an Empty airplane has a mass of 30400 kg and a takeoff acceleration of 1.20 metre per second square what is the net force accelerating the plane​

Answer:

a) The distance travelled by the the locomotive is 30 meters, b) The final displacement of the locomotive is 6 meters westwards.

Explanation:

a) The distance travelled is the sum of magnitudes of distances covered by the train during its motion. That is to say:

[tex]s_{T} = 18\,m + 12\,m[/tex]

[tex]s_{T} = 30\,m[/tex]

The distance travelled by the the locomotive is 30 meters.

b) The displacement is the vectorial distance of the train with respect to a point of reference, since west and east are antiparallel to each other, calculations can be simplified to a scalar form. Let suppose that movement to the east is positive. The calculations are presented below:

[tex]s_{P} = -18\,m +12\,m[/tex]

[tex]s_{P} = -6\,m[/tex]

The final displacement of the locomotive is 6 meters westwards.

Answer:

the dispalcement is 6m and distance travelled is 30m

Explanation:

Answer:

A police car with its siren on is driving towards you, and you perceive the pitch of the siren to increase.

Explanation:

In Physics, Doppler effect can be defined as the change in frequency of a wave with respect to an observer in motion and moving relative to the source of the wave.

Simply stated, Doppler effect is the change in wave frequency as a result of the relative motion existing between a wave source and its observer.

The term "Doppler effect" was named after an Austrian mathematician and physicist known as Christian Johann Doppler while studying the starlight in relation to the movement of stars.

The phenomenon of Doppler effects is generally applicable to both sound and light.

An example of the Doppler effect is a police car with its siren on is driving towards you, and you perceive the pitch of the siren to increase. This is so because when a sound object moves towards you, its sound waves frequency increases, thereby causing a higher pitch. However, if the sound object is moving away from the observer, it's sound waves frequency decreases and thus resulting in a lower pitch.

Other fields were the Doppler effects are applied are; astronomy, flow management, vibration measurement, radars, satellite communications etc.

Answer:

A police car with its siren on is driving towards you, and you perceive the pitch of the siren to increase.

Explanation:

This should be the correct answer.

Answer:

El avión recorrió 45 km en los 180 s.

Explanation:

La relación entre velocidad, distancia y tiempo se da de la siguiente manera;

[tex]Velocidad= \dfrac{Distancia}{Hora}[/tex]

Por lo cual los parámetros dados son los siguientes;

Velocidad = 900 km/h = 250 m / s

Tiempo = 180 s

Estamos obligados a calcular la distancia recorrida

De la ecuación para la velocidad dada arriba, tenemos;

Distancia recorrida = Velocidad pf viaje × Tiempo de viaje

Distancia recorrida = 900 km/h × 180 s = 900

Distancia recorrida = 900 km/h × 1 h/60 min × 1 min/60 s × 180 s = 45 km

Por lo tanto, el avión viajó 45 km en 180 s.

Answer:

La fuerza promedio que ejerce la resistencia del aire sobre el planeador cuando ingresa con un ángulo de planeo constante de 12 ° con respecto a la Tierra es 9613.8 N

Explanation:

Los parámetros dados son;

Masa del planeador = 980 kg.

Velocidad de lanzamiento horizontal = 480 km / h

Altitud de lanzamiento = 3500 m.

Velocidad de aterrizaje final = 210 km / h

Ángulo de deslizamiento, θ = 12 °

La ecuación para el equilibrio de fuerzas.

L · cos (θ) + D · sin (θ) = W

L · sin (θ) + D · cos (θ)

Dónde:

L = fuerza de elevación

D = fuerza de arrastre

W = Peso del "planeador"

W = Masa del "planeador" × Aceleración debido a la gravedad = 980 kg × 9.81 m / s²

W = 9613.8 N

Por lo tanto, tenemos;

L · cos (12) + D · sin (12) = 9613.8 ,,,,,,,,,,,,, (1)

L · sin (12) = D · cos (12) ,,,,,,,,,,,,, (2)

De la ecuación (2), tenemos;

L = D · (cos (12)/sin (12))

Sustituyendo el valor de L = D · (cos (12) / sin (12)) en la ecuación (1), se obtiene;

D · (cos (12)/sin (12)) · cos (12) + D · sin (12) = 9613.8

D = 9613.8/((cos (12)/sin (12)) · cos (12) + sin (12)) = 1998.82 N

L = D · (cos (12)/sin (12)) = 1998.82 · (cos (12)/sin (12)) = 9403.72 N

Por lo tanto, la fuerza da la fuerza promedio que la resistencia del aire ejerce sobre el planeador cuando ingresa con un ángulo de planeo constante de 12 ° con respecto a la Tierra;

F = √ (L² + D²) = √ (9403.72² + 1998.82²) = 9613.8 N.

