A plane has a mass of 360,000 kg takes-off at a speed of 300 km/hr. i) What should be the minimum acceleration to take off if the length of the runway is 2.00 km.ii) At this acceleration, how much time would the plane need from starting to takeoff. iii) What force must the engines exert to attain this acceleration

Answers

Answer 1

Answer:

i) the minimum acceleration to take off is 22500 km/h²

ii) the required time needed by the plane from starting to takeoff is 0.0133 hrs

iii) required force that the engine must exert to attain acceleration is 625 kN

Explanation:

Given the data in the question;

mass of plane m = 360,000 kg

take of speed v = 300 km/hr = 83.33 m/s

i)

What should be the minimum acceleration to take off if the length of the runway is 2.00 km

from Newton's equation of motion;

v² = u² + 2as

we know that a plane starts from rest, so; u = 0

given that distance S = 2 km

we substitute

(300)² = 0² + ( 2 × a × 2 )

90000 = 4 × a

a = 90000 / 4

a = 22500 km/h²

Therefore,  the minimum acceleration to take off is 22500 km/h²

ii) At this acceleration, how much time would the plane need from starting to takeoff.

from Newton's equation of motion;

v = u + at

we substitute

300 = 0 + 22500 × t

t = 300 / 22500

t = 0.0133 hrs

Therefore, the required time needed by the plane from starting to takeoff is 0.0133 hrs

iii) What force must the engines exert to attain this acceleration

we know that;

F = ma

acceleration a = 22500 km/hr² = 1.736 m/s²

so we substitute

F = 360,000 kg × 1.736 m/s²

F =  624960 N

F = 625 kN

Therefore, required force that the engine must exert to attain acceleration is 625 kN


Related Questions

A spring with a 10-kg mass and a damping constant 15 can be held stretched 2 meters beyond its natural length by a force of 6 newtons. Suppose the spring is stretched 4 meters beyond its natural length and then released with zero velocity. Suppose the spring is stretched 4 meters beyond its natural length and then released with zero velocity.

Required:
Find the position of the mass at any time t.

Answers

Answer:

Explanation:

Given that:

mass = 10kg

damping constant C = 15 kg/s

length = 2 m

Force F = 6N

Using the Hooke's law:

F = kx

6 = 15x

k = 6 N /2 m

spring constant k = 3 N/m

For the critical damping

C² - 4k*m= 0

m = C²/4k

m = (15)²/4(3) kg

m = 225/12 kg

m = 18.75 kg

The peak value of the electric field component of an electromagnetic wave is E. At a particular instant, the intensity of the wave is of 0.020 W/m2. If the electric field were increased to 5E, what would be the intensity of the wave?

Answers

Answer:

[tex]I_2=0.50 w/m^2[/tex]

Explanation:

From the question we are told that:

initial Intensity [tex]I_1=0.020 w/m^2[/tex]

Final Electric field [tex]E_2=5E[/tex]

Generally the equation for Relation ship between intensity and Electric field is mathematically given by

 [tex]\frac{I_1}{I_2}= \frac{E_1^2}{E_2^2}[/tex]

Therefore

 [tex]I_2=\frac{I_1}{ \frac{E_1^2}{E_2^2}}[/tex]

 [tex]I_2=\frac{0.020}{ \frac{E^2}{5E^2}}[/tex]

 [tex]I_2=0.50 w/m^2[/tex]

URGENT

The component of solid-state physics that works with and/or heats metals and alloys to give them certain desired
shapes or properties is..

Answers

Answer:

the is metallurgy .....

The steps to determine the sum are shown. (6.74x104)+(8.95 x 104) Step 1. Rearrange the expression: (6.74+8.95) 104 Step 2. Add the coefficients: (15.69) 104 Step 3. Write in scientific notation: 1.569x 10 What is the value of k in Step 3? =​

Answers

Answer:

We want to solve the sum:

6.74*10⁴ + 8.95*10⁴

first, we take the common factor 10⁴ out, so we get:

(6.74 + 8.95)*10⁴

Now we solve the sum:

(15.66)*10⁴

Now we want to rewrite it in exponential form, wo we can rewrite it as:

(15.66)*10⁴ = (1.566*10)*10⁴ = (1.566)*10*10⁴ = (1.566)*10⁴⁺¹ = 1.566*10⁵

k = 5.

