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arlik [135]
3 years ago
15

An airplane moves through the air at a constant speed. The engines’ thrust applies a force in the direction of motion, and this

force is equal in magnitude and opposite in direction to the drag force. Reducing thrust will cause the plane to fly at a slower—but still constant—speed. Explain why this is so.
Physics
1 answer:
lord [1]3 years ago
8 0

Explanation:

This is because the drag force suffered by the aircraft is proportional to the speed at which it travels. The thrust of the engines prints a speed to the plane and this speed prints a drag force, always reaching an equilibrium point of these two forces where the speed of the plane is constant and the acceleration is equal to zero.

Therefore, by reducing the thrust, the drag force is greater and the plane begins to decrease its speed, until it reaches the point where the new drag force is matched with the new thrust force, giving it a new final speed , without acceleration.

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An electric dipole consists of a particle with a charge of 6 x 10–6 c at the origin and a particle with a charge of –6 x 10–6 c
mote1985 [20]

An electric dipole consists of a particle with a charge of 6 x 10⁻⁶ c at the origin and a particle with a charge of –6 x 10⁻⁶ c on the x axis at x = 3 x 10⁻³ m. Its dipole moment is 18 x 10⁻⁹ Cm

Dipole moment of a dipole is dependent on the charge of the dipole and the distance between the two charges.

Electric Dipole consists of two charges which are equal and opposite in charge i.e. positive and negative charges.

Given,

Dipole moment, p = ?

Charge, q = 6 x 10⁻⁶C

Distance between charges, d = 3 x 10⁻³ m

Dipole moment (p) is given by:

p = charge x distance between the two charges

p = 6 x 10⁻⁶ x 3 x 10⁻³ Cm

p = 18 x 10⁻⁹ Cm

The dipole moment for the given charge configuration is 18 x 10⁻⁹ Cm

Learn more about Dipole moment here, brainly.com/question/14058533

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2 years ago
A 100 kg cart is moving at 3 m/s. Calculate the cart’s kinetic energy.
STALIN [3.7K]

Answer:

450

Explanation:

Given,

Mass= 100kg

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= 1/2× 100× 3^2

= 1/2× 900

= 450.

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3 years ago
A projectile is fired into the air from the top of a 200-m cliff above a valley as shown below. Its initial velocity is 60 m/s a
anastassius [24]

a) y(max)  = 337.76 m

b) t₁ = 5.30 s  the time for y maximum

c)t₂ =  13.60 s  time for y = 0 time when the fly finish

d) vₓ = 30 m/s        vy = - 81.32 m/s

e)x = 408 m

Equations for projectile motion:

v₀ₓ = v₀ * cosα          v₀ₓ = 60*(1/2)     v₀ₓ = 30 m/s   ( constant )

v₀y = v₀ * sinα           v₀y = 60*(√3/2)     v₀y = 30*√3  m/s

a) Maximum height:

The following equation describes the motion in y coordinates

y  =  y₀ + v₀y*t - (1/2)*g*t²      (1)

To find h(max), we need to calculate t₁ ( time for h maximum)

we take derivative on both sides of the equation

dy/dt  = v₀y  - g*t

dy/dt  = 0           v₀y  - g*t₁  = 0    t₁ = v₀y/g

v₀y = 60*sin60°  = 60*√3/2  = 30*√3

g = 9.8 m/s²

t₁ = 5.30 s  the time for y maximum

And y maximum is obtained from the substitution of t₁  in equation (1)

y (max) = 200 + 30*√3 * (5.30)  - (1/2)*9.8*(5.3)²

y (max) = 200 + 275.40 - 137.64

y(max)  = 337.76 m

Total time of flying (t₂)  is when coordinate y = 0

y = 0 = y₀  + v₀y*t₂ - (1/2)* g*t₂²

0 = 200 + 30*√3*t₂  - 4.9*t₂²            4.9 t₂² - 51.96*t₂ - 200 = 0

The above equation is a second-degree equation, solving for  t₂

t =  [51.96 ±√ (51.96)² + 4*4.9*200]/9.8

t =  [51.96 ±√2700 + 3920]/9.8

t =  [51.96 ± 81.36]/9.8

t = 51.96 - 81.36)/9.8         we dismiss this solution ( negative time)

t₂ =  13.60 s  time for y = 0 time when the fly finish

The components of the velocity just before striking the ground are:

vₓ = v₀ *cos60°       vₓ = 30 m/s  as we said before v₀ₓ is constant

vy = v₀y - g *t        vy = 30*√3  - 9.8 * (13.60)

vy = 51.96 - 133.28         vy = - 81.32 m/s

The sign minus means that vy  change direction

Finally the horizontal distance is:

x = vₓ * t

x = 30 * 13.60  m

x = 408 m

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Reika [66]

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