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3 years ago

Flapping flight is very energy intensive. A wind tunnel test

Physics

1 answer:

steposvetlana [31]3 years ago

4 0

Answer:

The metabolic power for starting flight=134.8W/kg

Explanation:

We are given that

Mass of starling, m=89 g=89/1000=0.089 kg

1 kg=1000 g

Power, P=12 W

Speed, v=11 m/s

We have to find the metabolic power for starting flight.

We know that

Metabolic power for starting flight= $\frac{P}{m}$

Using the formula

Metabolic power for starting flight= $\frac{12}{0.089}$

Metabolic power for starting flight=134.8W/kg

Hence, the metabolic power for starting flight=134.8W/kg

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If 0.035pC of charge is transferred via the movement of Al3+ ions, how's many of these must be transferred in total? Please add

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Each $Al^+^3$ ion contains three extra protons. Hence, the extra charge on each $Al^+^3$ = $3 \times 1.6 \times 10^-^1^9$ C

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According to question:

$n \times 3 \times 1.6 \times 10^-^1^9 =0.035 \times 10^-^1^2$

$n = \frac{0.035 \times 10^-^1^2}{3 \times 1.6 \times 10^-^1^9}$

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Suppose an automobile has 2000-joules of kinetic energy. when it moves at twice the speed, what will be its kinetic energy? what

NeX [460]

Answer:

K.Eₓ = 4 K.E

K.Eₓ = 9 K.E

Explanation:

Th formula for the kinetic energy of a body is given as follows:

$K.E = \frac{1}{2}mv^2\\$ ---------------equation (1)

where,

K.E = Kinetic Energy of Automobile

m = mass of automobile

v = speed of automobile

For twice speed:

vₓ = 2v

then,

$K.E_{x} = \frac{1}{2}mv_{x}^2\\K.E_{x} = \frac{1}{2}m(2v)^2\\K.E_{x} = 4\frac{1}{2}mv^2\\$

using equation (1):

K.Eₓ = 4 K.E

For thrice speed:

vₓ = 3v

then,

$K.E_{x} = \frac{1}{2}mv_{x}^2\\K.E_{x} = \frac{1}{2}m(3v)^2\\K.E_{x} = 9\frac{1}{2}mv^2\\$

using equation (1):

K.Eₓ = 9 K.E

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