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2 years ago

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A plane initially traveling at 200 m/s due west experiences a 10 m/s head wind coming from the opposite direction. A). What will

be the relative velocity of the plane from the frame of reference of a bystander in the ground looking at the plane when it experiences the head wind current? B) Will the plane be late, on time or early after the wind current?

1 answer:

Hitman42 [59]2 years ago

8 0

Ok so it would be late and the relative velocity would be 190 m/s because 200 m/s - 10 m/s is 190 m/s. Hope this helps.

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Calculate the mass of the air contained in a room that measures 2.50 m x 5.50 m x 3.00 m if the density of air is 1.29 g/dm3.53.

Law Incorporation [45]

Answer:

$5.32\cdot 10^4 g$

Explanation:

First of all, we need to find the volume of the room, which is given by

$V=2.50 m \cdot 5.50 m \cdot 3.00 m =41.3 m^3$

Now we can find the mass of the air by using

$m=dV$

where

$d=1.29 g/dm^3$ is the density of the air

$V=41.3 m^3 = 41,300 dm^3$ is the volume of the room

Substituting,

$m=(1.29)(41300)=5.32\cdot 10^4 g$

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2 years ago

Two horizontal rods are each held up by vertical strings tied to their ends. Rod 1 has length L and mass M; rod 2 has length 2L

antiseptic1488 [7]

Answer:

Rod 1 has greater initial angular acceleration; The initial angular acceleration for rod 1 is greater than for rod 2.

Explanation:

For the rod 1 the angular acceleration is

$\tau_1 = I_1\alpha _1 \\\\\alpha_1 = \dfrac{\tau_1}{I_1}$

Similarly, for rod 2

$\alpha_2 = \dfrac{\tau_2}{I_2}.$

Now, the moment of inertia for rod 1 is

$I_1 = \dfrac{1}{3}ML^2$ ,

and the torque acting on it is (about the center of mass)

$\tau_1 = Mg\dfrac{L}{2};$

therefore, the angular acceleration of rod 1 is

$\alpha_1 = \dfrac{Mg\dfrac{L}{2}}{\dfrac{1}{3}ML^2},$

$\boxed{\alpha_1 = \dfrac{3g}{2L} }$

Now, for rod 2 the moment of inertia is

$I_2 = \dfrac{1}{3}(2M)(2L)^2$

$I_2 = \dfrac{8}{3} ML^2,$

and the torque acting is (about the center of mass)

$\tau _2 = (2M)g \dfrac{(2L)}{2}$

$\tau _2 = 2MgL;$

therefore, the angular acceleration $\alpha_2$ is

$\alpha_2 = \dfrac{2MgL;}{\dfrac{8}{3} ML^2,}.$

$\boxed{\alpha_2 = \dfrac{3g}{4L}}$

We see here that

$\dfrac{3g}{2L} > \dfrac{3g}{4L}$

therefore

$\boxed{\alpha_1 > \alpha_2.}$

In other words , the initial angular acceleration for rod 1 is greater than for rod 2.

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1 year ago

A helicopter travels west at 80 mph. It is moving above a car traveling on a highway at 80 mph. Given this information, you can

gavmur [86]

Answer:

d. at the same velocity

Explanation:

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Armand is monitoring a large sealed tank by way of a sensor that records the liquid level over time on a graph. He looks at the

timofeeve [1]

Answer:

i need ppoints

Explanation:

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2 years ago

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Energy conservation with conservative forces: Two identical balls are thrown directly upward, ball A at speed v and ball B at sp

MatroZZZ [7]

Answer:

E) True. Ball B will go four times as high as ball A because it had four times the initial kinetic energ

Explanation:

To answer the final statements, let's pose the solution of the exercise

Energy is conserved

Initial

Em₀ = K

Em₀ = ½ m v²

Final

Emf = U = mg h

Em₀ = emf

½ m v² = mgh

h = v² / 2g

For ball A

h_A = v² / 2g

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h_B = (2v)² / 2g

h_B = 4 (v² / 2g) = 4 h_A

Let's review the claims

A) False. The neck acceleration is zero, it has the value of the acceleration of gravity

B) False. Ball B goes higher

C) False has 4 times the gravitational potential energy than ball A

D) False. It goes 4 times higher

E) True.

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1 year ago