 4500/(1000 + 2000) = 4500/3000 = 1.5 m/s²

Now, when we know acceleration, let's calculate force on the truck:

Ft = mt · a = 2000 · 1.5 = 3000 N

Explanation:

Complete the first and second sentences, choosing the correct answer from the given ones.
1. The water temperature in the dish depends on the A / B / C / D.
A. average kinetic energy of water molecules
B. total kinetic energy of water molecules
C. water mass. D. potential energy of the container with water
2. The internal energy of the water in the vessel is E / F / G.
E. potential energy of the vessel with water
F. average kinetic energy of water molecules
G. sum of kinetic energy and potential water molecules

Answers

Answer:

Hope this helps :)

Explanation:

1. A

2. G (because the basic definition of internal energy is, the sum of kinetic and potential energies of water molecules)

Two vectors having magnitudes of 5.00 and 9.00 respectively. If the value of their dot product is 12.0, find the angle between the two vectors. ​

Answers

Answer:

C = 74.53°

Explanation:

Let the magnitudes of 5.00 and 9.00 be vectors A and B respectively, hence the dot product of this vector is defined as

A.B = |A||B|cosC; let C be the angle between the vectors

12 = 5×9 cos C

Hence cos C = 12/45

C = cos^-1(12/45)

C = 74.53°

An LC circuit has a 6.00 mH inductor. The current has its maximum value of 0.570 A at t =0s. A short time later the capacitor reaches its maximum potential difference of 66.0 V. What is the value of the capacitance?

Answers

Answer:

C = 44.75 x 10⁻⁸ F

Explanation:

Assuming no loss of energy between capacitor and inductor

energy in inductor initially = 1/2 Li₀² where L is inductance and i₀ is peak current .

= .5 x 6 x 10⁻³ x .57²

= .97 x 10⁻³ J .

This energy is transferred to capacitor .

energy of capacitor = 1/2 CV²

= .5 x C x 66²

= 2178 C

2178C = .97 x 10⁻³

C = 44.75 x 10⁻⁸ F .

The magnetic energy stored in the inductor is transformed into electrical energy stored in the capacitor. The value of capacitance for the given circuit is 44.75 x 10⁻⁸ F

Finding the capacitance:

According to the law of conservation of energy, the magnetic energy stored in the inductor will be gradually lost and this energy will be stored in the capacitor as electrical energy.

Initially, the energy in the inductor is:

E = 1/2 Li₀²

where L is inductance

and i₀ is peak current.

E = 0.5 × 6 × 10⁻³ × (0.57)²

E = 0.97 × 10⁻³J

This energy is transformed into electrical energy stored in the capacitor.

So the capacitor energy is:

E = 1/2 CV²

where C is the capacitance

E = 0.5 × C × 66²

E = 2178 C

0.97 x 10⁻³ = 2178 C

C = 44.75 x 10⁻⁸ F

Learn more about capacitance:

https://brainly.com/question/12356566?referrer=searchResults

g it as been suggested that solar powered space ships could get a boost from a laser either on earth or in orbit around earth. the laser would have to be very powerful to give any measurable benefit to the ship. if the laser produces a 0.18-m diameter beam of 490-nm light, what is the minimum angular spread of the beam?

Answers

Answer:

The minimum angular spread of the laser beam is 3.32 × [tex]10^{-6}[/tex] radians (or 3.32 μrad).

Explanation:

The minimum angular spread of a wave is the ratio of its narrowest diameter to its wavelength.

From Rayleigh's formula,

Angular spread = 1.22 (wavelength ÷ diameter)

                          = 1.22 (λ ÷ D)

Given that:

diameter, D = 0.18 m and wavelength, λ = 490 nm, then;

Angular spread of the laser beam = 1.22 (λ ÷ D)

                         = 1.22[tex](\frac{490*10^{-9} }{0.18})[/tex]

                         = 1.22× 2.7222 × [tex]10^{-6}[/tex]

                        = 3.3211 × [tex]10^{-6}[/tex] rad

The minimum angular spread of the laser beam is 3.32 × [tex]10^{-6}[/tex] radians.