Answer:

4.24 km/h

Explanation:

When the man walks at a rate of 3 km/h, the rain appears to fall vertically, which means it has the same horizontal rate, 3 km/h.

When the man walks at a rate of 6 km/h, the rain's horizontal velocity relative to the man is now 6 km/h − 3 km/h = 3 km/h.  The rain meets the man at a 45 degree angle, so the vertical velocity is also 3 km/h.

Therefore, the total speed is:

v = √((3 km/h)² + (3 km/h)²)

v = 4.24 km/h

Answer:

Explanation:

Kinetic energy is 1/2mv²

where m is the mass of the object

v is the velocity

From the question

m = 200kg

v = 2.7m/s

Kinetic energy is

1/2 × 200 × 2.7²

= 729 Joules

Hope this helps you

Answer:

(1) The orbits are ellipses, with focal points ƒ1 and ƒ2 for the first planet and ƒ1 and ƒ3 for the second planet. The Sun is placed in focal point ƒ1.

(2) The two shaded sectors A1 and A2 have the same surface area and the time for planet 1 to cover segment A1 is equal to the time to cover segment A2.

(3) The total orbit times for planet 1 and planet 2 have a ratio a13/2 : a23/2

Answer:

g the day off and I will be there at work and I will be there at work and I will

Explanation:

you are not going to be able to make it to the meeting tonight but I can tomorrow

Answer:

By nuclear fission

Explanation:

The sun generates enormous energy through the process of nuclear fusion.

The core or the innermost part of the sun is characterized by high temperature and pressure. These two factors cause the separation of nuclei from electrons and the fusion of hydrogen nuclei to form a  helium atom.

During the fusion process, energy is released.

Explanation:

[tex]a = \frac{change \: in \: velocity}{change \: in \: time} [/tex]

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

[tex]v = 7 {ms}^{ - 1} [/tex]

[tex]u = 15 {ms}^{ - 1} [/tex]

[tex]t = 3 \: seconds[/tex]

[tex]a = \frac{7 - 15}{3} [/tex]

[tex]a = \frac{ - 8}{3} [/tex]

[tex]a = - 2.67 \: {ms}^{ - 2} [/tex]

Answer:

The lightbulbs were exploding after several mins of use

Explanation:

Answer:

the lightbulbs were exploding after several minutes of use

Explanation:

a p e x :))

Answer:

Medium B is denser than medium C

When light travels from B to C light bends away from the normal .

Refractive index of C with respect to B is 8/9

Ratio of speed of light in medium B to the speed of light in medium A = 4/3

Explanation:

Refractive index of medium A /  Refractive index of medium B =  4/3

Refractive index of medium A /  Refractive index of medium C = 3/2  

Dividing ,

Refractive index of medium B /  Refractive index of medium C

= 3/2  x  3/4

= 9 / 8

Refractive index of medium B  >   Refractive index of medium C

Refractive index of C with respect to B is 8/9

 speed of light in medium B /  the speed of light in medium A

= refractive index of A / refractive index of B = 4 / 3

Answer: The image is at a distance of 24 cm on the right of the lens, it will be an inverted, real image and has a height of 7 cm.

Explanation: To determine the position of the image of the object, use the Lens Equation, which is described by:

[tex]\frac{1}{f} = \frac{1}{u} +\frac{1}{v}[/tex]

where

f is focal length;

u is distance of the object from the lens;

v is distance of the image of the pbject from the lens;

Calculating the position of the image will be:

[tex]\frac{1}{v} =\frac{1}{f} - \frac{1}{u}[/tex]

[tex]\frac{1}{v} = \frac{1}{8} - \frac{1}{12}[/tex]

[tex]\frac{1}{v} = \frac{1}{24}[/tex]

v = 24 cm

Note that the distance of the image from the lens is positive. This means that the formed image is positioned at the right of the lens and is inverted.

