A remote controlled airplane moves 7.2 m in 2.5seconds what is the plane’s velocity

Answer:

[tex]0.05\ \text{kJ/kg}[/tex]

[tex]3141.6\ \text{kW}[/tex]

Explanation:

v = Velocity of wind = 10 m/s

A = Swept area of blade = [tex]\dfrac{\pi}{4}d^2[/tex]

d = Diameter of turbine = 80 m

[tex]\rho[/tex] = Density of air = [tex]1.25\ \text{kg/m}^3[/tex]

Wind energy per unit mass of air is given by

[tex]E=\dfrac{v^2}{2}\\\Rightarrow E=\dfrac{10^2}{2}\\\Rightarrow E=50\ \text{J/kg}[/tex]

The mechanical energy of air per unit mass is [tex]0.05\ \text{kJ/kg}[/tex]

Power is given by

[tex]P=\rho AvE\\\Rightarrow P=1.25\times \dfrac{\pi}{4}\times 80^2\times 10\times 50\\\Rightarrow P=3141592.65=3141.6\ \text{kW}[/tex]

The power generation potential of the wind turbine is [tex]3141.6\ \text{kW}[/tex].

Answer:

The constant rate of acceleration required in order to accomplish this is 1.921 feet per square second.

Explanation:

Let suppose that car accelerates uniformly in a rectilinear motion. Given that initial and final speeds and travelled distances are known, then the acceleration needed by the vehicle ([tex]a[/tex]), measured in feet per square second, is determined by the following kinematic formula:

[tex]a = \frac{v_{f}^{2}-v_{o}^{2}}{2\cdot \Delta x }[/tex] (1)

Where:

[tex]v_{o}[/tex], [tex]v_{f}[/tex] - Initial and final speeds, measured in feet per second.

[tex]\Delta x[/tex] - Travelled distance, measured in feet.

If we know that [tex]v_{o} = 34.907\,\frac{ft}{s}[/tex], [tex]v_{f} = 67.027\,\frac{ft}{s}[/tex] and [tex]\Delta x = 852\,ft[/tex], then acceleration needed to accomplish the task is:

[tex]a = 1.921\,\frac{ft}{s^{2}}[/tex]

The constant rate of acceleration required in order to accomplish this is 1.921 feet per square second.

Answer:

1.4999

Explanation:

Efficiency can be calculated using below expresion

Efficiency = W/Q.............eqn(1)

Where W= work = 50 J

Q= thermal energy= 150 J

But

W/Q= (Th-Tc)/Th ...........Eqn(2)

Th= temperature of the hot

Tc= temperature of the cold

Where Th/ Tc= ratio of the temperature hot and cold reservoirs?

If we simplify eqn(2) we have

W/Q = 1-Tc/Th.........eqn(3)

If we make the ratio subject of the formula we have

Tc/Th = 1-(W/Q)

Th/Tc = 1/(1-W/Q )

Then substitute the values

= 1/(1-50/150) = 1.4999

Hence, the smallest possible ratio of the temperatures (in kelvin) of the hot and cold reservoirs is 1.4999

Answer:

argon (Ar)

Explanation:

Argon is the element from the given choices with a complete valence electron shell.

The valence electron shell is the outermost shell of an atom.

Answer:

argon (Ar)

Explanation:

Answer: C

Explanation: 60 divided by 4 =15

Velocity can be defined as the rate of change of distance with time

Given data

Acceleration = 4/ms^2

Distance = 60m

Initial Velocity U= 0

Final Velocity V= ?

The expression for velocity is given by

V^2= U^2+2as

Let us substitute our given data into the expression

V^2 = 0^2 + 2*4*60

V^2 = 480

Square both sides

V= √480

V= 21.9 meters/second

V= 22 meters/seconds Approx.

The correct answer is option B

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Answer:

[tex]\triangle U=-e (V_2-V_1)[/tex]

[tex]\triangle U=130eV[/tex]

[tex]V_2=\sqrt{ \frac{2}{me}(\frac{1}{2}meV_1^2+e(V_2-V_1)}[/tex]

Explanation:

From the question we are told that

The potential at point 1, [tex]V_1 = 24V[/tex]

The potential at point 2, [tex]V_2 = 154V[/tex]

a)Generally work done by proton is given as

 [tex]w=-\triangle U[/tex]

 [tex]e\triangle V=-\triangle U[/tex]

 [tex]\triangle U=-e (V_2-V_1)[/tex]  

Generally the Equation for the change of electric potential energy AU of the proton, in terms of the symbols given and the charge of the proton e is mathematically given as