An electromagnetic wave is propagating towards the west. At a certain moment the direction of the magnetic field vector associated with this wave points vertically up. The direction of the electric field vector of this wave is:___________

Answers

Answer:

either +z direction or -z direction.

Explanation:

The direction of the electric field, in an electromagnetic wave always is perpendicular to the direction of the magnetic field and the direction of propagation of the wave.

You assume a system of coordinates with the negative x axis as the west direction, and the y axis as the up direction

In this case, the wave is propagating toward the west (- x direction), and the magnetic field vector points up (+ y direction), then, it is mandatory that the electric field vector points either +z direction or -z direction.

write the answer:
physics ... i need help ​

Answers

Answer:

6 gallons

Explanation:

At 30 mph, the fuel mileage is 25 mpg.

After 5 hours, the distance traveled is:

30 mi/hr × 5 hr = 150 mi

The amount of gas used is:

150 mi × (1 gal / 25 mi) = 6 gal

Richard is driving home to visit his parents. 150 mi of the trip are on the interstate highway where the speed limit is 65 mph . Normally Richard drives at the speed limit, but today he is running late and decides to take his chances by driving at 80 mph. How many minutes does he save?

Answers

Answer:

t = 25.5 min

Explanation:

To know how many minutes does Richard save, you first calculate the time that Richard takes with both velocities v1 = 65mph and v2 = 80mph.

[tex]t_1=\frac{x}{v_1}=\frac{150mi}{65mph}=2.30h\\\\t_2=\frac{x}{v_2}=\frac{150mi}{80mph}=1.875h[/tex]

Next, you calculate the difference between both times t1 and t2:

[tex]\Delta t=t_1-t_2=2.30h-1.875h=0.425h[/tex]

This is the time that Richard saves when he drives with a speed of 80mph. Finally, you convert the result to minutes:

[tex]0.425h*\frac{60min}{1h}=25.5min=25\ min\ \ 30 s[/tex]

hence, Richard saves 25.5 min (25 min and 30 s) when he drives with a speed of 80mph

Suppose that 7.4 moles of a monatomic ideal gas (atomic mass = 1.39 × 10-26 kg) are heated from 300 K to 500 K at a constant volume of 0.74 m3. It may help you to recall that CV = 12.47 J/K/mole and CP = 20.79 J/K/mole for a monatomic ideal gas, and that the number of gas molecules is equal to Avagadros number (6.022 × 1023) times the number of moles of the gas.
1) How much energy is transferred by heating during this process?2) How much work is done by the gas during this process?3) What is the pressure of the gas once the final temperature has been reached?4) What is the average speed of a gas molecule after the final temperature has been reached?5) The same gas is now returned to its original temperature using a process that maintains a constant pressure. How much energy is transferred by heating during the constant-pressure process?6) How much work was done on or by the gas during the constant-pressure process?

Answers

Answer:

Explanation:

1 ) Since it is a isochoric process , heat energy passed into gas

= n Cv dT , n is no of moles of gas , Cv is specific heat at constant volume and dT is rise in temperature .

= 7.4 x 12.47 x ( 500 - 300 )

= 18455.6 J.

2 ) Since there is no change in volume , work done by the gas is constant.

3 ) from  , gas law equation

PV = nRT

P = nRT / V

= 7.4 x 8.3 x 500 / .74

= .415 x 10⁵ Pa.