For the height of the image, it can be used the Magnification (M) of an image:

M = [tex]\frac{h_{i}}{h_{o}} = \frac{v}{u}[/tex]

where:

[tex]h_{i}[/tex] is the height of the image

[tex]h_{o}[/tex] is the height of the object

For this image:

[tex]\frac{h_{i}}{3.5} = \frac{24}{12}[/tex]

[tex]h_{i} =[/tex] 2 * 3.5

[tex]h_{i}[/tex] = 7 cm

The image formed for this lens is 24 cm on the right of the lens, inverted related to the object and has height of 7 cm.

Answer:

9.74 m/s²

Explanation:

g = GM / r²

g = (6.67×10⁻¹¹ m³/kg/s²) (6×10²⁴ kg) / (6.4×10⁶ m + 8848 m)²

g = 9.74 m/s²

Answer:

1120 N

Explanation:

The velocity with which he hits the water can be found with kinematics:

v² = v₀² + 2aΔy

v² = (0 m/s)² + 2 (-9.8 m/s²) (-9.00 m)

v = -13.3 m/s

Or it can be found with conservation of energy.

PE = KE

mgh = ½ mv²

v = √(2gh)

v = √(2 × -9.8 m/s² × -9.00 m)

v = -13.3 m/s

Sum of forces on the diver after he hits the water:

∑F = ma

F − mg = m Δv/Δt

F − (74.0 kg) (9.8 m/s²) = (74.0 kg) (0 m/s − (-13.3 m/s)) / (2.50 s)

F = 1120 N

Answer: B. Carbon dioxide and water

Explanation: Combustion simply refers to burning. A combustion reaction may thus be explained as a chemical process which involves the reaction between a substance and an oxidizer usually oxygen gas. The combustion reaction in most cases involves the reaction between hydrocarbons which are combustible compounds and oxygen gas. The reaction between the combustible compound and oxygen gas is then followed by the release of energy, with the product being carbondioxid and water. Below is an example of combustion reaction.

Combustion the Methane CH4.

Here, CH4 reacts with oxygen gas(O2) to produce carbondioxide and water.

CH4 + 2O2 ------> CO2 + 2H2O

Answer:

Correct answer B

Explanation:

Just took quiz

Given that,

Weight = 0.75 N

Acceleration due to gravity = 9.81 m/s²

Density of water = 1 g/ml

We need to calculate the volume of water

Using formula of weight

[tex]W=mg[/tex]

[tex]W=\rho Vg[/tex]

[tex]mg=\rho\times V\times g[/tex]

Where, V = volume

g = acceleration

m = mass

Put the value into the formula

[tex]0.076\times 9.8\times10^3=1\times V\times9.8[/tex]

[tex]V= \dfrac{0.076\times 9.8\times10^3}{1\times9.8}[/tex]

[tex]V=76\ ml[/tex]

Hence, The volume of water is 76 mL.

Answer:

76.5mL

Explanation:

w = mg

0.75N = m * 9.81

m = 0.75 / 9.81 = 0.0765 kg

The mass in grams is 0.0765 * 1000 = 76.5g

At a density of 1 g/mL,

76.5g ÷ 1g/ml = 76.5mL

Answer:

C.15m/s

Explanation:

60 divided by 4 is 15

Answer:

The distance covered by the body is 14.4 m.

Solution,

Initial velocity(u)= 0 m/s( according to Question, the body starts from rest.that's why initial velocity(u) becomes zero.

Acceleration (a)= 0.2 m/s

Time (t):

[tex] \frac{1}{5} minute = \frac{1}{5 } \times 60 \: seconds[/tex]

[tex] = 12 \: seconds[/tex]

Now,

Applying third equations of motion:

[tex]s = ut + \frac{1}{2} a {t}^{2} \\ s = 0 \times 12 + \frac{1}{2} \times 0.2 \times {(12)}^{2} \\s = 0 + 14.4 \\ s = 14.4 \: metre[/tex]

Thus, the distance covered by the body is 14.4 metre.