A certain light truck can go around a flat curve having a radius of 150m with a max speed if 26.5. What is the max speed it can go around a curve having a radius of 76.0m

Answers

Answer:

The maximum speed is 18.86 m/s.

Explanation:

initial  radius, r = 150 m

maximum speed, v = 26.5 m/s

new radius, r' = 76 m

Let the new maximum speed is v'.

The formula of the maximum speed is

[tex]tan\theta = \frac{v^2}{rg}[/tex]

So,

[tex]\frac{v'^2}{v^2}=\frac{r'}{r}\\\\\frac{v'^2}{26.5\times 26.5}=\frac{76}{150}\\\\v=18.86 m/s[/tex]

what advantage does hovercraft have over a boar or a road vehicle?​

Answers

Answer:

The advantages of Hovercraft:

They can travel over almost any non-porous surface.

They can operate to and from any unprepared beach or slipway.

They take fast, direct routes compared to a conventional marine vessel.

According to Coulomb's law, rank the interactions between charged particles from highest potential energy to lowest potential energy.

a. 1+ charge and 1- charge seperated by 200pm
b. 1+ charge and 1+ charge seperated by 100pm
c. 1+ charge and 1- charge seperated by 100pm
d. 2+ charge and 1- charge seperated by 100pm

Answers

According to Coulomb's law, rank the interactions between charged particles from highest potential energy to lowest potential energy.

Highest potential energy to lowest potential energy.

b. 1+ charge and 1+ charge seperated by 100 pm

a. 1+ charge and 1- charge seperated by 200 pm

c. 1+ charge and 1- charge seperated by 100 pm

d. 2+ charge and 1- charge seperated by 100 pm

[tex]\circ \: \: { \underline{ \boxed{ \sf{ \color{green}{Happy\:learning.}}}}}∘[/tex]

Some radar systems detect the size and shape of objects such as aircraft and geological terrain. What is the frequency of such a system which can detect objects as small as 19.1 cm?

Answers

Answer:

[tex]f=1.57\times 10^9\ Hz[/tex]

Explanation:

Given that,

A system can detect objects as small as 19.1 cm i.e. 0.191 m. It is the wavelength.

We know that,

Frequency, [tex]f=\dfrac{c}{\lambda}[/tex]

So,

[tex]f=\dfrac{3\times 10^8}{0.191}\\\\=1.57\times 10^9\ Hz[/tex]

So, the frequency of such a system is equal to[tex]1.57\times 10^9\ Hz[/tex].

Lab: Energy Transfer Instructions Click the links to open the resources below. These resources will help you complete the assignment. Once you have created your file(s) and are ready to upload your assignment, click the Add Files button below and select each file from your desktop or network folder. Upload each file separately. Your work will not be submitted to your teacher until you click Submit. Documents Descriptive Lab Report Guide Descriptive Lab Report Rubric

Answers

Answer:

The second one says cant open file............

A 3.7-kg object is acted on by two forces. One of the forces is 11 N acting toward the
east. What is the other force if the acceleration of the object is 1.0 m/s2 toward the east?

Answers

Answer:

F₂ = -7.3 N

Explanation:

Given that,

The mass of an object, m₁ = 3.7 kg

First force, F₁ = 11 N

The net acceleration of the object is 1 m/s².

We know that,

F₁+F₂ = ma

11+F₂ = (3.7)(1)

F₂ = 3.7-11

F₂ = -7.3 N

so, the other force is 7.3 N and it is acting in west direction.

Kinetic energy is the energy an object has due to its
die hele

Answers

Answer:

There are a couple of interesting things about kinetic energy that we can see from the equation.

Kinetic energy depends on the velocity of the object squared. This means that when the velocity of an object doubles, its kinetic energy quadruples. A car traveling at 60 mph has four times the kinetic energy of an identical car traveling at 30 mph, and hence the potential for four times more death and destruction in the event of a crash.

Kinetic energy must always be either zero or a positive value. While velocity can have a positive or negative value, velocity squared is always positive.

