A liquid fueled rocket is red on a test stand. The rocket nozzle has an exit diameter of 30 cm and the combustion gases leave the nozzle at a velocity of 3800 m/s and a pressure of 100 kPa, which is the same as the ambient pressure. The temperature of the gases in the combustion area is 2400 C. Find (a) the temperature of the gases at the nozzle exit plane, (b) the pressure in the combustion area, and (c) the thrust developed. Assume that the gases have a speci c heat ratio of 1.3, and a molar mass of 9. Assume that the ow in the nozzle is isentropic.

Answer:

The distance (in miles) by the bus up the hill when its speed decreased to 50 mph is approximately 1.353 miles

Explanation:

The parameters of the motion of the driver are;

The upgrade of the road, θ = 10°

The rate of constant acceleration of the bus driver = 5 ft./s²

The speed of the bus as it begins to go up the hill, v₁ = 80 mph = 117.3228 ft./s

The speed of the driver at a point on the hill, v₂ = 50 mph ≈ 73.32677 ft./s

The acceleration due to gravity, g ≈ 32.1740 ft./s²

Therefore, we have;

The acceleration due to gravity down the incline plane, gₓ = g·sinθ

∴ gₓ = g·sin(θ) ≈ 32.1740 ft./s² × sin(10°) ≈ 5.587 ft/s²

The net acceleration of the bus, on the incline plane, [tex]a_{Net}[/tex] = gₓ - a = 5.587 ft./s² -5 ft./s² = 0.587 ft./s²

The vertical component of the velocity, [tex]v_y[/tex] = v × sin(θ)

∴ [tex]v_y[/tex] = 117.3228 ft./s × sin(10°) ≈ 20.37289 ft./s

vₓ = 117.3228 ft./s × cos(10°) ≈ 115.5404 ft./s

The velocity of the car, v₂, on the inclined plane is given as follows;

v₂ = v₁ - [tex]a_{Net}[/tex] × t

∴ t = (v₁ - v₂)/[tex]a_{Net}[/tex]  = (117.3228 ft./s - 73.32677 ft./s)/(0.587 ft./s²) ≈ 74.95 s

The distance covered, 's', is given as follows;

s = v₁·t - 1/2·[tex]a_{Net}[/tex]·t²

∴ s = 117.3228 × 74.95 - 1/2 × 0.587 × 74.95² ≈ 7144.6069 ft.

The distance travelled up the hill, s ≈ 7144.6069 ft. ≈ 1.3531452 miles ≈ 1.353 miles

1). If 2.2 lb actually = 1 kg, then

(132lb) x (1 kg/2.2lb) = 60 kg.

(Sadly, 2.2 lb doesn't " = " 1 kg. The explanation, unfortunately, is beyond the scope of this discussion.)

2). Weight = (mass) x (gravity)

Moon weight = (60 kg)x(1.6 m/s^2)

Moon weight = 96 Newtons

This question is incomplete, the complete question is;

The loaded car of a roller coaster has mass M = 320 kg. It goes over the highest hill with a speed v of 21.4 m/s. The radius of curvature R of the hill is 15.8 m.

(a) What is the force (N) that the track must exert on the car? (positive is up)

(b) What must be the force (N) that the car exerts on a 61 kg passenger?

Answer:

a) the force (N) that the track must exert on the car is -6139.14 N

b) the force (N) that the car exerts on a 61 kg passenger is -1170.27 N

Explanation:

Given the data in the question;

Let N represent the force that the track must exerted on the car

Net force on the car Fnet = Mg + N

so

M × a = Mg + N

N = Ma - Mg

N = Ma - M(v²/R)

we substitute

N = (320kg × 9.8m/s²) - ( 320 × ((21.4m/s)² / 15.8 m) )

N = 3136 - ( 320 × 28.9848 )

N = 3136 - 9275.136

N = -6139.14 N

Therefore, the force (N) that the track must exert on the car is -6139.14 N

b) What must be the force (N) that the car exerts on a 61 kg passenger?