Further more information:

Answer:

d = 57.87 m

Explanation:

It is given that,

Acceleration of car due to applied brake is [tex]-5.4\ m/s^2[/tex]

Its initial speed was 25 m/s

Its final speed was 0 as its comes to rest.

Leat d is the distance covered by the car. It can be calculated using third equation of kinematics as :

[tex]v^2-u^2=2ad\\\\d=\dfrac{v^2-u^2}{2a}\\\\d=\dfrac{0^2-(25)^2}{2\times (-5.4)}\\\\d=57.87\ m[/tex]

So, the distance covered by the car is 57.87 m.

Answer:

d = 57.87 m

Explanation:  have a nice day :)

Answer:

Displacement

Explanation:

Displacement is a vector quantity i.e it has both magnitude and direction. The direction is indicated with compass bearing i.e North, South, East or West or you could have North East, North West, South West, South East.

Answer:

Height of the first projectile = 49.98 m

Height of the second projectile = 72.52 m

Explanation:

From the given information;

Two projectiles are thrown from the same point with the velocity of49m/s

First is  projected making an angle  θ with the horizontal

and the second at an angle of 90 - θ.

Thus; for the first height to the horizontal; we have;

[tex]H_1 = \dfrac{v^2 sin^2 \theta}{2g}[/tex]  ----- (1)

the second height in the vertical direction is :

[tex]H_2 = \dfrac{v^2 cos^2 \theta}{2g}[/tex]    -----(2)

However; the second is found to rise 22.5 m higher than the first; so , we have :

[tex]\dfrac{v^2 cos^2 \theta}{2g}= 22.5 + \dfrac{v^2 sin^2 \theta}{2g}[/tex]

Let's recall that :

Cos²θ = 1 - Sin²θ

Replacing it into above equation; we have:

[tex]\dfrac{v^2}{2g} - \dfrac{v^2 sin^2 \theta}{2g}= 22.5 + \dfrac{v^2 sin^2 \theta}{2g}[/tex]

[tex]\dfrac{v^2}{2g} - 22.5 = \dfrac{v^2 sin^2 \theta}{g}[/tex]

[tex]\dfrac{1}{2 } \dfrac{v^2}{g} - 22.5 = \dfrac{v^2 sin^2 \theta}{g}[/tex]

[tex]\dfrac{1}{2 } - \dfrac {9.8 \times 22.5}{(49)^2} = sin^2 \theta[/tex]

[tex]\dfrac{1}{2 } - \dfrac {220.5}{2401} = sin^2 \theta[/tex]

[tex]sin^2 \theta= 0.408[/tex]

From (1);

[tex]H_1 = \dfrac{v^2 sin^2 \theta}{2g}[/tex]

[tex]H_1 = \dfrac{49^2 \times 0.408}{2*9.8}[/tex]

[tex]H_1 = \dfrac{979.608}{19.6}[/tex]

[tex]\mathbf{H_1 =49.98 \ m }[/tex]

Height of the first projectile = 49.98 m

Similarly;

From(2)

[tex]H_2 = \dfrac{v^2 cos^2 \theta}{2g}[/tex]    

[tex]H_2 = \dfrac{v^2 (1-sin^2 \theta)}{2g}[/tex]

[tex]H_2 = \dfrac{49^2 (1-0.408 )}{2 \times 9.8}[/tex]

[tex]H_2 = \dfrac{2401 (0.592 )}{19.6}[/tex]

[tex]H_2 = \dfrac{1421.392}{19.6}[/tex]

[tex]\mathbf{H_2 = 72.52 \ m}[/tex]

Height of the second projectile = 72.52 m

Copper is a reddish orange, soft metal with a bright metallic luster. It is an excellent conductor of heat and electricity. Copper surfaces when exposed to air gradually turns to a dull, brownish color.

Answer:

20.5m/s

Explanation:

neglecting friction the amount of energy at the bottom equals to the amount of energy at top. Law of conservation of energy.