 [tex]\triangle U=-e (V_2-V_1)[/tex]

b)Generally the electric potential energy in electron volts (eV). is mathematically given as

 [tex]\triangle U=-e (154-24)V[/tex]

 [tex]|\triangle U| =|-e (130)V|[/tex]

 [tex]\triangle U=130eV[/tex]

c) Generally according to the law of conservation of energy

[tex](K.E+P.E)_1=(K.E+P.E)_2[/tex]

[tex]\frac{1}{2}meV_1^2+eV_1 =\frac{1}{2}mev_2^2+eV_2[/tex]

[tex]V_2^2=\frac{2}{me}(\frac{1}{2}meV_1^2+e(V_2-V_1)[/tex]

[tex]V_2=\sqrt{ \frac{2}{me}(\frac{1}{2}meV_1^2+e(V_2-V_1)}[/tex]

Answer:

Force = 2240 Newton.

Explanation:

Given the following data;

Mass A= 65kg

Mass B = 215kg

Acceleration = 8m/s²

To find the force;

Force is given by the multiplication of mass and acceleration.

Mathematically, Force is;

[tex] F = ma[/tex]

Where;

F represents force.

m represents the mass of an object.

a represents acceleration.

First of all, we would have to find the total mass.

Total mass = Mass A + Mass B

Total mass = 65 + 215

Total mass = 280kg

Substituting into the equation, we have

[tex] Force = 280 * 8 [/tex]

Force = 2240 Newton.

This question is incomplete, the complete question is;

According to information found in an old hydraulics book, the energy loss per unit weight of fluid flowing through a nozzle connected to a hose can be estimated by the formula; h= (0.04 to 0.09)(D/d)⁴V²/2g

where h is the energy loss per unit weight, D the hose diameter, d the nozzle tip diameter, V the fluid velocity in the hose, and g the acceleration of gravity.

Do you think this equation is valid in any system of units

Answer:

YES, the equation is a general equation that is valid in any system of units

Explanation:

Given the data in the question;

h = (0.04 to 0.09)(D/d)⁴ × [tex]\frac{V^{2} }{2g}[/tex]

so

[ N.m/N ] = (0.04 to 0.09) ( m/m)² × (m²/s²)1/2 × (s²/m)

[ N.L/N ] = (0.04 to 0.09) ( L⁴/L⁴) × (L²/T²)1/2 × (T²/L)

∴ [ L ] = (0.04 to 0.09) [L]

So as each term in the equation must have the same dimensions, the constant term (0.04 to 0.09) must be without dimension.

Therefore, YES, the equation is a general equation that is valid in any system of units

Why words are more important than numbers: ... Words on the other hand are harder to manipulate, they also tell you why someone voted a particular way and to improve your delivery and thus your customer satisfaction you need to understand the why's...

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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:

D = -4/7 = - 0.57

C = 17/7 = 2.43

Explanation:

We have the following two equations:

[tex]3C + 4D = 5\ --------------- eqn (1)\\2C + 5D = 2\ --------------- eqn (2)[/tex]

First, we isolate C from equation (2):

[tex]2C + 5D = 2\\2C = 2 - 5D\\C = \frac{2 - 5D}{2}\ -------------- eqn(3)[/tex]

using this value of C from equation (3) in equation (1):

[tex]3(\frac{2-5D}{2}) + 4D = 5\\\\\frac{6-15D}{2} + 4D = 5\\\\\frac{6-15D+8D}{2} = 5\\\\6-7D = (5)(2)\\7D = 6-10\\\\D = -\frac{4}{7}[/tex]

D = - 0.57

Put this value in equation (3), we get:

[tex]C = \frac{2-(5)(\frac{-4}{7} )}{2}\\\\C = \frac{\frac{14+20}{7}}{2}\\\\C = \frac{34}{(7)(2)}\\\\C = \frac{17}{7}\\[/tex]

C = 2.43

Answer:

The magnitude of the average acceleration of the ball during this time interval is 1.238 x 10⁴ m/s².

Explanation:

Given;

mass of the super ball, m = 50 g = 0.05 kg

initial velocity of the ball, u = 29.5 m/s

final velocity of the ball, v = -20.0 m/s (negative because it rebounds)

time of contact of the ball and the wall, t = 4 ms = 4 x 10⁻³ s

The force exerted on the brick wall by the ball is given as;

[tex]F = ma\\\\ma = \frac{m(v-u)}{t} \\\\a = \frac{v-u}{t} \\\\a = \frac{(-20) - 29.5}{4.0 \ \times \ 10^{-3}} \\\\a = \frac{-49.5}{4.0 \ \times \ 10^{-3}} \\\\a = -1.238 \times 10^4 \ m/s^2\\\\|a| = 1.238 \times 10^4 \ m/s^2[/tex]

Therefore, the magnitude of the average acceleration of the ball during this time interval is 1.238 x 10⁴ m/s².

