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Andreyy89
3 years ago
11

Justify why does a bird flap its wing to fly higher??​

Physics
1 answer:
SpyIntel [72]3 years ago
8 0

Answer:

they use thermals and air currents to glide.

Explanation:

when they flap higher they use thermals and air currents  because flapping takes a lot of fuel,energy

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Light from a helium-neon laser (? = 633 nm) is used to illuminate two narrow slits. The interference pattern is observed on a sc
Sladkaya [172]

Answer:

0.3376 mm

Explanation:

The computation of the spacing in mm between the slits is shown below:

As we know that

d = \frac{m\lambda L}{\Delta y}

where,

\lambda = wavelength

L = distance from the scrren

\Delta y = spanning distance

As there are 11 bright fingers seen so m would be

= 11 - 1

= 10

Now placing these values to the above formula

So, the spacing is

= \frac{(10)(633 \times 10^{-9})(3.2m)}{60 \times 10^{-3}}

= 0.3376 mm

We simply applied the above formula.

6 0
3 years ago
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Which characteristics is common in mature rivers river
amm1812
It channels erode wider fed by many tributaries and has more discharge and is less steep
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3 years ago
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A bin is given a push across a horizontal surface. The bin has a mass m, the push gives it an initial speed of 1.60 m/s, and the
emmasim [6.3K]

Answer:

The bin moves 0.87 m before it stops.

Explanation:

If we analyze the situation and apply the law of conservation of energy to this case, we get:

Energy Dissipated through Friction = Change in Kinetic Energy of Bin (Loss)

F d = (0.5)(m)(Vi² - Vf²)

where,

F = Frictional Force = μR    

but, R = Normal Reaction = Weight of Bin = mg

Therefore, F = μmg

Hence, the equation becomes:

μmg d = (0.5)(m)(Vi² - Vf²)

μg d = (0.5)(Vi² - Vf²)

d = (0.5)(Vi² - Vf²)/μg

where,

Vf = Final Velocity = 0 m/s (Since, bin finally stops)

Vi = Initial Velocity = 1.6 m/s

μ = coefficient of kinetic friction = 0.15

g = 9.8 m/s²

d = distance moved by bin before coming to stop = ?

Therefore,

d = (0.5)[(1.6 m/s)² - (0 m/s)²]/(0.15)(9.8 m/s²)

<u>d = 0.87 m</u>

5 0
3 years ago
In each of two coils the rate of change of the magnetic flux in a single loop is the same. The emf induced in coil 1, which has
irinina [24]

Answer:

 Coil 2 have  235  loops

Explanation:

Given  

The number of loops in coil 1 is n ₁= 159

The emf induced in coil 1 is  ε ₁ = 2.78 V

The emf induced in coil 2 is  ε ₂ = 4.11 V

Let

n ₂  is the number of loops in coil 2.

Given, the emf in a single loop in two coils are same. That is,

ϕ ₁/n ₁= ϕ ₂ n ₂⟹ 2.78/159 = 4.11/ n ₂

n₂=\frac{159 * 4.11}{2.78}

n₂=235

Therefore, the coil 2 has  n ₂= 235  loops.

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