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Andrews [41]
3 years ago
15

During takeoff, an airplane climbs with a speed of 130 m/s at an angle of 47 degrees above the horizontal. The speed and directi

on of the airplane constitute a vector quantity known as the velocity. The sun is shining directly overhead. How fast is the shadow of the plane moving along the ground? (That is, what is the magnitude of the horizontal component of the plane's velocity?)
Physics
1 answer:
Reptile [31]3 years ago
5 0
130cos(47deg) = 88.6598 m/s
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the orion nebula (at least the part we can see) is not very old (yet). while several hot, massive stars have had a chance to for
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3 0
1 year ago
1 point
Yuliya22 [10]
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8 0
3 years ago
In the doorknob shown above, when the handle is rotated a distance of 189 millimeters, the spindle is rotated a distance of 27 m
Alisiya [41]

If my math is right its A) 7

because 189 divided by 27 is 7

7 0
3 years ago
Read 2 more answers
How much work would be needed to raise the payload from the surface of the moon (i.e., x = r) to an altitude of 5r miles above t
Cloud [144]

Let the data is as following

mass of payload = "m"

mass of Moon = "M"

now we know that we place the payload from the position on the surface of moon to the position of 5r from the surface

So in this case we can say that change in the gravitational potential energy is equal to the work done to move the mass from one position to other

so it is given by

W = U_f - U_i

we know that

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U_i = -\frac{GMm}{r}

now from above formula

W = -\frac{GMm}{6r} + \frac{GMm}{r}

W = \frac{5GMm}{6r}

so above is the work done to move the mass from surface to given altitude

7 0
3 years ago
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