Using equations, determine the temperature, pressure and density of the air for a aircraft flying at 19.5 km. Is this aircraft subsonic or supersonic

Answer:

6N

Explanation:

Given parameters:

Mass of object  = 6kg

Initial velocity  = 5m/s

Final velocity  = 25m/s

Time  = 30s

Unknown:

Net force acting on the object  = ?

Solution:

From Newton's second law of motion:

   Force  = mass x acceleration

Acceleration is the rate of change of velocity with time

  Acceleration  = [tex]\frac{Final velocity - Initial velocity }{time}[/tex]  

  Force  = mass x  [tex]\frac{Final velocity - Initial velocity }{time}[/tex]  

So;

 Force  = 6 x [tex]\frac{25 - 5}{30}[/tex]    = 6N

Explanation:

Given data:

mass of the car = 1500 kg

maximum friction force = 7000 N

initial velocity v_i = 20 m/s ( it is not given in the question just an assumption)

final velocity v_f = 0 m/s

[tex]\begin{array}{l}

\sum F_{y}=M g-F_{n}=0 \\

\sum F_{x}=-F_{s}=m a_{x} \\

-F_{s}=m a_{x}

\end{array}[/tex]  

[tex]a_{x}=\frac{-F_{s}}{m}=\frac{-7000}{1500}[/tex]

[tex]a_{x}=-4.7 \mathrm{~m} / \mathrm{s}^{2}[/tex]

Now we can find the distance from this formula:

[tex]v_{f x}^{2}=v_{i x}^{2}+2 a_{x}(\Delta x)[/tex]

[tex]0=20^{2}+(2 \times-4.7 \times \Delta x)[/tex]

[tex]20^{2}=9.4 \Delta x[/tex]

[tex]\Delta x=\frac{20^{2}}{9.4}=42.55 \mathrm{~m}[/tex]

So, the shortest distance in which the car can stop safely without kidding

=42.55 m

Answer:

a. Photogates placed at the beginning, end, and at various locations along the track that the car travels on.

b. A meterstick to measure the distance of the track that the car travels on.

Explanation:

Physics can be defined as the field or branch of science that typically deals with nature and properties of matter, motion and energy with respect to space, force and time.

In this scenario, a student is provided with a battery-powered toy car that the manufacturer claims will always operate at a constant speed. The student must design an experiment in order to test the validity of the claim.

Therefore, to test the validity of the claim, the student should use the following measuring tools;

a. Photogates placed at the beginning, end, and at various locations along the track that the car travels on. This device is typically used to measure time with respect to the rate of change of the interruption or block of an infra-red beam.

b. A meterstick to measure the distance of the track that the car travels on.

Hence, with these two devices the student can effectively measure or determine the validity of the claim.

Answer:

i dont know i am right but here Luminous intensity is defined as dI=dΨλ / dΩ, where dΨλ is the luminous flux (light energy flux in watts per m2) emitted within a solid angle dΩ. The light energy flux may be expressed in terms of the incident x-ray energy flux and the x-ray absorption and conversion properties of the scintillator(7,8,9).

Explanation:

Answer:

the second time there is a gas between you and the star,

Explanation:

When you observe the star for the first time you do not have a given between you and the star, therefore you observe the emission spectrum of the same that is formed by lines of different intensity and position that indicate the type and percentage of the atoms that make up the star.

 When you observe the same phenomenon for the second time there is a gas between you and the star, this gas absorbs the wavelengths of the star that has the same energies and the atomisms and molecular gas, therefore these lines are not observed by seeing a series of dark bands,

The information obtained from the two spectra is the same, the type of atoms that make up the star

Answer:

[tex]Vm=0.894m/s[/tex]

Explanation:

From the question we are told that

Velocity if travel [tex]v=4m/s[/tex]

Diameter of  prototype [tex]d_1=20m[/tex] and [tex]d_2=110m[/tex]

Scale ratio=[tex]\frac{1}{20}[/tex]

Generally Velocity of of the model using Froud's model is mathematically given as

[tex]Fm=Fp[/tex]

[tex]\frac{Vm}{\sqrt{Lmg}} =\frac{Vp}{\sqrt{Lpg}}[/tex]

[tex]Vm=Vp*\frac{Vp}{\sqrt{Lpg} }[/tex]

[tex]Vm=4*\frac{1}{\sqrt{20}}[/tex]

[tex]Vm=0.894m/s[/tex]

Answer:

form 1 question??????????

