A high-speed dart is shot from ground level with a speed of 150 m/s at an angle 30° above the horizontal. What is the vertical c
1 answer:
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(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)
(6) (17, -12, -6)
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)
Answer:
Explanation:
Since the front and back of the rocket simultaneously line up with forward and backward end of the platform respectively .
Then length of the platform = length of the train rocket .
A )
Time to cross a particular point on the platform
= length of rocket train / .96 x 3 x 10⁸
= 90 / .96 x 3 x 10⁸
= 31.25 x 10⁻⁸ s
B) Rest length of the rocket = length of platform = 90 m
C ) length of platform as viewed by moving observer =
=
= 321 m
D ) For the observer on platform time taken = 31.25 x 10⁻⁸ s
for the observer in the rocket , time will be dilated so time recorded by observer in motion ,
8.75 x 10⁻⁸ s .
The ball was dropped from a height 20 meters
Explanation:
The given is
1. A ball is dropped from the top of a cliff
2. By the time it reaches the ground, all the energy in its gravitational
potential energy store has been transferred into its kinetic energy
store, that mean K.E = P.E
3. The ball is travelling at 20 m/s when it hits the ground
4. The gravitational field strength is 10 N/kg
We need to find the height that the ball dropped from it
The ball dropped from the top of a cliff means the initial speed is 0
→ K.E =
where v is the final speed, in the initial speed and m
is the mass
→ v = 20 m/s and = 0 m/s
→ K.E =
→ K.E =
→ K.E = 200 m joules ⇒ when the ball hits the ground
→ P.E = m g h
where g is the gravitational field strength, m is the mass and h is
the height
→ g = 10 N/kg
→ P.E = m(10)(h)
→ P.E = 10 m h joules
→ P.E = K.E
→ 10 m h = 200 m
Divide both sides by 10 m
→ h = 20 meters
The ball was dropped from a height 20 meters
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You can learn more about gravitational potential energy in brainly.com/question/1198647
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