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:

Explanation:

From the question we are told that

Work done [tex]W=2820[/tex]

a)Generally the heat flow for an adiabatic process is 0 (zero)

[tex]Q= U + W =>0[/tex]

[tex]Q=0[/tex]

b)Generally Change in internal energy of gas is mathematically given by

Since [tex]W=-2820J[/tex]

Therefore

[tex]U=2820[/tex]

Giving

[tex]Q= 2820 -2820[/tex]

[tex]Q=O[/tex]

c)With increases in internal energy brings increase in temperature

Therefore

Energy increases

Answer:

the speed of electron is 5.021 x 10⁶ m/s

the speed of proton is 2733.91 m/s

Explanation:

Given;

magnitude of electric field, E = 554 N/C

charge of the particles, Q = 1.6 x 10⁻¹⁹ C

time of motion, t = 51.6 ns = 51.6 x 10⁻⁹ s

The force experienced by each particle is calculated as;

F = EQ

F = (554)(1.6 x 10⁻¹⁹)

F = 8.864 x 10⁻¹⁷ N

The speed of the particles are calculated as;

[tex]F = ma\\\\F = \frac{mv}{t} \\\\v = \frac{Ft}{m} \\\\v_e = \frac{Ft}{m_e}\\\\v_e = \frac{(8.864 \times 10^{-17})(51.6\times 10^{-9})}{9.11 \times \ 10^{-31}}\\\\v_e = 5.021 \times 10^{6} \ m/s[/tex]

[tex]v_p = \frac{Ft}{m_p}\\\\v_p = \frac{(8.864 \times 10^{-17})(51.6\times 10^{-9})}{1.673 \times \ 10^{-27}}\\\\v_p = 2733.91 \ m/s[/tex]

Assuming the particle is in free fall once it is shot up, its vertical velocity v at time t is

v = 30 m/s - g t

where g = 9.8 m/s² is the magnitude of the acceleration due to gravity, and its height y is given by

y = (30 m/s) t - 1/2 g t ²

(a) At its maximum height, the particle has 0 velocity, which occurs for

0 = 30 m/s - g t

t = (30 m/s) / g ≈ 3.06 s

at which point the particle's maximum height would be

y = (30 m/s) (3.06 s) - 1/2 g (3.06 s)² ≈ 45.9184 m ≈ 46 m

(b) It takes twice the time found in part (a) to return to 0 height, t ≈ 6.1 s.

(c) The particle falls 35 m below its starting point when

-35 m = (30 m/s) t - 1/2 g t ²

Solve for t to get a time of about t ≈ 7.1 s

Answer:

the inertia of the crate is (49.67 kg)r²

Explanation:

Given the data in the question;

First; we will use the law of conservation of momentum to determine the mass of the crate;

m₁v₁ - m₂v₂ = 0

given that; m₁ = 0.60 kg and v₂ = 0.058 m/s

we substitute

0.60 × v₁ = m₂ × 0.058 = 0

m₂ = 0.60v₁ / 0.058 ----------- EQU 1

Next, we use the energy conservation relation to find the velocity

According to conservation of energy;

1/2m₁v₁² + 1/2m₂v₂² = 7 J

we substitute

1/2×0.60×v₁² + 1/2×m₂×(0.058)² = 7 J

0.3v₁² + 0.001682m₂ = 7 J ----- EQU 2

substitute value of m₂ form equ 1 into equ 2

0.3v₁² + 0.001682(0.60v₁ / 0.058) = 7 J

0.3v₁² + 0.0174v₁ = 7 J

0.3v₁² + 0.0174v₁ - 7 J = 0

we solve the quadratic equation;

{  x =  [-b±√( b² - 4ac)] / 2a   }

v₁  =  [-0.0174 ±√( 0.0174² - 4×0.3×-7)] / 2×0.3

=  [-0.0174 ±√(8.4003)] / 0.6

= [-0.0174 ± 2.8983 ] / 0.6  

= -4.8595 or 4.8015     but{ v₁ ≠ - }

so v₁ = 4.8015 m/s ≈ 4.802 m/s

next we input value of  v₁ into equation 1

m₂ = (0.60×4.8015) / 0.058

m₂ =  2.8809 / 0.058

m₂ =  49.67 kg

So, the moment of inertia of the crate will be;

I₂ = m₂r²

we substitute value of m₂

I₂ = (49.67 kg)r²

Therefore, the inertia of the crate is (49.67 kg)r²

Answer:

difficulty concentrating

I believe the answer would be protentional because they have the potential energy in them to lift the hammer.