Let N represent the force that the car exerts on 61kg passengers

so

Net force of passengers Fnet = mg + N

Ma = Mg + N

N = Ma - Mg

N = Ma - M(v²/R)

N = (61kg × 9.8m/s²) - ( 61 × ((21.4m/s)² / 15.8 m) )

N = 597.8 - ( 61 × 28.9848)

N = 597.8 - 1768.0728

N = -1170.27 N

Therefore, the force (N) that the car exerts on a 61 kg passenger is -1170.27 N

The centripetal force of the track on the car moving in the circular path is [tex]1.465 \times 10^6 \ N[/tex].

The force (N) that the car exerts on a 61 kg passenger is 597.8 N.

The centripetal force of the track on the car moving in the circular path is calculated as follows;

[tex]F_c = \frac{mv^2}{r}\\\\ F_c = \frac{320 \times 21.4^2}{0.1} \\\\F_c = 1.465 \times 10^6 \ N[/tex]

The force (N) that the car exerts on a 61 kg passenger is equal to the force the passenger exerts on the car based on Newton's third law of motion.

F = mg

F = 61 x 9.8

F = 597.8 N

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Answer:

Actions

Explanation:

if u mean what are these called then it's actions

Answer: Regular Physical Activity

Explanation:

Answer:

5.7375 seconds

Explanation:

The computation of the time required to reach that maximum velocity is shown below:

Given that

Mass = m =  5100 kg

Net upward force F = 8 × times 10^5 N

Initial speed = V_i = 0

Maximum speed = V = 900 m.s

Based on the above information

Impluse J = m(V - V_i)

= 5100 (900 - 0)

= 459 × 10^4 kg m.s

As we know that

J = FT

So

T = J ÷ F

= (459 × 10^4) ÷ (8 × 10^5)

= 5.7375 seconds

Here is a copy+pasted definition...

Transactive memory is a psychological hypothesis first proposed by Daniel Wegner in 1985 as a response to earlier theories of "group mind" such as groupthink.[1] A transactive memory system is a mechanism through which groups collectively encode, store, and retrieve knowledge. Transactive memory was initially studied in couples and families where individuals had close relationships but was later extended to teams, larger groups, and organizations to explain how they develop a "group mind",[1] a memory system that is more complex and potentially more effective than that of any of its individual constituents. A transactive memory system includes memory stored in each individual, the interactions between memory within the individuals, as well as the processes that update this memory. Transactive memory, then, is the shared store of knowledge.

In simple terms...

A transactive memory is a psychological hypothesis. It is a system in the brain which allows human's to encode, store, or get information.

For example...

let's say I learn 5 colors when I'm younger. My "transactive memory" will store it and will allow my brain to "retrieve" the information to allow me to remember colors I previously "stored".

Please mark brainliest!

Answer:

a) True

Explanation:

There are several possible explanations for this positive charge

* The explanation of the small positive charge in the plant when the bee approaches is like a defense system of the plants,

to prevent the bees from taking the pollen, but the flowers need the bees to transport the pollen for fertilization, so this possibility is not correct

* The air is conductive so the bee indexes a charge in the nearby air, this charge must be negative and this charge induced in the air induces a charge on the flower that must be positive.

When reviewing the different statements we have

a) True, it agrees with the second explanation of the phenomenon

b) False. The earth is a deposit of negative charge

c) false. If this is the case the charge should be negative

d) False. This residual charge from the other bees is quickly neutralized by the charge from the Earth.

Answer:

Explanation:

.

Answer:

a.If we increase the wind velocity, the maximum vertical dispersal height will decrease, while the rate of diffusion will increase

b.If we increase the humidity, the maximum vertical dispersal height will increase after 24 hours.

c.If we increase the lapse rate, the maximum vertical dispersal height of the pollutants will increase

Explanation:

a.If we increase the wind velocity, the maximum vertical dispersal height will decrease, while the rate of diffusion will increase

b.If we increase the humidity, the maximum vertical dispersal height will increase after 24 hours.

c.If we increase the lapse rate, the maximum vertical dispersal height of the pollutants will increase

Answer:

a decrease

b increase

c increase

Answer:

They are equal.

Explanation:

Answer:

220 bastanyan sagot ko yawa

Answer:

Explanation:

Water.

Carbohydrates.

Protein-Amino acids.

Fat.

Vitamins.

Minerals.

Omega-3 fatty acids.