Kinetic energy is not a vector. So a tennis ball thrown to the right with a velocity of 5 m/s, has the exact same kinetic energy as a tennis ball thrown down with a velocity of 5 m

: Ánh nắng mặt trời có cường độ đồng đều với bước sóng nằm trong vùng khả khiến 430nm-690nm đến đập vuông góc với một bản mỏng nước có bề dày 320nm, chiết suất 1,33 lơ lửng trong không khí. Tìm bước sóng thích hợp để ánh sáng phản từ bản mỏng là sáng nhất đối với người quan sát

Answers

Sorry I don’t know this language

A 1.2-kg mass suspended from a spring of spring constant 22 N.m-1 executes simple harmonic motion of amplitude 5 cm. Assuming that the mass is at the equilibrium posiiton at t = 0, what is its displacement at t = 1.0 s?

Answers

Answer:

[tex]d =3.7*10^{-3} m[/tex]

Explanation:

From the question we are told that:

Mass [tex]m=1.2kg[/tex]

Spring constant [tex]\mu=22Nm^{-1}[/tex]

Amplitude [tex]A=5cm=0.05m[/tex]

Generally the equation for displacement d is mathematically given by

 [tex]d = Asin(\omega t)[/tex]

Where

 [tex]\omega=angular\ velocity[/tex]

 [tex]\omega=\sqrt{k/m}[/tex]

 [tex]\omega=\sqrt{22/1.2}[/tex]

 [tex]\omega=4.2817rads^{-1}[/tex]

Therefore  

 [tex]d = 0.05*sin(4.2817*1)[/tex]  

 [tex]d =3.7*10^{-3} m[/tex]

How does gravity affect your ability to live on a planet?

Answers

If we didn’t have it, we might just float into space. Or it would be hard to live with everything floating around
If it were too light, we would float off, too heavy, and then our bones would have to be extremely dense lest we want to get crushed by just existing

Effects of global warming is

A-decrease in temperature
B-melting of polar ice caps
C-breathing problems

Answers

Answer:

B- the melting of polar ice caps

Explanation:

As the world's temperature increases, polar ice caps will no longer be able to remain solid.

An LC circuit is built with a 20 mH inductor and an 8.0 PF capacitor. The capacitor voltage has its maximum value of 25 V at t = 0 s.(a)How much time does it take until the capacitor is fully discharged for the first time? (b)What is the inductor current at that time?

Answers

Answer:

a) the required time is 0.6283 μs

b) the inductor current is 0.5 mA

Explanation:

Given the data in the question;

The capacitor voltage has its maximum value of 25 V at t = 0

i.e V[tex]_m[/tex] = V₀ = 25 V

we determine the angular velocity;

ω = 1 / √( LC )

ω = 1 / √( ( 20 × 10⁻³ H ) × ( 8.0 × 10⁻¹² F) )

ω = 1 / √( 1.6 × 10⁻¹³  )

ω = 1 / 0.0000004

ω = 2.5 × 10⁶ s⁻¹

a) How much time does it take until the capacitor is fully discharged for the first time?

V[tex]_m[/tex] =  V₀sin( ωt )

we substitute

25V =  25V × sin( 2.5 × 10⁶ s⁻¹ × t )

25V =  25V × sin( 2.5 × 10⁶ s⁻¹ × t )

divide both sides by 25 V

sin( 2.5 × 10⁶ × t ) = 1

( 2.5 × 10⁶ × t ) = π/2

t = 1.570796 / (2.5 × 10⁶)

t = 0.6283 × 10⁻⁶ s

t = 0.6283 μs

Therefore, the required time is 0.6283 μs

b) What is the inductor current at that time?

[tex]I[/tex](t) = V₀√(C/L) sin(ωt)

{ sin(ωt) = 1 )

[tex]I[/tex](t) = V₀√(C/L)

we substitute

[tex]I[/tex](t) = 25V × √( ( 8.0 × 10⁻¹² F ) / ( 20 × 10⁻³ H ) )

[tex]I[/tex](t) = 25 × 0.00002

[tex]I[/tex](t) = 0.0005 A

[tex]I[/tex](t) = 0.5 mA

Therefore, the inductor current is 0.5 mA

The time taken for the capacitor to fully discharge is 6.28 x 10⁻⁷ s.

