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Goshia [24]
4 years ago
8

In 1986, the first flight around the globe without a single refueling was completed. The aircraft’s average speed was 186 km/h.

If the airplane landed at this speed and accelerated at −1.5 m/s2 , how long did it take for the airplane to stop?
Physics
1 answer:
Whitepunk [10]4 years ago
3 0

Answer: 34.4 sec

Explanation: Assuming that the desacceleration was constant, we have  a = -1.5m/s²

vi = 186 km/h → 51.6 m/s

vf = 0 (because it stopped)

Vf = Vi + at

0 = 51.6 - 1.5t

1.5t = 51.6

t = 34.4 s

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The wavelengths for visible light rays correspond to which of these options?
ryzh [129]

Answer:

sorry about the other person but its b

Explanation:

5 0
3 years ago
Points a and b lie in a region where the y-component of the electric field is Ey=α+β/y2. The constants in this expression have t
Drupady [299]

Answer:

V_{a} - V_{b} = 89.3

Explanation:

The electric potential is defined by

         V_{b} - V_{a} = - ∫ E .ds

In this case the electric field is in the direction and the points (ds) are also in the direction and therefore the angle is zero and the scalar product is reduced to the algebraic product.

         V_{b} - V_{a} = - ∫ E ds

We substitute

         V_{b} - V_{a} = - ∫ (α + β/ y²) dy

We integrate

          V_{b} - V_{a} = - α y + β / y

We evaluate between the lower limit A  2 cm = 0.02 m and the upper limit B 3 cm = 0.03 m

           V_{b} - V_{a} = - α (0.03 - 0.02) + β (1 / 0.03 - 1 / 0.02)

            V_{b} - V_{a} = - 600 0.01 + 5 (-16.67) = -6 - 83.33

            V_{b} - V_{a} = - 89.3 V

As they ask us the reverse case

             V_{b} - V_{a} = - V_{b} - V_{a}

             V_{a} - V_{b} = 89.3

3 0
4 years ago
A certain car can accelerate from 0 to 60 mph in 7.9 s. What is the car's average acceleration in mph/s?
Anna007 [38]

Answer:

<em>a = 7.6\ mph/s</em>

Explanation:

<u>Motion With Constant Acceleration </u>

It's a type of motion in which the velocity of an object changes uniformly in time.

The equation that describes the change of velocities is:

v_f=v_o+at

Where:

a   = acceleration

vo = initial speed

vf  = final speed

t    = time

Solving the equation [for a:

\displaystyle a=\frac{v_f-v_o}{t}

The car accelerates from vo=0 to vf=60 mph in t=7.9 s, thus the acceleration is:

\displaystyle a=\frac{60 \ mph-0}{7.9}

a = 7.6\ mph/s

7 0
3 years ago
A 67 kg soccer player uses 5100 kJ of energy during a 2.0 h match. What is the
Oksanka [162]

The  average power produced by the soccer player is  710 Watts.

Given the data in the question;

  • Mass of the soccer player; m = 67kg
  • Energy used by the soccer player; E = 5100KJ = 5100000J
  • Time; t = 2.0h = 7200s

Power; P =\ ?

Power is simply the amount of energy converted or transferred per unit time. It is expressed as:

Power = \frac{Energy\ converted }{time}

We substitute our given values into the equation

Power = \frac{5100000J}{7200s}\\\\Power = 708.33J/s \\\\Power = 710J/s \ \ \ \ \ [ 2\ Significant\ Figures]\\\\Power = 710W

Therefore, the  average power produced by the soccer player is  710 Watts.

Learn more: brainly.com/question/20953664

8 0
3 years ago
1. What is the kinetic energy of a 1.75 kg ball travelling at a speed of 54 m/s?
Over [174]

Answer:

We conclude that the kinetic energy of a 1.75 kg ball traveling at a speed of 54 m/s is 2551.5 J.

Explanation:

Given

  • Mass m = 1.75 kg
  • Velocity v = 54 m/s

To determine

Kinetic Energy (K.E) = ?

We know that a body can possess energy due to its movement — Kinetic Energy.

Kinetic Energy (K.E) can be determined using the formula

K.E=\frac{1}{2}mv^2

where

  • m is the mass (kg)
  • v is the velocity (m/s)
  • K.E is the Kinetic Energy (J)

now substituting m = 1.75, and v = 54 in the formula

K.E=\frac{1}{2}mv^2

K.E=\frac{1}{2}\left(1.75\right)\left(54\right)^2

K.E=1458\times 1.75

K.E=2551.5 J

Therefore, the kinetic energy of a 1.75 kg ball traveling at a speed of 54 m/s is 2551.5 J.

7 0
3 years ago
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