ME= P.E+k.E

ME= mgh+ 1/2mv*v

ME = (m* 10 *49)+(1/2m(4.6*4.6))

MEtop=500.98m

MEbottom=MEtop: in order to obey the law of conservation of energy

MEbottom=mgh+ 1/2mv*v

500.98m=m*10*29+1/2m v* v

500.98m= m(290+1/2v*v) ...cancel m by m

500.98-290=1/2v*v criss cross multiple

210.98*2=v*v

v=√421.96

v=20.5 m/s

Answer:

I take the thread and wrap it round the cylinder until it goes round the circumference of the cylinder.

Explanation:

I take the thread and wrap it round the cylinder until it goes round the circumference of the cylinder.

I then cut the thread at this length and extend it out on a table or floor.

I then place the length of the thread along the meter rule and mark of the position of its tip and bottom along the meter rule.

The difference between the bottom and top points is the circumference of the cylinder.

If the thread is longer than the meter rule, i mark off where the the length of the meter rule stops along the thread and then place the meter rule at that point and then mark off where the length of the thread stops.

The circumference of the cylinder is then 1 meter plus the extra measurement along the meter rule.

Answer:

C

Explanation:

It is directly propotional to the amount of charge and inversly propotional to distance.

Answer:

C. Distance and amount of charge

Explanation:

The magnitude of the electric potential depends on which factors?

Distance and amount of charge

Answer:

The velocity of the marble relative to the water is 3.63 m/s

Explanation:

The information given are;

The velocity of the ferry relative to the water = 3.5 m/s

Direction of motion of the ferry = East

The velocity of the marble (on the ferry) relative to the ferry = 0.95 m/s

The direction of the marble = North

The (resultant) velocity of the marble relative to the water, [tex]v_R[/tex], is found as follows;

Resolution of the velocities;

Velocity of the ferry = 3.5·i + 0·j

Velocity of the marble = 0·i + 0.95·j

[tex]v_R[/tex] = Combined (resultant) velocity = √(3.5² + 0.95²) = 3.63 m/s

The velocity of the marble relative to the water = 3.63 m/s.

Resultant force is basically the force left after everything is added.

if a ball is being pushed one one side with 180N, and being pushed on teh opposite side with 84N (I added friction and air resistance since they're acting on the same side), then the resultant force would be:

180N - 84N = 96N   (you can determine whether it's positive or negative based on the direction of the vector)

Answer:

The average force has a magnitude 6524 N due north.

Explanation:

The average net force F = ma where m = mass of car = 1400 kg and a = acceleration.

a = (v - u)/t where u = initial velocity of car = 0 m/s (since it starts from rest)

v = final velocity of car = 27 m/s due north and t = time of motion = 5.8 s

a = (27 m/s - 0 m/s)/5.8 s = 27 m/s ÷ 5.8 s = 4.66 m/s

Since the direction of the velocity change is the direction of the acceleration, the acceleration is 4.66 m/s due north.

The average force, F = ma = 1400 kg × 4.66 m/s = 6524 N

Since the acceleration is due north, the average force takes the direction of the acceleration.

So the direction of the average force is due north

The average force has a magnitude 6524 N due north.

Answer:

79%

Explanation:

The following data were obtained from the question:

Input temperature = 100°C

Output temperature = 22°C

Efficiency =.?

Next, we shall convert the temperature from celsius to Kelvin temperature. This is illustrated below:

T (K) = T (°C) + 273

Input temperature = 100°C

Input temperature = 100°C + 273 = 373K

Output temperature = 22°C

Output temperature = 22°C + 273 = 295K

The efficiency of the locomotive can be obtained as follow:

Input temperature = 373K

Output temperature = 295K

Efficiency =.?

Efficiency = output /input x 100

Efficiency = 295/373 x 100

Efficiency = 79%

Therefore, the efficiency of the locomotive is 79%.

Answer:

Remember the equation:

F = m*a

where F is force, m is mass and a is acceleration.

If we have F constant. and we want that increases, then we can have the case where m decreases.

The mass can decrease in cases like a rocket, where as the fuel of the rocket consumes, the mass of the rocket decreases and the acceleration increases.

b) The cases where a constant force results in a constant acceleration of zero, is when the force is canceled, an example of this is the constant force of the gravity in all the objects. The objects that are in the ground are being affected by this force, but the gravitational force is canceled with the normal force of the ground. Then we have a constant force that does not cause any acceleration.