Answer:

c. yes, 7.7 cm from the 5-A wire

Explanation:

Given;

distance between the two wires, r = 20 cm = 0.2 m

first current, I₁ = 5.0 A

second current, I₂ = 8.0 A

The magnetic field due to the two wires occurs in different directions and it can be zero at this region.

Let x be the distance from 5 A wire where the magnetic field is zero.

[tex]B = \frac{\mu_o}{2\pi} [\frac{I_1}{x} -\frac{I_2}{r-x} ] = 0\\\\ \frac{\mu_o}{2\pi} [\frac{I_1}{x} -\frac{I_2}{r-x} ] = 0\\\\ \frac{\mu_o}{2\pi} [\frac{5}{x} -\frac{8}{0.2-x} ] = 0\\\\\frac{5}{x} -\frac{8}{0.2-x} = 0\\\\\frac{5}{x} = \frac{8}{0.2-x}\\\\5(0.2-x) = 8x\\\\1 -5x = 8x\\\\1 = 5x \ + \ 8x\\\\1 = 13x\\\\x = \frac{1}{13} \\\\x = 0.077 \ m\\\\x = 7.7 \ cm[/tex]

Therefore, the correct option is c. yes, 7.7 cm from the 5-A wire

Answer:

a protractor

Explanation:

because protractors measure angles

Answer:

[tex]1.25\ \mu\text{g/m}^3[/tex]

Explanation:

v = Velocity of the breeze = 4 m/s

w = Width of the valley = 5000 m

h = Height of the valley = 1000 m

Volumetric flow rate is given by

[tex]\dot{V}=vwh\\\Rightarrow \dot{V}=4\times 5000\times 1000\\\Rightarrow \dot{V}=2\times10^{7}\ \text{m}^3/\text{s}[/tex]

[tex]\dot{m}[/tex] = Mass flow rate of pollutant = 25 g/s = [tex]25\times 10^6\ \mu\text{g/s}[/tex]

Concentration is given by

[tex]C=\dfrac{\dot{m}}{\dot{V}}\\\Rightarrow C=\dfrac{25\times 10^6}{2\times 10^7}\\\Rightarrow C=1.25\ \mu\text{g/m}^3[/tex]

The steady state concentration of pollutants in the valley, is [tex]1.25\ \mu\text{g/m}^3[/tex].

Answer: gravitational potential energy

Explanation:

The Answer to your question is:

A.

The Earth's tectonic activity occurs by the movement of the fourteen main tectonic plates of the planet, which move over the mantle continuously, and result in the formation of mountains, earthquakes, tsunamis, volcanic activities, etc.

The decrease in the Earth's tectonic activity would therefore occur if the Earth's mantle became cooler, as the tectonic plates move over the magma, which is a paste formed by silicate of iron and magnesium and whose temperature reaches 600° and 1,200° Celsius , driven by forces from inside the planet.

Therefore the letter C is correct, as with the Earth's mantle cooler, the magma would become more solid which would decrease the Earth's tectonic activity.

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Answer:

Hello your question is poorly written below is the complete question

Suppose the battery in a clock wears out after moving Ten thousand coulombs of charge through the clock at a rate of 0.5 Ma how long did the clock run on does battery and how many electrons per second slowed?

answer :

a) 231.48 days

b) n = 3.125 * 10^15

Explanation:

Battery moved 10,000 coulombs

current rate = 0.5 mA

A) Determine how long the clock run on the battery. use the relation below

q = i * t ----- ( 1 )

q = charge , i = current , t = time

10000 = 0.5 * 10^-3 * t

hence  t = 2 * 10^7 secs

hence the time = 231.48  days

B) Determine how many electrons per second flowed

q = n*e ------ ( 2 )

n = number of electrons

e = 1.6 * 10^-19

q = 0.5 * 10^-3 coulomb ( charge flowing per electron )

back to equation 2

n ( number of electrons ) = q / e = ( 0.5 * 10^-3 ) / ( 1.6 * 10^-19 )

hence : n = 3.125 * 10^15

Answer:

$ 5.93

Explanation:

If we are to put into consideration the following factors even though it was not all stated in the question.

What is the cost of heating a hot tub containing 1500 kg of water from 10.0°C to 40.0°C, assuming 75.0% efficiency to take heat loss to surroundings into account? The cost of electricity is ¢8.50/kW·h.