Answer:

To measure work, you must multiply the force by the distance through which it acts.

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:

yes

Explanation:

yes

According to question this is a riddle and the doctor was the boy's mother so she could not operate on him.

A statement, question, or phrase that is presented as a problem to be solved and has a dual or disguised meaning is called a riddle. Enigmas, which are difficulties typically presented in metaphoric or allegorical language that call for inventiveness and careful thought to solve, and conundra, which are problems that rely on puns either in the question or the answer, are two different forms of riddles.

Across many nations and even entire continents, many riddles take on a similar format. Riddles might be borrowed close to home as well as long distances. A man and his son were rock climbing on a particularly dangerous mountain when they slipped and fell. the man was killed, but the son lived and was rushed to a hospital.

Therefore, According to question this is a riddle and the doctor was the boy's mother so she could not operate on him.

Learn more about riddle here:

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Answer:

Explanation:

75 N and 90 N are acting in opposite direction so net force = 90 - 75 = 15 N .

Friction force will act in the direction opposite to the direction of net force .

So friction force will act in the direction in which 75 N is acting .

Total force acting in the direction of 75 =  75 + 12 = 87 N

Net force acing on the third child = 90 - 87 = 3 N  

Its direction will be that in the direction of 90 N .

Answer:

I think the answers March 21

Answer:

Once a year, mean solar time and sidereal time will be the same.

Answer:

Follows are the solution to this question:

Explanation:

Please find the correct question in the attachment file.

Let:

[tex]\overrightarrow{R}= R_i\hat{i}+R_j\hat{j}+R_k\hat{k}\\\\\overrightarrow{S}= S_i\hat{i}+S_j\hat{j}+S_k\hat{k}\\\\[/tex]

Calculating the value of  [tex]\overrightarrow{R} \times \overrightarrow{S}:[/tex]

[tex]\to \left | \begin{array}{ccc}\hat{i}&\hat{j}&\hat{K}\\R_i&R_j&R_k\\S_i&S_j&S_k\end{array}\right | = \hat{i}[R_j S_k-S_jR_k]-\hat{j}[R_i S_k-S_iR_k]+\hat{k}[R_i S_j-S_iR_j][/tex]

Calculating the value of [tex]\overrightarrow{R} \cdot (\overrightarrow{R} \times \overrightarrow{S}):[/tex]

[tex]\to (R_i\hat{i}+R_j\hat{j}+R_k\hat{k}) \cdot ( \hat{i}[R_j S_k-S_jR_k]-\hat{j}[R_i S_k-S_iR_k]+\hat{k}[R_i S_j-S_iR_j])[/tex]

by solving this value it is equal to 0.

Answer:

Explanation:

Magnetic field due to current at a distance of 4.4 cm

B = 10⁻⁷ x 2 x 5.2 / 4.4 x 10⁻²           [ B = 10⁻⁷ x 2i / r = ]

= 2.36 x 10⁻⁵ T.

Force on moving electron = Bqv , B is magnetic field , q is charge and v is velocity of charge .

Force = 2.36 x 10⁻⁵  x 1.6 x 10⁻¹⁹ x 6.2 x 10⁴

= 23.41 x 10⁻²⁰ N .

This force will be perpendicular to the direction of current .

Answer:

x = 1.04866

Explanation:

Force can be defined from power energy by the expressions

          F = [tex]- \frac{ dU}{ dx}[/tex]

in this case we are the expression of the potential energy

          U = [tex]\frac{2.6}{x^{8} } - \frac{4.3}{ x^{4} }[/tex]

let's find the derivative

         dU / dx = 2.6 ( [tex]\frac{-8}{x^{9} }[/tex]) - 4.3 ([tex]\frac{-4}{ x^{5} }[/tex])

         dU / dx = [tex]- \frac{20.8}{ x^{9} } + \frac{17.2 }{ x^{5} }[/tex]

we substitute

          F = + \frac{20.8}{ x^{9} }  - \frac{17.2 }{ x^{5} }

at the equilibrium point the force is zero, so

           [tex]\frac{20.8}{ x^{9} } = \frac{17.2}{ x^{5} }[/tex]

           20.8 / 17.2 = x⁴

            x⁴ = 1.2093

             x = [tex]\sqrt[4]{ 1.2093}[/tex]

             x = 1.04866