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Leokris [45]
2 years ago
10

The apparent height of a building 10.5 km away is 0.02 radians. What is the approximate height of the building to the nearest me

ter
Physics
1 answer:
Ksenya-84 [330]2 years ago
8 0

Answer:

Approximate height of the building is 23213 meters.

Explanation:

Let the height of the building be represented by h.

0.02 radians = 0.02 × \frac{180^{o} }{\pi }

                     = 0.02 x (180/\frac{22}{7})

0.02 radians  = 1.146°

10.5 km = 10500 m

Applying the trigonometric function, we have;

Tan θ = \frac{opposite}{adjacent}

So that,

Tan 1.146° = \frac{h}{10500}

⇒ h = Tan 1.146° x 10500

      = 2.21074 x 10500

      = 23212.77

h = 23213 m

The approximate height of the building is 23213 m.

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A plane designed for vertical takeoff has a mass of 8.0 × 10³ kg. Find the net work done by all forces on the plane as it accele
artcher [175]

Answer:

<em>the net work done after starting from rest is =  2.4 × 10⁵ J</em>

Explanation:

Work: Work can be defined as the product of force and distance. The fundamental unit of work is Joules (J),  The unit of Energy is Joules (J), as such Energy and work are interchangeable during calculation, This is illustrated below

E = W = 1/2mv².......................... Equation 1

Where m = mass of the plane, v = velocity of the plane, E = Energy, W = work done.

v² = u² + 2as ................................. Equation 2.

Where v = final velocity of the plane, u = initial velocity of the plane, a = acceleration of the plane, distance of the plane.

<em>Given: a = 1.0 m/s², s = 30 m, u = 0 m/s (at rest)</em>

<em>Substituting these values into equation 2</em>

<em>v² = 0² +2×1×30</em>

<em>v² = 60</em>

<em>v = √60</em>

<em>v = 7.75 m/s</em>

Also given: m = 8.0 × 10³ kg, and v = 7.75 m/s

<em>Substituting these values into equation 1,</em>

<em>W = 1/2(8.0×10³)(7.75)²</em>

<em>W = (4.0×10³)(60)</em>

W = 240 × 10³ J

<em>W = 2.4 × 10⁵ J</em>

<em>Therefore the net work done after starting from rest is =  2.4 × 10⁵ J</em>

4 0
3 years ago
Notice that in each conversion factor the numerator equals the denominator when units are taken into account. A common error in
navik [9.2K]

Answer:

he factor for the temporal part 1.296 107 s² = h²

 m / s² = 12960 km / h²

Explanation:

This is a unit conversion exercise.

In the unit conversion, the size of the object is not changed, only the value with respect to which it is measured is changed, for this reason in the conversion the amount that is in parentheses must be worth one.

In this case, it is requested to convert a measure km/h²

Unfortunately, it is not clearly indicated what measure it is, but the most used unit in physics is   m / s² , which is a measure of acceleration. Let's cut this down

the factor for the distance is 1000 m = 1 km

the factor for time is 3600 s = 1 h

let's make the conversion

        m / s² (1km / 1000 m) (3600 s / 1h)²

note that as time is squared the conversion factor is also squared

        m / s² = 12960 km / h²

the factor for the temporal part 1.29 107 s² = h²

6 0
3 years ago
The diagram below shows a model of the Earth's orbit around the Sun.
Sladkaya [172]
I would say that I and IV because summer is I and fall is IV
7 0
3 years ago
Read 2 more answers
Hitungkan pecutan bagi blok di bawah: / Cal<br>(a)<br>m= 2 kg<br>F= 8.0 N​
ioda

Answer:

Acceleration = 4 m/s²

Explanation:

Given the following data;

Force = 8 N

Mass = 2 kg

To find the acceleration of the block;

Newton's Second Law of Motion states that the acceleration of a physical object is directly proportional to the net force acting on the physical object and inversely proportional to its mass.

Mathematically, it is given by the formula;

Acceleration = \frac {Net \; force}{mass}

Substituting into the formula, we have;

Acceleration = \frac {8}{2}

Acceleration = 4 m/s²

4 0
3 years ago
Question 20
oksian1 [2.3K]

Answer:

The image distance is 17.56 cm

Explanation:

We have,

Height of light bulb is 3 cm.

The light bulb is placed at a distance of 50 cm. It means object distance is, u =-50 cm

Focal length of the lens, f = +13 cm

Let v is distance between image and the lens. Using lens formula :

\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}\\\\\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}\\\\\dfrac{1}{v}=\dfrac{1}{13}+\dfrac{1}{(-50)}\\\\v=17.56\ cm

So, the image distance is 17.56 cm.

5 0
2 years ago
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