Answer:

0.000153DaL  

Explanation:

We have been given:

         15.3mL to convert to DaL

DaL is a unit of volume which indicates a decaliter.

 This implies that;

             1 Da L  = 1 x 10²L

So:

               1 mL  = 1 x 10⁻³L

       So 15.3mL will give 15.3 x 10⁻³L

So;

           1 x 10²L   =  1 DaL  

      15.3 x 10⁻³L  will give [tex]\frac{15.3 x 10^{-3} }{1 x 10^{2} }[/tex]   = 15.3 x 10⁻⁵DaL

Therefore, this is 0.000153DaL  

Answer:

48 rev

Explanation:

a) we can calculate the distance covered by the tube using the formula:

θ = (ω + ωo)t/2

where ω is the final angular speed, θ is the distance covered, t is the time taken, ωo is the initial angular speed.

Firstly, we calculate the distance covered from 0 to 9 s then from 9s to 24 s.

within 9s, the tub runs from rest (0) to 4 rev/s, hence:

t = 9s, wo = 0, w = 4 rev/s = (4 * 2π) rad/s = 8π rad/s. Hence:

θ = (ω + ωo)t/2 = (0 + 8π)9 / 2 = 36π rad

θ = 36π rad = (36π)/2π rev = 18 rev

Also, within 15 s, the tub runs from 4 rev/s to rest, hence:

t = 15 s, wo = 4 rev/s = 8π rad/s, w = 0 rad/s. Hence:

θ = (ω + ωo)t/2 = (8π + 0)15 / 2 = 60π rad

θ = 60π rad = (60π)/2π rev = 30 rev

Therefore the total revolutions by the tube during 24 s interval = 30 rev + 18 rev = 48 rev

Answer:

Ultraviolet radiation

Explanation:

The type of electromagnetic radiation most likely to cause skin cancer as a result of sun exposure overtime is the ultraviolet radiation.

Electromagnetic radiation occurs with a broad spectrum starting from gamma rays to the radio waves.

From one end to the other, their energy decreases as the wavelength increases.

Within this broad spectrum, the ultraviolet rays which are before the visible rays are very energetic and can cause skin cancer.

Answer:

ultraviolet i think sorry if im wrong :/

Explanation:

Answer:

The Gravitational potential energy at large distances is directly proportional to the masses and inversely proportional to the distance between them. The gravitational potential energy increases as r increases.

Examples of Gravitational Energy

A raised weight.

Water that is behind a dam.

A car that is parked at the top of a hill.

A yoyo before it is released.

Answer:

58.8J

Explanation:

Given parameters;

Mass of ball  = 4kg

Height above the floor  = 1.5m

g  = 9.8n/kg

Unknown:

Potential energy  = ?

Solution:

The potential energy of a body is the energy due to the position of the body.

It is mathematically expressed as:

  Potential energy = mass x acceleration due to gravity x height

  Potential energy  = 4 x 9.8 x 1.5  = 58.8J

Answer:

C. Is traveling on a curve in the road

    Hope this helps :3

The pick up truck has a changing velocity because, it is travelling on a curve in the road. A change in direction results in its change in velocity because, velocity is a vector quantity.

Velocity is a physical quantity that measures the distance covered by an object per unit time. It is a vector quantity, thus having magnitude as well direction.

The rate of change in velocity is called acceleration of the object. Like velocity, acceleration also is a vector quantity. Thus, a change in magnitude or direction or change in both for velocity make the object to accelerate.

Here, all the three vehicles  are travelling with the same velocity. But, the truck is moving to a curve on the road. The curvature in the path will make a change in its velocity.

Find more on velocity:

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The image related with this question is attached below:

Answer:

the correct one is the first,   the refractive index of the two materials must be the same

Explanation:

When a beam of light passes through two materials, it must comply with the law of refraction

         n₁ sin θ₁ = n₂ sin θ₂

where n₁ and n₂ are the refractive indices of each medium.

