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3 years ago

As an airplane is flying 20.0 degrees north of east at a speed of 750 km/h. How fast is it moving to the north

Physics

2 answers:

dedylja [7]3 years ago

7 0

<span>in this case velocity of airoplane will be 750*sin20'

</span>

Gwar [14]3 years ago

3 0

Answer:

$v_y=71.25\ m/s$

Explanation:

Given that,

Speed of an airplane, v = 750 km/hr = 208.33 m/s

An airplane is flying 20.0 degrees north of east, $\theta=20^{\circ}$

To find,

Velocity in north direction.

Solution,

Let $v_y$ is the speed of the airplane to the north. The y component of any motion is given by :

$v_y=v\ sin\theta$

$v_y=208.33\ m/s\ sin(20)$

$v_y=71.25\ m/s$

So, the airplane is moving with a speed of 71.25 m/s to the north direction.

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Can you label the parts of the eye?

photoshop1234 [79]

Answer:

Iris
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Anterior chamber
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Optic nerve
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Hope it helps :)

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3 years ago

If the coefficient of static friction is 0.357, and the same ladder makes a 58.0° angle with respect to the horizontal, how far

zavuch27 [327]

Answer: d= 0.57* l

Explanation:

We need to check that before ladder slips the length of ladder the painter can climb.

So we need to satisfy the equilibrium conditions.

So for ∑Fx=0, ∑Fy=0 and ∑M=0

We have,

At the base of ladder, two components N₁ acting vertical and f₁ acting horizontal

At the top of ladder, N₂ acting horizontal

And Between somewhere we have the weight of painter acting downward equal to= mg

So, we have N₁=mg

and also mg*d*cosФ= N₂*l*sin∅

So,

d= $\frac{N2}{mg}*l$ * tan∅

Also, we have f₁=N₂

As f₁= чN₁

So f₁= 0.357 * 69.1 * 9.8

f₁= 241.75

Putting in d equation, we have

d= $\frac{241.75}{69.1*9.8} *l$ * tan 58

d= 0.57* l

So painter can be along the 57% of length before the ladder begins to slip

3 0

3 years ago

Read 2 more answers

A 2-kg wheel rolls down the road with a linear speed of 15m/s. Find its trwansitional and rotational kinetic energies.

Elza [17]

Answer:

The translational kinetic energy is 225 J

The rotational kinetic energy is 225 J

Explanation:

Given;

mass of the wheel, m = 2-kg

linear speed of the wheel, v = 15 m/s

Transnational kinetic energy is calculated as;

E = ¹/₂MV²

where;

M is mass of the moving object

V is the velocity of the object

E = ¹/₂ x 2 x (15)²

E = 225 J

Rotational kinetic energy is calculated as;

E = ¹/₂Iω²

where;

I is moment of inertia

ω is angular velocity

$E = \frac{1}{2} I \omega^2\\\\E = \frac{1}{2} *mr^2*(\frac{v}{r})^2\\\\E = \frac{1}{2} *mr^2*\frac{v^2}{r^2} \\\\E = \frac{1}{2}mv^2$

E = ¹/₂ x 2 x (15)²

E = 225 J

Thus, the translational kinetic energy is equal to rotational kinetic energy

6 0

3 years ago

Type the correct answer in the box. Spell the word correctly.

Veronika [31]

Answer:

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6 0

3 years ago

A particle cannot generally be localized to distances much smaller than its de Broglie wavelength. This fact can be taken to mea

Vlada [557]

The De broglie wavelength of a thermal neutron at room temperature 300K = 1.5 × A°

<h3>How is the De broglie wavelength of a thermal neutron at room temperature calculated?</h3>

Temperature, T = 300K

Momentum, p = mv

Therefore v = p/m

Energy, E= 1/2 m( p/m) ²

Boltzman Energy= 3/2 KT

3/2KT = 1/2 m(p/m)²

Therefore p = $\sqrt{3RTm}$

According to De broglie hypothesis, P = h ÷ λ

Therefore, λ = h ÷ $\sqrt{3RTm}$

= 6.6× $10^{-34}$ ÷ $\sqrt{3 \times 1.38 \times 10^{-23} \times 300 \times 1.6 \times 10^{-27} }$

= 0.15 × $10^{-9}$

Therefore the De broglie wavelength of a thermal neutron at room temperature 300K = 1.5 × A°

To learn more about De broglie wavelength, refer: <u>https://brainly.in/question/6131028</u>

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2 years ago