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mel-nik [20]
2 years ago
15

An airplane travels horizontally at a constant velocity v. An object is dropped from the plane and one second later another obje

ct is dropped from the plane. If air resistance is negligible, what happens to the vertical distance between the two objects while they are both falling?
Physics
1 answer:
Delvig [45]2 years ago
8 0

Answer:

the vertical distance between the two object will increase uniformly when they are dropped after a fixed interval of time

Explanation:

Since airplane is moving horizontally with constant speed v

so when object is dropped from the plane then the speed of the object will be same as that of the speed of the airplane

so we can say that two object when dropped after some interval of time then they always lie in same vertical line

now we know that they both have same acceleration in vertical line so the motion of two objects relative to each other in vertical direction is always uniform motion because they have no acceleration with respect to each other

So the vertical distance between the two object will increase uniformly when they are dropped after a fixed interval of time

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At a certain instant after jumping from the airplane A, a skydiver B is in the position shown and has reached a terminal (consta
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Answer:

a=2330

b= 0.223secs

Explanation:

pb=2330m

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6 0
1 year ago
In general it is best to conceptualize vectors as arrows in space, and then to make calculations with them using their component
Dimas [21]

(1) A - B

(2) B - C

(3) - A + B - C

(4) 3A - 2C

(5) - 2A + 3B - C

(6) 2A - 3 (B - C)

Answer:

(1)  (3,-5,-4)

(2) (-5, 4, 0)

(3) (-6, 4, 3)

(4) (-3, -2, -11)

(5) (-11, 14, 8)

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Explanation:

A⃗ =(1,0,−3)

B⃗ =(−2,5,1)

C⃗ =(3,1,1)

Vector additions and subtraction are done on a component by component basis, that is, only data from component î can be added to or subtracted from another Vector's component î. And so on for components j and k.

1) (A - B) = (1,0,−3) - (−2,5,1) = (1-(-2), 0-5, -3-1) = (3,-5,-4)

2)  (B - C) = (−2,5,1) - (3,1,1) = (-2-3, 5-1, 1-1) = (-5, 4, 0)

3) -A + B - C = -(1,0,−3) + (−2,5,1) - (3,1,1) = (-1-2-3, 0+5-1, 3+1-1) = (-6, 4, 3)

4) 3A - 2C = 3(1,0,−3) - 2(3,1,1) = (3,0,-9) - (6,2,2) = (3-6, 0-2, -9-2) = (-3, -2, -11)

5) -2A + 3B - C = -2(1,0,−3) + 3(−2,5,1) - (3,1,1) = (-2,0,6) + (-6,15,3) - (3,1,1) = (-2-6-3, 0+15-1, 6+3-1) = (-11, 14, 8)

6) 2A - 3 (B - C) = 2(1,0,−3) - 3[(−2,5,1) - (3,1,1)] = (2,0,-6) - 3(-5,4,0) = (2+15, 0-12, -6-0) = (17, -12, -6)

3 0
1 year ago
Calculate the amount of work done to draw a current of 8A from a point at 100V to a point at 120V in 2 seconds?
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Given:
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A goat enclosure is in the shape of a right triangle. One leg of the enclosure is built against the side of the barn. The other
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Answer:

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Or

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ira [324]

Answer:

B

Explanation:

A square of frequency of consists of the infinite sum of sine waves, whose frequencies are the odd multiples of the main frequency f i.e f, 3f,5f, 7f, ... . Given that the range of frequencies, to which the system responds is 20-40 kHz, for a square wave of frequency 10kHz we need to look for the harmonics whose frequencies are in the systems' respond range, which are the  harmonics of 20, 30 and 40 kHz

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