4 ) Average kinetic energy  of gas molecules after attainment of final temperature

= 3/2 x R/ N x T

= 1.5 x 1.38  x 10⁻²³ x 500

= 1.035 x 10⁻²⁰ J

1/2 m v² = 1.035 x 10⁻²⁰

v² = 2 x 1.035 x 10⁻²⁰ / 1.39 x 10⁻²⁶

= 1.49 x 10⁶

v = 1.22 x 10³ m /s

5 )  In this process , pressure remains constant

gas is cooled from 500 to 300 K

heat will be withdrawn .

heat withdrawn

= n Cp dT

= 7.4 x 20.79 x 200

= 30769.2 J .

6 )

gas will have reduced volume due to cooling

reduced volume = .74 x 300 / 500

= .444 m³

change in volume

= .74 - .444

= .296 m³

work done on the gas

= P x dV

pressure x change in volume

= .415 x 10⁵ x .296

= 12284 J.

A visitor to a lighthouse wishes to determine the height of the tower. She ties a spool of thread to a small rock to make a simple pendulum, which she hangs down the center of a spiral staircase of the tower. The period of oscillation is 6.01 s. What is the height of the tower

Answers

Answer:

The height of the tower is 8.96 m.

Explanation:

We have, a visitor to a lighthouse wishes to determine the height of the tower. She ties a spool of thread to a small rock to make a simple pendulum, which she hangs down the center of a spiral staircase of the tower. The period of oscillation is 6.01 s.

It is required to find the height of the tower. Let it is l. The time period of a simple pendulum is given by :

[tex]T=2\pi \sqrt{\dfrac{l}{g}}[/tex]

l is length of pendulum, or height of tower

[tex]l=\dfrac{T^2g}{4\pi^2}\\\\l=\dfrac{(6.01)^2\times 9.8}{4\pi^2}\\\\l=8.96\ m[/tex]

So, the height of the tower is 8.96 m.

assume that the initial speed is 25 m/s and the angle of projection is 53 degree above the hroizontal. the cannon ball leaves the uzzel of the cannon at a highet of 200 m.( the cannon is at the edge of the cliff) A: find the horizontal distance the cannon travles. B: when does the cannon ball reach the ground? C: find the maximum highet the cannon ball reaches.

Answers

Answer:

A.  xmax = 131.49 m

B.  t = 8.74 s

C.  ymax = 220.33 m

Explanation:

A. In order to find the horizontal distance which cannon travels you first calculate the flight time. The flight time can be calculated by using the following formula:

[tex]y=y_o+v_osin\theta-\frac{1}{2}gt^2[/tex]      (1)

yo: height from the projectile is fired = 200m

vo: initial velocity of the projectile = 25m/s

g: gravitational acceleration = 9.8 m/s^2

θ: angle between the direction of the initial motion of the ball and the horizontal = 53°

t: time

You need the value of t when the projectile hits the ground. Then, in th equation (1) you make y = 0m.

When you replace the values of all parameters in the equation (1), you obtain the following quadratic formula:

[tex]0=200+(25)sin53\°t-\frac{1}{2}(9.8)t^2\\\\0=200+19.96t-4.9t^2[/tex] (2)

You use the quadratic formula to obtain the value of t:

[tex]t_{1,2}=\frac{-19.96\pm\sqrt{(19.96)^2-4(-4.9)(200)}}{2(-4.9)}\\\\t_{1,2}=\frac{-19.96\pm65.71}{-9.8}\\\\t_1=8.74s\\\\t_2=-4.66s[/tex]

You use the positive value because it has physical meaning.

Now, you can calculate the horizontal range of the projectile by using the following formula:

[tex]x_{max}=v_ocos\theta t[/tex]      

[tex]x_{max}=(25m/s)(cos53\°)(8.74s)=131.49m[/tex]

The cannon ball travels a horizontal distance of 131.49 m

B. The cannon ball reaches the canon for t = 8.74s

C. The maximum height is obtained by using the following formula:

[tex]y_{max}=y_o+\frac{v_o^2sin^2\theta}{2g}[/tex]     (3)

By replacing in the equation (3) the values of all parameters you obtain:

[tex]y_{max}=200m+\frac{(25m/s)^2(sin53\°)^2}{2(9.8m/s^2)}\\\\y_{mac}=200m+20.33m=220.33m[/tex]

The maximum height reached by the cannon ball is 220.33m

When Marcel finds the distance L from the previous part, it turns out to be greater than Lend, the distance from the pivot to the end of the seesaw. Hence, even with Jacques at the very end of the seesaw, the twins Gilles and Jean exert more torque than Jacques does. Marcel now elects to balance the seesaw by pushing sideways on an ornament (shown in red) that is at height h above the pivot. (Figure 3)With what force in the rightward direction, Fx, should Marcel push? If your expression would give a negative result (using actual values) that just means the force should be toward the left.Express your answer in terms of W, Lend, w, L2, L3, and h.

Answers

Answer:

Fx = - (1/h)( wL2 + wL3 - wLend )

Explanation:

Assuming The twins Gilles and Jean has a weight ( w ) each

The torque that would balance the equation would be = wL2 + wL3 -------- 1

THEREFORE the ccw torques are = wLend + Fh ----------- 2

hence equation 2 equals equation 1

= wLend + Fh = wL2 + wL3 --------- 3

equation 3 can as well be represented as

F = ( 1/h) ( wL2 + wL3 - wLend )---------- 4

From equation 4 it can be seen that F is on the left hand side therefore the value of Fx is negative

therefore equation 4 is represented as

 Fx = - (1/h)( wL2 + wL3 - wLend )

An object is thrown vertically and has an upward velocity of 18 m/s when it reaches one fourth of its maximum height above its launch point. What is the initial (launch) speed of the object

Answers

Answer:

v = 25.45 m/s

Explanation:

In order to calculate the initial speed of the object, you take into account the formula for the maximum height reaches by the object. Such a formula is given by:

[tex]h_{max}=\frac{v_o^2}{g}[/tex]   (1)

vo: initial speed of the object = 18 m/s

g: gravitational acceleration = 9.8 m/s²

Furthermore you use the following formula for the final speed of the object:

[tex]v^2=v_o^2-2gh[/tex]       (2)

h: height

You know that the speed of the object is 18m/s when it reaches one fourth of the maximum height. You use this information, and you replace the equation (1) in to the equation (2), as follow:

[tex]v^2=v_o^2-2g(\frac{h_{max}}{4})=v_o^2-\frac{1}{2}g(\frac{v_o^2}{g})\\\\v^2=v_o^2-\frac{1}{2}v_o^2=\frac{1}{2}v_o^2[/tex]

Then, you solve the previous result for vo:

[tex]v_o=\sqrt{2}v=\sqrt{2}(18m/s)=25.45\frac{m}{s}[/tex]

The initial speed of the object was 25.45 m/s

a vector has components x=6 m and y=8 m. what is its magnitude and direction?

Answers

Answer: 10m

Explanation:

The magnitude of the vector would be 10

[tex]\sqrt{6^{2}+8^{2} } =10[/tex]

Inside a stereo speaker, you will find two permanent magnets: one on the cone and one near the cone. True of false?

Answers

Answer:

false

Explanation:

A standing wave on a string that is fixed at both ends has frequency 80.0 Hz. The distance between adjacent antinodes of the standing wave is 12.0 cm. What is the speed of the waves on the string, in m/s

Answers

Answer:

v = 19.2 m/s

Explanation:

In order to find the speed of the string you use the following formula:

[tex]f=\frac{v}{2L}[/tex]          (1)

f: frequency of the string = 80.0Hz

v: speed of the wave = ?

L: length of the string = 12.0cm = 0.12m

The length of the string coincides with the wavelength of the wave for the fundamental mode.