The current in the inductor at the given time is 0.0005 A.

Angular velocity of circuit

The angular velocity of the circuit is calculated as follows;

[tex]\omega = \frac{1}{\sqrt{LC} } \\\\\omega = \frac{1}{\sqrt{20 \times 10^{-3} \times 8 \times 10^{-12} } } \\\\\omega = 2.5 \times 10^6 \ rad/s[/tex]

Time for the capacitor to fully discharge

V = V₀(sinωt)

sinωt = V/V₀

sinωt = = 25/25

sin(ωt) = 1

ωt = sin⁻¹ (1)

ωt = π/2

2.5 x 10⁶ t = π/2

t = 1.57 / (2.5 × 10⁶)

t = 6.28 x 10⁻⁷ s

Inductor current at the given time

The current in the inductor at the given time is calculated as follows

[tex]I(t) = V_0 \sqrt{\frac{C}{L} } \\\\I = 25 \times \sqrt{\frac{8\times 10^{-12}}{20 \times 10^{-3}} } \\\\I = 0.0005 \ A[/tex]

Learn more about inductor current here: https://brainly.com/question/4425414

Why can’t we see the molecules moving in a solid or liquid?

Answers

Answer:

Because molecules are to small for us people to see with the bare eye, unless you use a telescope.

Explanation:

Explanation:

Because in solid form, the molecules stood in place, not moving. In liquid state, the molecules move slowly but not as fast as air molecules.

If you exert a force of 5 N into a nutcracker, and it outputs a force of 20 N, what is the mechanical advantage of the nutcracker. Show formula PLSSS HELPPPP!!! i'll make you brainliest

Answers

Answer: 4

Explanation:

MA = output force / input force

MA = 20 / 5

MA = 4

Hope this helps.  Please mark brainliest.

sanaysay tungkol sa pangangalaga ng mga endangered animals​

Answers

What details do you need in this essay exactly?

Friction is a (1)______
that (2)______
motion. It acts in a direction (3)______
to the motion of
the object. There is (4)______
exerted on an object that
moves on a (5)_____
than on a (6)_____
Friction opposes the motion of an object as it moves across a
surface. An object slows down (7)______
on a rough
surface than on a smooth surface.​

Answers

Answer:

friction is motion heat that collect hot vibration

Hunter is studying a new form of plastic chip that can be added onto a credit card, linking it to a bank account
separate from the main card. This chip works by passing an electric current through a sensing coil. What application
of solid-state physics is Hunter most likely working with?
O electromagnetism
O metallurgy
O quantum mechanics
O crystallography

Answers

Answer:

electromagnetism....

The charge that passes a cross-sectional area A=10-4 m2 varies with time according
to the relation
Q=4 + 2t + t2, where Q is in coulombs and t is in seconds. (a) Find the relation
that gives the instantaneous current at time t =2 s is​

Answers

Answer:

Current = dQ/dt

or I = dQ/dt

Where I represents current.

Which is the rate of flow of charge.

Q=4 + 2t + t²

dQ/dt = 2 + 2t --- This is the relation that gives the instantaneous current.

At time t=2sec

dQ/dt = I = 2 + 2t

= 2 + 2(2)

=2 + 4

= 6A.

Consider the following possibilities and select the correct choice.
1. Tx Ty > Tz
2. Tx Ty < Tz
3. Tx Ty = Tz

Answers

Answer:

Tx not but mybe

Explanation:

for that reason its just trying to help

When a condenser discharges electricity, the instantaneous rate of change of the voltage is proportional to the voltage in the condenser. Suppose you have a discharging condenser and the instantaneous rate of change of the voltage is -0.01 of the voltage (in volts per second). How many seconds does it take for the voltage to decrease by 90 %?

