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liq [111]
2 years ago
11

A pilot heads his jet due east. The jet has a speed of 475 mi/h relative to the air. The wind is blowing due north with a speed

of 30 mi/h. (Assume that the i vector points east, and the j vector points north.)
a. Express the velocity of the wind as a vector in component form.
b. Express the velocity of the jet relative to the air as a vector in component form.
c. Find the true velocity of the jet as a vector.
d. Find the true speed of the jet. (Round your answer to the nearest integer.)
e. Find the direction of the jet.
Physics
1 answer:
oksian1 [2.3K]2 years ago
7 0

Explanation:

a. The velocity of the wind as a vector in component form will be represented as v vector:

    v=30j

b.The velocity of the jet relative to the air as a vector in component form will be represented as u vector

    u=475i

c. The true velocity of the jet as a vector will be represented as w:

  w=u+v

  w=475i+30j

d.  The true speed of the jet will be calculated as:

    IwI=\sqrt{(475)^2+(30)^2}

    IwI=\sqrt{225625+900}

    IwI=\sqrt{226525}

    IwI=476 mi/h

e. The direction of the jet will be:

tita=tan^{-1}\frac{30}{475}

tita=tan^{-1}(0.0632)

tita=3.62degrees,or,N86.38degreesS

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Two objects of the same size are both perfect blackbodies. One has a temperature of 3000 K, so its frequency of maximum emission
bija089 [108]

Answer:

a) The colder body (3000k), b) hearter body c) 12000K body

Explanation:

This exercise should know the power emitted by the objects and the distribution of this emission in the energy spectrum, for this we will use Stefan's laws and that of Wien's displacement

Stefan's Law                     P = σ A e T⁴

Wien displacement law   λ T = 2,898 10⁻³ m K

Let's calculate the power emitted for each object.

As they are perfect black bodies e = 1, they also indicate that they have the same area

T = 3000K

       P₁ = σ A T₁⁴

T = 12000K

       P₂ = σ A T₂⁴

       P₂ / P₁ = T₂⁴ / T₁⁴

       P₂ / P₁ = (12000/3000)⁴

       P₂ / P₁ = 256

This indicates that the hottest body emission is 256 times the coldest body emission.

Let's calculate the maximum emission wavelength

Body 1

T = 3000K

       λ T = 2,898 10-3

       λ₁ = 2.89810-3 / T

       λ₁ = 2,898 10-3 / 3000

       λ₁ = 0.966 10-6 m

      λ₁ = 966 nm

T = 12000K

      λ₂ = 2,898 10-3 / 12000

      λ₂ = 0.2415 10-6 m

      λ₂ = 214 nm

a) The colder body (3000k) emits more light in the infrared, since the emission of the hot body is at a minimum (emission tail)

b) The two bodies have emission in this region, the body of 3000K in the part of rise of the emission and the body to 12000K in the descent of the emission even when this body emits 256 times more than the other, so this body should have the highest broadcast in this area

c) The emission of the hottest 12000K body is mainly in UV

d) The hottest body emits more energy in UV and visible

e) No body has greater emission in all zones

5 0
3 years ago
Calculate the acceleration of a 270000-kg jumbo jet just before takeoff when the thrust on the aircraft is 160000 N .
Radda [10]

Answer:

<h3>The answer is 0.59 m/s²</h3>

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

a =  \frac{f}{m}  \\

f is the force

m is the mass

From the question we have

a =  \frac{160000}{270000}  =  \frac{16}{27}  \\  = 0.592592...

We have the final answer as

<h3>0.59 m/s²</h3>

Hope this helps you

7 0
3 years ago
50 points Answer fast please
erma4kov [3.2K]

Answer:

22 hours

Explanation:

6 0
3 years ago
Which 3 components must be used in order for a light bulb to light?
crimeas [40]

Answer:

a,e,d

Explanation:

i got it right when i was doing study island

6 0
3 years ago
A 150-N box is being pulled horizontally in a wagon accelerating uniformly at 3.00 m/s2. The box does not move relative to the w
Zepler [3.9K]

Answer:

Frictional force, F = 45.9 N

Explanation:

It is given that,

Weight of the box, W = 150 N

Acceleration, a=3\ m/s^2

The coefficient of static friction between the box and the wagon's surface is 0.6 and the coefficient of kinetic friction is 0.4.  

It is mentioned that the box does not move relative to the wagon. The force of friction is equal to the applied force. Let a is the acceleration. So,

m=\dfrac{W}{g}

m=\dfrac{150}{9.8}

m=15.3\ kg

Frictional force is given by :

F=ma

F=15.3\times 3

F = 45.9 N

So, the friction force on this box is closest to 45.9 N. Hence, this is the required solution.

8 0
3 years ago
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