Then, you solve for v in the equation (1), and replace the values of the other parameters:

[tex]v=2Lf=2(0.12m)(80.0Hz)=19.2\frac{m}{s}[/tex]

The speed of the wave is 19.2 m/s

Some runners train with parachutes that trail behind them to provide a large drag force. These parachutes are designed to have a large drag coefficient. One model expands to a square 1.8 m on a side, with a drag coefficient of 1.4. A runner completes a 200 m run at 5.0 m/s with this chute trailing behind. Part A How much thermal energy is added to the air by the drag force

Answers

Answer:

13.9 kJ

Explanation:

Given that

Length of the side, l = 1.8 m

Drag coefficient, C(d) = 1.4

Distance of run, d = 200 m

Velocity of run, v = 5 m/s

Density, ρ = 1.23

Using the Aerodynamics Drag Force formula. We have

F(d) = 1/2.ρ.A.C(d).v²

The Area, A needed is 1.8 * 1.8 = 3.24 m². So that,

F(d) = 1/2 * 1.23 * 3.24 * 1.4 * 5²

F(d) = 139.482/2

F(d) = 69.74

recall that, energy =

W = F * d

W = 69.74 * 200

W= 13948

W = 13.9kJ

Therefore, the thermal energy added to the air by the drag force is 13.9kJ

a) Write the names of the materials used in the ohm law according to the Figure 1?
b) If the voltage of a circuit is 12 V and the resistance is 40 , What is the generated power?

Answers

Answer:

a. i. conducting wire

ii high-pass and low-pass filters

iii. Cobra-4 Xpert-link

iii. voltage source

b. Power generated is 3.6 W.

Explanation:

Ohm's law state that the current passing through a metallic conductor, e.g wire is directly proportional to the potential difference across its ends, provided temperature is constant.

i.e      V = IR

i. conducting wire

ii high-pass and low-pass filters

iii. Cobra-4 Xpert-link

iii. voltage source

b. Given that; V = 12 V and R = 40 Ohm's.

P = IV

From Ohm's law, I = [tex]\frac{V}{R}[/tex]

So that;

P = [tex]\frac{V^{2} }{R}[/tex]

   = [tex]\frac{12^{2} }{40}[/tex]

  = [tex]\frac{144}{40}[/tex]

 = 3.6 W

The power is 3.6 W.

A small rock with mass 0.12 kg is fastened to a massless string with length 0.80 m to form a pendulum. The pendulum is swinging so as to make a maximum angle of 45 ∘ with the vertical. Air resistance is negligible. Part A What is the speed of the rock when the string passes through the vertical position

Answers

Answer:

v = 3.33 m/s

Explanation:

In the position of 45 degrees, all the energy of the rock is gravitational, then we have:

E = m*g*L*cos(angle)

and in the vertical position of the string, all the energy is kinetic, so we have:

E = m*v^2/2

If there is no dissipation, both energies are equal, so we have:

m*g*L*cos(45) = m*v^2/2

9.81 * 0.8 * 0.7071 * 2 = v^2

v^2 = 11.0986

v = 3.33 m/s

50 points!! please help :((

Answers

for decrease: it’s the first and last one and for increase it’s the middle two

Answer:

Loudness: decreases

Amplitude: decreases

Pitch: stays the same

Frequency: stays the same

Explanation:

1.

An oscilloscope measures how much the microphone is vibrating, or how much electricity it is sending. This means that a louder noise will register higher on the oscilloscope. Since the size of the waves at Y is lower than at X, the loudness of the sound has decreased.

2.

Similarly to loudness, amplitude measures how far the crests of the waves are from the nodes. Since Y is closer to the center line than X, it has a lower amplitude.

3 and 4.

The pitch and frequency, for our purposes, are essentially the same thing here. They are dependent on how close together the waves on the oscilloscope are, or how quickly the microphone is vibrated. Since this stays the same throughout the entire sound, they both stay the same.

Hope this helps!