Q = mc∆T .........1

Where m is the mass of water = 1500kg

C is the specific heat capacity of water = 4.184

∆T = temperature change

= 40° - 10°

= 30°

Now, substitute the values into equation 1

= 1500×4.184×30

= 188280

Q in kw/s

= 188280/3600

= 52.3kw/hr

Efficiency = 75%

= 75/100

= 0.75

Hence we have

52.3/0.75

= 69.73kw/hr

C = Q × cost per kw/hr

C = 69.73× $8.5/100

C = 5.92705

C =$ 5.93

Hence, the cost of heating the hot tub is $ 5.93

Answer:

please put pic of the questions

Answer:

g ±Δg = (9.8 ± 0.2) m / s²

Explanation:

For the calculation of the acceleration of gravity they indicate the equation of the simple pendulum to use

          T = [tex]2\pi \sqrt{ \frac{L}{g} }[/tex]

          T² =  [tex]4\pi ^2 \frac{L}{g}[/tex]4pi2 L / g

          g = [tex]4\pi ^2 \frac{L}{T^2}[/tex]

They indicate the average time of 20 measurements 1,823 s, each with an oscillation

let's calculate the magnitude

           g = [tex]4\pi ^2 \frac{0.823}{1.823^2}[/tex]4 pi2 0.823 / 1.823 2

            g = 9.7766 m / s²

now let's look for the uncertainty of gravity, as it was obtained from an equation we can use the following error propagation

for the period

             T = t / n

             ΔT = [tex]\frac{dT}{dt}[/tex] Δt + [tex]\frac{dT}{dn}[/tex] ΔDn

In general, the number of oscillations is small, so we can assume that there are no errors, in this case the number of oscillations of n = 1, consequently

              ΔT = Δt / n

              ΔT = Δt

now let's look for the uncertainty of g

             Δg = [tex]\frac{dg}{dL}[/tex] ΔL + [tex]\frac{dg}{dT}[/tex]  ΔT

             Δg = [tex]4\pi ^2 \frac{1}{T2}[/tex]   ΔL + 4π²L  (-2  T⁻³) ΔT

a more manageable way is with the relative error

             [tex]\frac{\Delta g}{g} = \frac{\Delta L }{L} + \frac{1}{2} \frac{\Delta T}{T}[/tex]

we substitute

              Δg = g ( \frac{\Delta L }{L} + \frac{1}{2}  \frac{\Delta T}{T}DL / L + ½ Dt / T)

the error in time give us the stanndard deviation  

let's calculate

               Δg = 9.7766 ([tex]\frac{0.001}{0.823} + \frac{1}{2} \ \frac{0.671}{1.823}[/tex])

               Δg = 9.7766 (0.001215 + 0.0184)

               Δg = 0.19 m / s²

the absolute uncertainty must be true to a significant figure

                Δg = 0.2 m / s2

therefore the correct result is

               g ±Δg = (9.8 ± 0.2) m / s²

Consider the balloon and air inside the flask to be in a closed system. Using the First Law of Thermodynamics to explain what happened to the balloon as heat was added by the environment.

Balloon will burst, due to more heat and in another flask air gets heated.

"The first law of thermodynamics states that energy can neither be created nor destroyed, only altered in form. For any system, energy transfer is associated with mass crossing the control boundary, external work, or heat transfer across the boundary. These produce a change of stored energy within the control volume."

"Heat is the energy that is transferred from one body to another as the result of a difference in temperature. If two bodies at different temperatures are brought together, energy is transferred i.e, heat flows—from the hotter body to the colder."

Know more about heat here

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Answer:

Hello your question is incomplete attached below is the complete question

answer :

Total charge enclosed within the sphere : [tex]\frac{q_{r1} }{4\pi e_{0}R^3 } . r[/tex]

Total charge enclosed outside the sphere : [tex]\frac{q}{4\pi e_{0}r^2 } .r[/tex]

Explanation:

Given data:

Total charge of a uniformly charged sphere = Q

radius = R

first step : find the electric field inside and outside the uniformly charged sphere

2nd step : determine the total charge enclosed within and outside the sphere

make a sketch of the uniformly charged sphere

Attached below is a detailed solution

Answer:

R' = R/4

Explanation:

The resistance of a metal rod is R. It is given by the relation as follows :