In this case, it indicates that the light does not change direction, so the input and output angle of the interface must be the same,

       θ₁ = θ₂ = θ

substituting

          n₁ = n₂

therefore the refractive index of the two materials must be the same

When reviewing the answers, the correct one is the first

Answer:

T = 10010 N

Explanation:

To solve this problem we must use the translational equilibrium relation, let's set a reference frame

X axis

       Fₓ-Fₓ = 0

       Fₓ = Fₓ

whereby the horizontal components of the tension in the cable cancel

Y Axis  

        [tex]F_{y} + F_{y} - W =0[/tex]

        2[tex]F_{y}[/tex] = W

let's use trigonometry to find the angles

        tan θ = y / x

        θ = tan⁻¹ (0.30 / 0.50 L)

        θ = tan⁻¹ (0.30 / 0.50 15)

        θ = 2.29º

the components of stress are

         F_{y} = T sin θ

we substitute

       2 T sin θ = W

       T = W / 2sin θ

        T = [tex]\frac{ 800}{ 2sin 2.29}[/tex]

        T = 10010 N

Answer:I think it’s self monitoring sorry if wrong

Explanation:

Answer:

It self monitoring

Explanation:

I took the test

Answer:

A

Explanation:

Ke = 1/2 MV^2

Answer:

a. 10.5 s b. 6.6 s

Explanation:

a. The driver's perception/reaction time before drinking.

To find the driver's perception time before drinking, we first find his deceleration from

v² = u² + 2as where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 0 m/s (since he stops), a = deceleration of driver and s = distance moved by driver = 385 ft = 385 × 0.3048 m = 117.35 m

So, a = v² - u²/2s

substituting the values of the variables into the equation, we have

a = v² - u²/2s

a = (0 m/s)² - (22.35 m/s)²/2(117.35 m)

a =  - 499.52 m²/s²/234.7 m

a = -2.13 m/s²

Using a = (v - u)/t where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 0 m/s (since he stops), a = deceleration of driver = -2.13 m/s² and t = reaction time

So, t = (v - u)/a

Substituting the values of the variables into the equation, we have

t = (0 m/s - 22.35 m/s)/-2.13 m/s²

t = - 22.35 m/s/-2.13 m/s²

t = 10.5 s

b. The driver's perception/reaction time after drinking.

To find the driver's perception time after drinking, we first find his deceleration from

v² = u² + 2as where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 30 mi/h = 30 × 1609 m/3600 s = 13.41 m/s, a = deceleration of driver and s = distance moved by driver = 385 ft = 385 × 0.3048 m = 117.35 m

So, a = v² - u²/2s

substituting the values of the variables into the equation, we have

a = v² - u²/2s

a = (13.41 m/s)² - (22.35 m/s)²/2(117.35 m)

a = 179.83 m²/s² - 499.52 m²/s²/234.7 m

a = -319.69 m²/s² ÷ 234.7 m

a = -1.36 m/s²

Using a = (v - u)/t where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 30 mi/h = 30 × 1609 m/3600 s = 13.41 m/s, a = deceleration of driver = -1.36 m/s² and t = reaction time

So, t = (v - u)/a

Substituting the values of the variables into the equation, we have

t = (13.41 m/s - 22.35 m/s)/-1.36 m/s²

t = - 8.94 m/s/-1.36 m/s²

t = 6.6 s

Answer:

a) A:170572.5 J

   C: 55794.9J

b) 170572.5 J

c) 41.4413265306m

d) 2.7455874717m/s

Explanation:

a) Kinetic energy = 0.5*m*v²

KE at A = 0.5*420*28.5² = 170572.5 J

KE at C = 0.5*420*16.3² = 55794.9 J

b) Mechanical energy is the total kinetic energy plus potential energy at a point on the system. There is no potential energy at A.

ANSWER: 170572.5 J

c) v²=u²+2as

28.5²=2(9.8)s

812.25/19.6=s

s=41.4413265306m

d) h=height from part c, r=radius of loop

v²=u²+2as

v²=gr or a=v²/r

Ei=Ef

mgh=0.5mv²+mg(2r)

gh=0.5v²+2gr

h=0.5r+2r

h=5/2r

r=2/5h=(2/5)(41.4413265306)=16.5765306122

F=ma

mg=m(v²/r)

g=v²/r

v²=(9.8)(16.5765306122)

v=√162.45

=12.7455874717m/s

Answer:

A. 72000 C

B. 1100 W

C. 26.4 cents.

Explanation:

From the question given above, the following data were obtained:

Current (I) = 10 A

Voltage (V) = 110 V

Time (t) = 2 h

A. Determination of the charge.

We'll begin by converting 2 h to seconds. This can be obtained as follow:

1 h = 3600 s

Therefore,

2 h = 2 h × 3600 s / 1 h

2 h = 7200 s

Thus, 2 h is equivalent to 7200 s.

Finally, we shall determine the charge. This can be obtained as follow:

Current (I) = 10 A

Time (t) = 7200 s

Charge (Q) =?

Q = It

Q = 10 × 7200

Q = 72000 C

B. Determination of the power.

Current (I) = 10 A

Voltage (V) = 110 V

Power (P) =?