Answers

Answer:

460.52 s

Explanation:

Since the instantaneous rate of change of the voltage is proportional to the voltage in the condenser, we have that

dV/dt ∝ V

dV/dt = kV

separating the variables, we have

dV/V = kdt

integrating both sides, we have

∫dV/V = ∫kdt

㏑(V/V₀) = kt

V/V₀ = [tex]e^{kt}[/tex]

Since the instantaneous rate of change of the voltage is -0.01 of the voltage dV/dt = -0.01V

Since dV/dt = kV

-0.01V = kV

k = -0.01

So, V/V₀ = [tex]e^{-0.01t}[/tex]

V = V₀[tex]e^{-0.01t}[/tex]

Given that the voltage decreases by 90 %, we have that the remaining voltage (100 % - 90%)V₀ = 10%V₀ = 0.1V₀

So, V = 0.1V₀

Thus

V = V₀[tex]e^{-0.01t}[/tex]

0.1V₀ = V₀[tex]e^{-0.01t}[/tex]

0.1V₀/V₀ = [tex]e^{-0.01t}[/tex]

0.1 = [tex]e^{-0.01t}[/tex]

to find the time, t it takes the voltage to decrease by 90%, we taking natural logarithm of both sides, we have

㏑(0.01) = -0.01t

So, t = ㏑(0.01)/-0.01

t = -4.6052/-0.01

t = 460.52 s

Which of the following items is an example of a symbol Bernice bobs her hair

Answers

what? Is there a picture you can show us with this question

if a glass rod rubbed with Silk piece is taken to a ball pen rubeed with wool the ball pen_____​

Answers

Answer:

The glass rod remains charged due to electrification by the silk

To calculate the final enthalpy of the overall chemical equation, which step must occur?

Answers

Answer:

Explanation:

Reverse the second equation, and change the sign of the enthalpy

g:p:e = gravitational field strength x height

gravitational field strength = 9.8N/kg

Answers

Answer:

kdfihiyfxnlL chapter kar kaam

Explanation:

off influx teenon Palin

Give an example of a physical entity that is quantized. State specifically what the entity is and what the limits are on its values.

Answers

Answer:

A charge is a physical entity that has been quantized. The limits on its values are the value of a charged particle quantized in the state where 'n'...

Explanation:

One example of a physical entity that is quantized is:

The amount of money in your pocket.

The amount can't have any fraction of 1 cent.  

Its value must be an integer-multiple of cents, or 0.01 dollar.    

When it increases or decreases, it jumps from one integer number of cents to the next integer number.  It doesn't "slide" from one to the next.  It can never have a value between two integer numbers of cents.

In June 1985, a laser beam was sent out from the Air Force Optical Station on Maui and reflected back from the shuttle Discovery as it passed by 354 km overhead. The diameter of the central maximum of the beam at the shuttle position was 500 nm. What is the effective diameter of the circular laser aperture at the Maui ground station

Answers

This question is incomplete, the complete question is;

In June 1985, a laser beam was sent out from the Air Force Optical Station on Maui and reflected back from the shuttle Discovery as it passed by 354 km overhead. The diameter of the central maximum of the beam at the shuttle position was said to be 9.1 m, and the beam wavelength was 500 nm.

What is the effective diameter of the circular laser aperture at the Maui ground station

Answer:

the effective diameter of the circular laser aperture at the Maui ground station is 4.747 cm

Explanation:

Given the data in the question;

Separation between observer and point L = 354 km = 354000 m

Linear separation D = 9.1 m

wavelength λ = 500 nm = 500 × 10⁻⁹ m

Now, for small angles;

θ = D / L

θ = 9.1 m /  354000 m

θ = 2.57 × 10⁻⁵ rad

For a circular aperture;

sinθ = ( 1.22 × λ ) / d

for small angles;

θ = ( 1.22 × λ ) / d

so

θ = 2 × θ

θ = 2 × [( 1.22 × λ ) / d]

we substitute

2.57 × 10⁻⁵ = 2 × [( 1.22 × 500 × 10⁻⁹ ) / d]

2.57 × 10⁻⁵ = 0.00000122 / d

d = 0.00000122 / 2.57 × 10⁻⁵

d = 0.04747 m

d = ( 0.04747 × 100 )m

d = 4.747 cm

Therefore, the effective diameter of the circular laser aperture at the Maui ground station is 4.747 cm

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