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4 years ago

14

a plane being propelled due east is actually moving at 225mph an angle of 40 degrees north of east, how fast is the crosswind th

at is blowing due north (relative the ground?

1 answer:

Marianna [84]4 years ago

4 0

Answer:

144.63 mph

Explanation:

225 sin (40)

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The image produced by a convex mirror is always closer to the mirror than it would be in a plane mirror for the same object dist

rosijanka [135]

Answer:

True

Explanation:

The image produced a convex mirror is always virtual irrespective of location. The size of the image is always smaller than the object. In a plane mirror the distance of the object and the distance of the image is same. But in a convex the image distance is always less than the object distance.

So, this statement is true.

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4 years ago

Jack is testing how sound travels. Damon stood on the opposite side of the room and whispered a sentence to him. Then, Jack held

34kurt

D) Sound travels better through the compact molecules of air inside the balloon than the less compact molecules on the outside.

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3 years ago

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A flatbed truck is carrying a 20.0 kg crate along a level road. the coefficient of static friction between the crate and the bed

Dahasolnce [82]

<span>3.92 m/s^2 Assuming that the local gravitational acceleration is 9.8 m/s^2, then the maximum acceleration that the truck can have is the coefficient of static friction multiplied by the local gravitational acceleration, so 0.4 * 9.8 m/s^2 = 3.92 m/s^2 If you want the more complicated answer, the normal force that the crate exerts is it's mass times the local gravitational acceleration, so 20.0 kg * 9.8 m/s^2 = 196 kg*m/s^2 = 196 N Multiply by the coefficient of static friction, giving 196 N * 0.4 = 78.4 N So we need to apply 78.4 N of force to start the crate moving. Let's divide by the crate's mass 78.4 N / 20.0 kg = 78.4 kg*m/s^2 / 20.0 kg = 3.92 m/s^2 And you get the same result.</span>

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3 years ago

A person uses a constant force to push a 14.0 kg crate 1.80 m up a frictionless 10⁰ incline and to also increase its speed from

zubka84 [21]

a) Making h as the shorter cathetus, we have,

$h=1.8*sin10=0.312$

We proceed to calculate the work done by gravity

$W_g=-mgh$

$W_g=-140*0.312=-43.76J$

b) It's necessary to calculate the Kinetic Energy, we use the equation of KE,

$\Delta KE = \frac{1}{2}m(v_2^2-v_1^2)$

$\Delta KE = \frac{1}{2}14(1.5^2-0.5^2)=14J$

c) By work energy theorem

$W_{net}=\Delta KE \rightarrow W_net = 14J$

d) we calculate net work through the law of conservation

$W_{net}=W_f+W_g=14$

$\rightarrow W_f = 14-W_g=14-(-43.76)$

$W_f=57.76J$

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3 years ago

Two spherical objects at the same altitude move with identical velocities and experience the same drag force at a time t. If Obj

Anna35 [415]

Answer:

<em>The object with the twice the area of the other object, will have the larger drag coefficient.</em>

<em></em>

Explanation:

The equation for drag force is given as

$F_{D} = \frac{1}{2}pu^{2} C_{D} A$

where $F_{D}$ IS the drag force on the object

p = density of the fluid through which the object moves

u = relative velocity of the object through the fluid

p = density of the fluid

$C_{D}$ = coefficient of drag

A = area of the object

Note that $C_{D}$ is a dimensionless coefficient related to the object's geometry and taking into account both skin friction and form drag. The most interesting things is that it is dependent on the linear dimension, which means that it will vary directly with the change in diameter of the fluid

The above equation can also be broken down as

$F_{D}$ ∝ $P_{D}$ A

where $P_{D}$ is the pressure exerted by the fluid on the area A

Also note that $P_{D}$ = $\frac{1}{2}pu^{2}$

which also clarifies that the drag force is approximately proportional to the abject's area.

<em>In this case, the object with the twice the area of the other object, will have the larger drag coefficient.</em>

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3 years ago