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3 years ago

3) If an airplane is in flight cruising at constant velocity, what can be said about the net work

Physics

1 answer:

Crazy boy [7]3 years ago

6 0

Explanation:

<h3>In physics, constant velocity occurs when there is no net force acting on the object causing it to accelerate. In terms of airplane flight, the two main forces influencing its velocity forward are drag and thrust. At a constant altitude, when the force of thrust equals the opposing force of drag, then the airplane will experience uniform motion in one direction. This can be further explained by Newton’s First Law. </h3>

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Welcome to this IE. You may navigate to any page you've seen already using the IE Outline tab on the right. A particle beam is m

Genrish500 [490]

Answer:

the magnitude of a uniform electric field that will stop these protons in a distance of 2 m is 10143.57 V/m or 1.01 × 10⁴ V/m

Explanation:

Given the data in the question;

Kinetic energy of each proton that makes up the beam = 3.25 × 10⁻¹⁵ J

Mass of proton = 1.673 × 10⁻²⁷ kg

Charge of proton = 1.602 × 10⁻¹⁹ C

distance d = 2 m

we know that

Kinetic Energy = Charge of proton × Potential difference ΔV

Potential difference ΔV = Kinetic Energy / Charge of proton

we substitute

Potential difference ΔV = ( 3.25 × 10⁻¹⁵ ) / ( 1.602 × 10⁻¹⁹ )

Potential difference ΔV = 20287.14 V

Now, the magnitude of a uniform electric field that will stop these protons in a distance of 2 m will be;

E = Potential difference ΔV / distance d

we substitute

E = 20287.14 V / 2 m

E = 10143.57 V/m or 1.01 × 10⁴ V/m

Therefore, the magnitude of a uniform electric field that will stop these protons in a distance of 2 m is 10143.57 V/m or 1.01 × 10⁴ V/m

3 0

3 years ago

A top fuel dragster with a mass of 500.0 kg starts from rest and completes a quarter mile (402 m) race in a time of 5.0 s. The d

blsea [12.9K]

The average power needed to produce this final speed is 1069.1 hp.

Mass of the dragster, m = 500.0 kg,

Displacement travelled by the dragster, s = 402 m,

Time taken in this travel, t = 5.0 s,

Final velocity of the dragster, v = 130 m/s.

Let the initial velocity of the dragster be u and acceleration be a.

Using kinematical equation, s = ut + (1/2)at^2.

402 = u*5 + (1/2)*a*5^2

10*u + 25*a = 804. ...........(1)

Using kinematical equation, v = u +at.

130 = u + 5*a

5*u + 25*a = 650. .............(2)

Solving (1)and (2), we get,

u = 30.8 m/s.

According to work-energy theorem,

Work done = change in kinetic energy

W = (1/2)*m*(v^2 - u^2)

W = (1/2)*500*(130^2 - 30.8^2)

W = 3987840. J

Therefore power rating of the dragster is given by,

P ⇒ W/t. = 3987840/5 = 797568 watt.

P ⇒ 797568/746 = 1069.1 hp.

Learn more about Power rating here brainly.com/question/20137708

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5 0

2 years ago

Read 2 more answers

Ohm's law states which relationship between electrical quantities? a. Volts = current x resistance b. Volts = current divided by

miv72 [106K]

Answer:

a. Volts = current x resistance

Explanation:

Ohm's law states that at constant temperature, the current flowing in an electrical circuit is directly proportional to the voltage applied across the two points and inversely proportional to the resistance in the electrical circuit.

Mathematically, Ohm's law is given by the formula;

$V = IR$