A subatomic particle created in an experiment exists in a certain state for a time of before decaying into other particles. Apply both the Heisenberg uncertainty principle and the equivalence of energy and mass to determine the minimum uncertainty involved in measuring the mass of this short-lived particle.

Answers

Answer:

Δm Δt> h ’/ 2c²

Explanation:

Heisenberg uncertainty principle, stable uncertainty of energy and time, with the expressions

     ΔE Δt> h ’/ 2

     h’= h / 2π

to relate this to the masses let's use Einstein's relationship

      E = m c²

let's replace

     Δ (mc²) Δt> h '/ 2

the speed of light is a constant that we can condense exact, so

      Δm Δt> h ’/ 2c²

     

Three sheets of plastic have unknown indices of refraction. Sheet 1 is placed on top of sheet 2, and a laser beam is directed onto the sheets from above so that it strikes the interface at an angle of 26.50 with the normal. The refracted beam in sheet 2 makes an angle of 31.70 with the normal. The experiment is repeated with sheet 3 on top of sheet 2, and with the same angle of incidence, the refracted beam makes an angle of 36.70 with the normal. If the experiment is repeated again with sheet 1 on top of sheet 3, determine the expected angle of refraction in sheet 3? Assume the same angle of incidence.

Answers

Answer:

The angle of refraction of sheet 3 when sheet 1 is on top of it is [tex]\theta_{r_s } = 23.1 ^o[/tex]

Explanation:

From the question we are told that

     The angle of incidence is  [tex]\theta _i = 26.50 ^o[/tex]

      The angle of refraction angle for  sheet 1 is  [tex]\theta _{r_1}} = 31.70 ^o[/tex]

       The angle of refraction for sheet 3 is  [tex]\theta _{r_3}} = 36.70 ^o[/tex]

According to Snell's  law  

       [tex]\frac{n_2}{n_1} = \frac{sin (\theta_1)}{sin (\theta_{r_1})}[/tex]

Where  [tex]n_1 \ and \ n_2[/tex]  are refractive index of sheet 1  and  sheet 2  

       =>   [tex]n_2 = n_1 \frac{sin(\theta_i)}{sin (\theta _{r_1})}[/tex]

Also  when sheet 3 in on top of sheet 2

       [tex]\frac{n_2}{n_3} = \frac{sin \theta_i}{sin \theta_{r_3}}[/tex]

substituting for  [tex]n_2[/tex]

      [tex]n_1 \frac{sin(\theta_i)}{sin (\theta _{r_1})} = n_3 \frac{sin \theta_i}{sin \theta_{r_3}}[/tex]

      [tex]n_1 \frac{sin(\theta_i)}{sin (\theta _{r_1})} = n_3 \frac{sin \theta_i}{sin \theta_{r_3}}[/tex]

=>    [tex]n_3 = n_1 * \frac{sin(\theta_{r_3})}{sin(\theta_{r_1})}[/tex]

when sheet 1 in on top of sheet 3

        [tex]\frac{n_3}{n_1} = \frac{sin(\theta_i)}{\theta_{r_s}}[/tex]

where [tex]r_s[/tex] is the angle of refraction when sheet 1 is on top of sheet 3

substituting for  [tex]n_3[/tex]

         [tex]\frac{ n_1 * \frac{sin(\theta_{r_3})}{sin(\theta_{r_1})}}{n_1} = \frac{sin(\theta_i)}{\theta_{r_s}}[/tex]

=>   [tex]sin (\theta _{r_s}) = n_1 * sin (\theta_i) * \frac{sin (\theta_{r_1})}{ n_1 * sin(\theta_{r_3})}[/tex]

substituting values

      [tex]sin (\theta _{r_s}) = n_1 * sin (26.50) * \frac{sin (31.70)}{ n_1 * sin(36.70)}[/tex]

=>     [tex]\theta_{r_s } = sin^{-1} (0.3923)[/tex]

=>   [tex]\theta_{r_s } = 23.1 ^o[/tex]