P = IV

P = 10 × 110

P = 1100 W

C. Determination of the cost of operation.

We'll begin by converting 1100 W to KW. This can be obtained as follow:

1000 W = 1 KW

Therefore,

1100 W = 1100 W × 1 KW / 1000 W

1100 W = 1.1 KW

Thus, 1100 W is equivalent to 1.1 KW

Next, we shall determine the energy consumption of the range. This can be obtained as follow:

Power (P) = 1.1 KW

Time (t) = 2 h

Energy (E) =?

E = Pt

E = 1.1 × 2

E = 2.2 KWh

Finally, we shall determine the cost of operation. This can be obtained as follow:

1 KWh cost 12 cents.

Therefore, 2.2 KWh will cost = 2.2 × 12

= 26.4 cents.

Thus, the cost of operating the range for 2 h is 26.4 cents.

Answer:

LAW 1 :  For a given metal and frequency, the number of photoelectrons emitted is directly proportional to the intensity of the incident radiation.  

---------------------------------------------

LAW 2: For a given metal, there exists a certain frequency below which the photoelectric emission does not take place. This frequency is called threshold frequency.

-----------------------------------------------

LAW 3: For a frequency greater than the threshold frequency, the kinetic energy of photoelectrons is dependent upon frequency or wavelength but not on the intensity of light.

-----------------------------------------------

LAW 4: Photoelectric emission is an instantaneous process. The time lag between incidence of radiations and emission of electron is 10^-9 seconds.

Explanation:

Answer:

LAW 1 : For a given metal and frequency, the number of photoelectrons emitted is directly proportional to the intensity of the incident radiation. ... LAW 4: Photoelectric emission is an instantaneous process.

Answer:

displacement is the distance between the starting point and the ending point of an object's journey

Answer:

Explanation:

Properties of the sun

Introduction to Stars A. The Sun 1. ...

Layers of the sun (Atmosphere) a. Photosphere: Layer that emits the radiation we see. ...

Layers of the sun (Interior) a. ...

Solar Wind: a. ...

The sun in X-Rays: a. ...

Luminosity: The measure of the energy put out by the sun. ...

Luminosity continued: c. ...

Luminosity continued: e.

The heat of the Sun's surface is so large that neither solid nor liquid can reside there. The heat is the physical property of the sun.

The Sun is a billion star at the core of our universe, a hot. It is an incandescent ball of hydrogen and helium.

The Sun is millions of kilometers away from the land, and life is not possible without its energy.

The major properties of the sun are as follows;

1. The mass of the sun is 1.98892 x 10³⁰ kg.

2 The density of the sun is 1.622 x 10⁵ kg/m³

3. The surface gravity of the sun is 27.94 g.

4. The diameter is 1,391,000 kilometers.

5. The volume of the Sun is 1.412 x 1018 km³

The heat of the Sun's surface is so hot that neither solid nor liquid can reside there instead, the component materials are mostly gaseous atoms with a few molecules. As a result, no permanent surface exists.

Hence heat, density, and mass are the major properties of the sun.

To learn more about the sun refer to the link;

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Answer:

Explanation:

Well they told you the exact formula to use. Work is the force multiplied by  the distance through which its applied.

W = (20N)(35m)

= 700 Joules

13.) Power is the amount of work done over the time through which the work is being done.

P = W/t

= 10J/2s

= 5J/s

Answer:

(a) the unknown inertia is 0.388 kg

(b) the average acceleration of the heavier cart is 317 m/s²

(c) the average acceleration of the lighter cart is 539 m/s²

Explanation:

Given;

mass of the first cart, m₁ = 0.66 kg

initial speed of the first cart, u₁ = 1.85 m/s

let the mass of the cart with unknown inertia be m₂

initial velocity of the second cart, u₂ = 2.17 m/s to the left

velocity of the first cart after collision, v₁ = 1.32 m/s to the left

velocity of the second cart after collision, v₂ = 3.22 m/s

time of collision, t = 0.010 s

(a) What is the unknown inertia?

