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Norma-Jean [14]
3 years ago
8

How roofs are lifted off building in tornado?

Physics
1 answer:
bekas [8.4K]3 years ago
7 0

Explanation:

As a tornado or other storm system passes over a building, low pressure can tug a roof upward. When those forces surpass the force exerted by the weight of the roof, the structure flies up and is swept away by wind currents..

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A pilot in a small plane encounters shifting winds. He flies 26.0 km northeast, then 45.0 km due north. From this point, he flie
cluponka [151]

Answer:

a) v₃ = 19.54 km, b)  70.2º north-west

Explanation:

This is a vector exercise, the best way to solve it is finding the components of each vector and doing the addition

vector 1 moves 26 km northeast

let's use trigonometry to find its components

         cos 45 = x₁ / V₁

         sin 45 = y₁ / V₁

         x₁ = v₁ cos 45

         y₁ = v₁ sin 45

         x₁ = 26 cos 45

         y₁ = 26 sin 45

         x₁ = 18.38 km

         y₁ = 18.38 km

Vector 2 moves 45 km north

        y₂ = 45 km

Unknown 3 vector

          x3 =?

          y3 =?

Vector Resulting 70 km north of the starting point

           R_y = 70 km

we make the sum on each axis

X axis

      Rₓ = x₁ + x₃

       x₃ = Rₓ -x₁

       x₃ = 0 - 18.38

       x₃ = -18.38 km

Y Axis

      R_y = y₁ + y₂ + y₃

       y₃ = R_y - y₁ -y₂

       y₃ = 70 -18.38 - 45

       y₃ = 6.62 km

the vector of the third leg of the journey is

         v₃ = (-18.38 i ^ +6.62 j^ ) km

let's use the Pythagorean theorem to find the length

         v₃ = √ (18.38² + 6.62²)

         v₃ = 19.54 km

to find the angle let's use trigonometry

           tan θ = y₃ / x₃

           θ = tan⁻¹ (y₃ / x₃)

           θ = tan⁻¹ (6.62 / (- 18.38))

           θ = -19.8º

with respect to the x axis, if we measure this angle from the positive side of the x axis it is

          θ’= 180 -19.8

          θ’= 160.19º

I mean the address is

          θ’’ = 90-19.8

          θ = 70.2º

70.2º north-west

3 0
3 years ago
A 3.0-kg object moves to the right with a speed of 2.0 m/s. It collides in a perfectly elastic collision with a 6.0-kg object mo
Zinaida [17]

Answer:

The kinetic energy of the system after the collision is 9 J.

Explanation:

It is given that,

Mass of object 1, m₁ = 3 kg

Speed of object 1, v₁ = 2 m/s

Mass of object 2, m₂ = 6 kg

Speed of object 2, v₂ = -1 m/s (it is moving in left)

Since, the collision is elastic. The kinetic energy of the system before the collision is equal to the kinetic energy of the system after the collision. Let it is E. So,

E=\dfrac{1}{2}m_1v_1^2+\dfrac{1}{2}m_2v_1^2

E=\dfrac{1}{2}\times 3\ kg\times (2\ m/s)^2+\dfrac{1}{2}\times 6\ kg\times (-1\ m/s)^2

E = 9 J

So, the kinetic energy of the system after the collision is 9 J. Hence, this is the required solution.

3 0
3 years ago
A small sphere has a harge of 9uC and other small sphere has a charge of 4uC.
Helga [31]

Answer:

Electrical force, F = 90 N

Explanation:

It is given that,

Charge on sphere 1, q_1=9\ \mu C=9\times 10^{-6}\ C

Charge on sphere 2, q_1=4\ \mu C=4\times 10^{-6}\ C

Distance between two spheres, d = 6 cm = 0.06 m

Let F is the electrical force between them. It is given by the formula of electric force which is directly proportional to the product of charges and inversely proportional to the square of distance between them such that,

F=k\dfrac{q_1q_2}{d^2}

F=9\times 10^9\times \dfrac{9\times 10^{-6}\times 4\times 10^{-6}}{(0.06)^2}

F = 90 N

So, the electrical force between them is 90 N. Hence, this is the required solution.

7 0
3 years ago
Who was the carthaginian general who used elephants to cross the alps in the second punic war
goldfiish [28.3K]
<span>the answer is Hannibal</span>
3 0
3 years ago
Read 2 more answers
A mass is placed at the end of a spring. It has starting velocity of V &amp; allowed to oscillate freely. If the mass has a star
LiRa [457]

Answer:

Equation for SHM can be written

V = w A cos w t        where w is the angular frequency and the velocity is a                                         maximum at t = 0

V1 = w1  A cos w1 t

V2 = w2 A cos w2 t

V2 / V1 = w2 / w1     since cos X t = 1 if t = zero

V2 / V1 = 2 pi f2 / (2 pi f1) = f2 / f1 = T1 / T2

If the velocity is twice as large the period will be 1/2 long

8 0
2 years ago
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