An industrial flywheel (a solid disk) of mass 10.0 kg and radius 17.3 cm is rotating at an angular speed of 22.0 rad/s. Upon being switched to a slower setting, the flywheel uniformly slows down to 13.5 rad/s after rotating through an angle of 13.8 radians. Calculate the angular acceleration of the flywheel in the process of slowing down

Answers

Answer:

Explanation:

During slowing down , initial angular velocity ω₁ = 22 rad /s

final angular velocity ω₂ = 13.5 rad /s

using the law's of motion formula for rotation

ω₂² =  ω₁² + 2 αθ  , α is angular acceleration and θ is angle in radian rotated during this period

13.5² = 22² - 2xα x 13.8

2xα x 13.8 = 484 - 182.25

α  =  10.93 rad / s²

When a fuel is burned in a cylinder fitted with a piston, the volume expands from an initial value of 0.250 L against an external pressure of 2.00 atm. The expansion does 288 J of work on the surroundings. What is the final volume of the cylinder

Answers

Answer:

Vf = 0.0017 m³ = 1.7 L

Explanation:

The work done by the system on the surrounding at constant pressure is given by the following formula:

W = PΔV

W = P(Vf - Vi)

where,

W = Work done = 288 J

P = Constant Pressure = (2 atm)(101325 Pa/atm) = 202650 Pa

Vf = Final Volume f Cylinder = ?

Vi = Initial Volume of Cylinder = (0.25 L)(0.001 m³/ 1 L) = 0.00025 m³

Therefore,

288 J = (202650 Pa)(Vf - 0.00025 m³)

Vf = 288 J/202650 Pa + 0.00025 m³

Vf = 0.0017 m³ = 1.7 L

If you jumped out of a plane, you would begin speeding up as you fall downward. Eventually, due to wind resistance, your velocity would become constant with time. While your velocity is constant, the magnitude of the force of wind resistance is

Answers

Answer:

Mg or your weight.

Explanation:

When your velocity is constant, the net force acting on you is 0. That means the upwards force of air resistance must fully balance the downwards force of gravity on you, which is Mg.

A space ship traveling east flies directly over the head of an inertial observer who is at rest on the earth's surface. The speed of the space ship can be found from this relationship: . The navigator's on-board instruments indicate that the length of the space ship is 20 m. If the length of the ship is measured by the inertial earth-bound observer, what value will be obtained

Answers

Answer:

10 metres

Explanation:

So, we are given the following data or parameters or information which is going to assist us in solving this particular problem or Question efficiently.

=> The speed of the space ship can be found from this relationship: ✓(1 - [v^2/c^2] ) = 1/2.

=> The length of the space ship = 20 m.

=> Assumption = '' If the length of the ship is measured by the inertial earth-bound observer".

Thus, from the speed of the space ship can be found from this relationship we can determine the value;

✓(1 - [v^2/c^2] ) = 1/2.

V = 20 × 1/2 = 10 metres.

Note that we use the contraction formula to solve for V.

A ball is thrown upward from the ground with an initial speed of 19.2 m/s; at the same instant, another ball is dropped from a building 18 m high. After how long will the balls be at the same height above the ground?

Answers

Answer:

0.938 seconds

Explanation:

For the ball thrown upwards, we use the formula below to solve it:

[tex]s = ut - \frac{1}{2}gt^2[/tex]

where s = distance moved

u = initial speed = 19.2 m/s

t = time taken

g = acceleration due to gravity = 9.8 [tex]m/s^2[/tex]

Let x be the height at which both balls are level, this means that:

=> [tex]x = 19.2t - 4.9t^2[/tex]________(1)

For the ball dropped downwards, we use the formula below:

[tex]s = ut + \frac{1}{2}gt^2[/tex]

u = 0 m/s

At the point where both balls are level:

s = 18 - x

=> [tex]18 - x = 0 + 4.9t^2[/tex]

