3 years ago

A passenger plane is flying above the ground.Describe the two components of its mechanical enserfy

Physics

1 answer:

Goryan [66]3 years ago

3 0

The force and the air resistance depends on the mechanical enserfy.

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Which energy transfer occurs when a cube of ice is placed in a glass of<br> water?

Daniel [21]

Acetic acid is a weak acid and sodium hydroxide is strong base. Salts of the two will hydrolyse to give basic solution. So, at neutral point, pH of the solution will be greater than 8.

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2 years ago

A light ray moving at a 54.3 deg

MrMuchimi

Answer:

$45.25^{\circ}$

Explanation:

Applying Snell's law

$n_1sin\theta_1 = n2*sin\theta_2\\1.33*sin(54.3) = 1.52*sin\theta_2\\sin\theta_2 = 1.33*0.81208353/1.52 = 0.71057\\\theta_2=sin^{-1} (0.71057)\\\theta_2 = 45.25^{\circ}$

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2 years ago

An electric motor consumes 8.40 kJ of electrical energy in 1.00 min. Part A If one-third of this energy goes into heat and other

Salsk061 [2.6K]

Answer:

The torque is 0.31 Nm.

Explanation:

Electrical energy, E = 8400 J

time, t = 1 min

Angular speed, w = 2900 rpm = 303.53 rad/s

efficiency = 2/3 of input power

The toque is given by

$P =\tau w\\\\\frac{2}{3}\times \frac{E}{t}=\tau w\\\\\frac{2}{3}\times \frac{8400}{60}=\tau \times 303.53\\\\\tau =0.31 Nm$

3 0

3 years ago

the distriict record holding sprinter stars in the blocks and accelerates to 10/s in 20s what was the acceleration of the sprint

Mumz [18]

Acc= vf-vi/time=10-0/20= 0.5 m/s/s

7 0

3 years ago

A student was asked to push a large box down the hallway about 35 meters. the task took the student five minutes and a force of

zaharov [31]

Rate of work or Power done by the student can be computed with below formulas.
1. Formula for Work
Work = Force x distance
Force or F = Force applied to the object
distance or d = displacement

2. Formula for Power
P = Work done / Time it was done or completed

Substituting the values we have,
Work or Work done = 150 N x 35 meters = 5250 J (unit of work).

Lastly computing for the rate we have,
Power or rate of work = 5250 J / 300 seconds (or 5 min) = 17.5 Joule/S or Watts

Answer is 17.5 watts

5 0

3 years ago