Apply the principle of conservation of linear momentum, to determine the unknown inertia.

let leftward direction be negative direction

let rightward direction be positive direction

m₁u₁ + m₂u₂ = m₁v₁  + m₂v₂

0.66(1.85) + m₂(-2.17) = 0.66(-1.32) + m₂(3.22)

1.221 - 2.17m₂ = -0.8712 + 3.22m₂

1.221 + 0.8712 = 3.22m₂ + 2.17m₂

2.0922 = 5.39m₂

m₂ = 2.0922 / 5.39

m₂ = 0.388 kg

The unknown inertia is 0.388 kg

(b) the average acceleration of the heavier cart

the heavier cart has a mass of 0.66 kg

[tex]a = \frac{v_1 - u_1}{t} \\\\a = \frac{-1.32 - 1.85}{0.01} \\\\a = -317 \ m/s^2\\\\|a| = 317 \ m/s^2[/tex]

(c) the average acceleration of the lighter cart;

the lighter cart has a mass of 0.388 kg

[tex]a = \frac{v_2 - u_2}{t} \\\\a = \frac{3.22 - (-2.17)}{0.01} \\\\a =\frac{3.22 \ +\ 2.17}{0.01} \\\\a= 539\ m/s^2[/tex]

Answer:

ieces A and B must also have the same type of charges

Explanation:

In electrostatics, charges of the same sign repel and charges of different signs attract.

If we apply this to our case, we have that part A and C repel each other, therefore they have the same type of charge.

Also part A and C repel each other, therefore they have the same type of charge.

If we use the transitive property of mathematics, pieces A and B must also have the same type of charges

Static friction (magnitude Fs) keeps the car on the road, and is the only force acting on it parallel to the road. By Newton's second law,

Fs = m a = W a / g

(a = centripetal acceleration, m = mass, g = acceleration due to gravity)

We have

a = v ² / R

(v = tangential speed, R = radius of the curve)

so that

Fs = W v ² / (g R)

Solving for v gives

v = √(Fs g R / W)

Perpendicular to the road, the car is in equilibrium, so Newton's second law gives

N - W = 0

(N = normal force, W = weight)

so that

N = W

We're given a coefficient of static friction µ = 0.4, so

Fs = µ N = 0.4 W

Substitute this into the equation for v. The factors of W cancel, so we get

v = √((0.4 W) g R / W) = √(0.4 g R) = √(0.4 (9.80 m/s²) (105 m)) ≈ 20.3 m/s

Answer:

3.65 x mass

Explanation:

Given parameters:

Time  = 20s

Initial velocity  = 0m/s

Final velocity  = 73m/s

Unknown:

Force the ball experience  = ?

Solution:

To solve this problem, we apply the equation from newton's second law of motion:

    F  =  m [tex]\frac{v - u}{t}[/tex]  

m is the mass

v is the final velocity

u is the initial velocity

 t is the time taken

So;

  F  = m ([tex]\frac{73 - 0}{20}[/tex] )  = 3.65 x mass

Answer:

The higher the amplitude, the higher the energy. To summarise, waves carry energy. The amount of energy they carry is related to their frequency and their amplitude. The higher the frequency, the more energy, and the higher the amplitude, the more energy

Explanation:

The main difference between a wave and energy is: wave is oscillation of energy whereas energy is ability of doing work.

A wave is an energetic disturbance in a medium that doesn't include any net particle motion. Elastic deformation, a change in pressure, an electric or magnetic intensity, an electric potential, or a change in temperature are a few examples.

The capacity to do work is energy. Energy can only be changed from one form to another; it cannot be created or destroyed. Energy is measured in Joules, the same unit used to measure work. There are several sorts of energy since it is present in many different things.

There are two types of energy: kinetic and potential. Kinetic energy is the energy that is in motion, whereas potential energy is the energy that is stored in an object and is determined by the amount of work that is required.

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Answer:

the charge on the object is 71.043×10^-20 C and the number of electron is 4.44

Explanation:

from coulumbs law, The force that is acting over both charge can be computed as

F=( kq1q2)/r^2..............eqn(1)

Where

F=electrostatic force= 3.89 x 10-21 N

k= column constant= 9 x 10^9 Nm^2/C^2

q1 and q2= magnitude of the charges

r= distance between the charges= 1.08 x 10-3 m.

Since both charges are experiencing the same force, eqn(1) can be written as

F=( kq^2)/r^2.

We can make q subject of the formula

q= √(Fr^2)/k

= √[(3.89 x 10^-21× (1.08 x 10^-3)^2]/8.99 x 10^9

q= 71.043×10^-20 C

Hence, the charge is 71.043×10^-20 C

From quantization law, the number of electron can be computed as

N=q/e

Where

N= number of electron

q= charges

=1.6×10^-19C

N=71.043×10^-20/1.6×10^-19

=4.44

Hence, the charge on the object is 71.043×10^-20 C and the number of electron is 4.44