=> [tex]x = 18 - 4.9t^2[/tex]__________(2)

Equating both (1) and (2):

[tex]19.2t - 4.9t^2 = 18 - 4.9t^2\\\\=> 19.2t = 18\\\\t = 18/19.2 = 0.938 secs[/tex]

They will be level after 0.938 seconds

A dart is inserted into a spring-loaded dart gun by pushing the spring in by a distance . For the next loading, the spring is compressed a distance . How much faster does the second dart leave the gun compared with the first

Answers

Complete question is;

A dart is inserted into a spring - loaded dart gun by pushing the spring in by a distance x. For the next loading, the spring is compressed a distance 2x. How much faster does the second dart leave the gun compared to the first?

Answer:

The second dart leaves the gun two times faster than the first one.

Explanation:

If we assume there was no energy loss during the spring - dart energy transfer, we can easily apply the principle of conservation of energy. So;

Potential energy = kinetic energy

Thus;

½kx² = ½mv²

Making velocity "v" the subject, we have;

v = √(kx²/m)

Since the initial distance is "x", thus initial launching velocity is;

v1 = √(kx²/m)

Since next distance is 2x, thus, second launch velocity is;

v2 = √(k(2x)²/m)

Expanding, we have;

v2 = √(4kx²/m)

v2 = 2√(kx²/m)

Comparing this to the one gotten for v1 earlier, we can see that it is double v1.

So, v2 = 2v1

Hence, The second dart leaves the gun two times faster than the first one.

Water flows at 0.850 m/s from a hot water heater, through a 450-kPa pressure regulator. The pressure in the pipe supplying an upstairs bathtub 3.70m above the heater is 414-kPa. What's the flow speed in this pipe?

Answers

Answer:

The velocity is  [tex]v_2= 0.45 \ m/s[/tex]

Explanation:

From the question we are told that

      The initial speed of the hot water is  [tex]v_1 = 0.85 \ m/s[/tex]

     The pressure from the heater  [tex]P_1 = 450 \ KPa = 450 *10^{3} \ Pa[/tex]

      The height of the hot water before flowing is  [tex]h_1 = 0 \ m[/tex]

      The height of bathtub above the heater is [tex]h_2 = 3.70 \ m[/tex]

       The pressure in the pipe is [tex]P_2 = 414 KPa = 414 *10^{3} \ Pa[/tex]

       The density of water is [tex]\rho = 1000 \ kg/m^3[/tex]

Apply Bernoulli equation

      [tex]P_1 + \rho gh_1 +\frac{1}{2} \rho v_1^2 = \rho g h_2 + \frac{1}{2}\rho v_2 ^2[/tex]

Substituting values

     [tex](450 *10^{3}) + (1000 * 9.8 * 0) + (0.5 * 1000 * 0.85^2) = (1000 * *9.8*3.70) + (0.5*1000*v_2^2 )[/tex]

=>   [tex]v_2^2 = \frac{ (450 *10^{3}) + (1000 * 9.8 *0 ) + (0.5 * 1000 * 0.85^2) -[ (1000 * *9.8*3.70) ]}{0.5*1000}[/tex]

=>   [tex]v_2= \sqrt{ \frac{ (450 *10^{3}) + (1000 * 9.8 * 0) + (0.5 * 1000 * 0.85^2) -[ (1000 * *9.8*3.70) ]}{0.5*1000}}[/tex]

=>    [tex]v_2= 0.45 \ m/s[/tex]

Consider a weather balloon floating in the air. There are three forces acting on this balloon: the force of gravity is FG, the force from lift towards balloon is FL, and the force from the wind is labeled Fw. The orientation of these forces along with a coordinate system is given below:

Assume that || FG || = 20 N, || FL ||= 25 N, and || Fw ll = 15 N.

Required:
Find the magnitude of the resultant force acting on the weather balloon and round your answer to two decimal places.

Answers

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