[tex]R=\rho\dfrac{l}{A}[/tex]

Where

l is the length and A is the area of cross-section

[tex]A=\pi r^2=\pi (\dfrac{d}{2})^2[/tex]

If both its length and its diameter are quadrupled, it means,

l' = 4l

and d'= 4d

It means,

[tex]A'=\pi (\dfrac{4d}{2})^2[/tex]

Let new resistance be R'. So,

[tex]R'=\rho\dfrac{l'}{A'}\\\\R'=\rho\dfrac{4l}{\pi (\dfrac{4d}{2})^2}\\\\=\rho \dfrac{4l}{\pi \dfrac{16d^2}{2}}\\\\=\dfrac{4}{16}\times \dfrac{\rho l}{\pi \dfrac{d^2}{2}}\\\\=\dfrac{1}{4}\times \dfrac{\rho l}{\pi \dfrac{d^2}{2}}\\\\R'=\dfrac{R}{4}[/tex]

So, the correct option is (B) "R/4".

Answer:

density = mass/volume

so . . .  

density = (36 g)/(15 cm³) = 2.4 g/cm³

Explanation:

Answer:

i think rocks towards is correct answer

Answer:

It is gently on

Explanation:

I'm am not a 100 percent sure but try this o took that before

This question is incomplete, the complete question is;

A monatomic gas fills the left end of the cylinder in the following figure. At 300 K , the gas cylinder length is 14.0 cm and the spring is compressed by65.0 cm . How much heat energy must be added to the gas to expand the cylinder length to 16.0 cm ?

Answer:

the required heat energy is 16 J

Explanation:

Given the data in the question;

Lets consider the ideal gas equation;

PV = nRT

from the image, we calculate initial pressure;

Pi = ( 2000N/M × 0.06m) / 0.0008 m²

Pi = 15 × 10⁴ Pa

next we find Initial velocity

Vi = (0.0008 m²)(0.14) = 1.1 × 10⁻⁴ m²

now we find the number of moles

n = [(15 × 10⁴ Pa)(1.1 × 10⁻⁴ m²)] / 8.31 J/molK × 300K

N = 6.6 × 10⁻³ mol

next we calculate the final temperature;

Pf = ( 2000N/m × 0.08) / 0.0008 m²

Pf = 2 × 10⁵ Pa

Calculate the final Volume

Vf = (0.0008 m² × 0.16 m = 1.28 × 10⁻⁴ m³

we also determine the final temperature

[tex]T_{f}[/tex] =  (2 × 10⁵ Pa × 1.28 × 10⁻⁴ m³) / 6.6 × 10⁻³ × 8.31 J/molK

[tex]T_{f}[/tex]  = 466.8 K

so change in temperature ΔT

ΔT =  466.8 K - 300K = 166.8 K

we then calculate the change in thermal energy

ΔU = nCΔT

ΔU = ( 6.6 × 10⁻³ mol ) × 12.5 × 166.8K

ΔU = 13.761 J

C is the isochoric molar specific heat which is equal to 3R/2 for monoatomic

now we calculate the work done;

W = 1/2 × K( x[tex]_{i\\}[/tex]² - x[tex]_{f\\}[/tex]² )

W = 1/2 × ( 2000 N/m) ( 0.06² - 0.08² )

= - 2.8 J

and we then calculate the heat energy using the following expression;

Q = ΔU - W

we substitute

Q = 13.761 - (- 2.8 J)

Q = 13.761 + 2.8 J)

Q =  16 J

Therefore, the required heat energy is 16 J

Answer:

hello

Explanation:

For scientists, they offer a unique opportunity to study aspects of the sun like its corona, the layer of plasma surrounding the star. ... By studying the inner regions of the corona that we can generally only see and photograph well during a total solar eclipse, scientists can learn more about how weather works in space.

have a nice day

hope it helps

byee

Answer:

For scientists,they offer a unique opportunity to study aspects of the sun like it's Corona, the layer of plasma surrounding the star......By studying the inner regions of the Corona that we can generally only see and photograph well during a total solar eclipse, scientists can learn more about how weather works in space

Answer:

8 units

Explanation:

Answer:

Probably (C), but don't forget the argument that:

Roses are red, and roses are flowers; therefore all flowers are red.