If you blow air between a pair of closely-spaced Ping-Pong balls suspended by strings, the balls will swing
2 answers:
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Answer:
The target's velocity is about 1320 m/s in the direction 265.7°.
Explanation:
In order for there to be a collision between missile and target, we must have ...
(target starting position) + (target movement) = (missile movement)
assuming the missile starts from the origin of all measurements. The missile moves 10.2 seconds before impact, so moves a distance of ...
(10.2 s)(1350 m/s) = 13,770 m
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We are interested in the target movement, so we can solve for that:
(target movement) = (missile movement) - (target starting position)
In terms of meters, this is ...
(target movement) = 13770∠25° - 23500∠55° ≈ 13467.74∠-94.3°
The target covers this distance in the same 10.2 seconds before collision, so its speed is (13467.74 m)/(10.2 s) ≈ 1320.4 m/s.
As a positive angle, the target's direction is ...
-94.3° +360° = 265.7°
The direction of the target's velocity is 265.7°.
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If you're calculating this by hand, there are a couple of ways you can do it. You can convert to rectangular coordinates and back (perhaps least confusing), or you can use the law of cosines to solve the triangle, then translate angles back to the x-y coordinate plane.
Using rectangular coordinates, we have ...
13770∠25° = 13770(cos(25°), sin(25°)) ≈ (12479.9, 5819.45)
23500∠55° = 23500(cos(55°), sin(55°)) ≈ (13479.0, 19250.1)
Then the difference is ...
(12479.9, 5819.45) -(13479.0, 19250.1) ≈ (-999.188, -13430.6)
and the (3rd-quadrant) angle is ...
target direction = arctan(-13430.6/-999.188) ≈ -94.3° = 265.7°
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The target's speed is found by dividing the distance it covers by the time it takes.
√(13430.6² +999.188²)/10.2 ≈ 1320.36 . . . m/s
Answer:
Groceries stay in the bag.
Explanation:
Given:
Maximum force = 250 N
Bag filled with = 20 kg
Lifted acceleration =
Solution:
We need to calculate the exerted force on the grocery bag by using Newton's second law.
Where:
F = Exerted force on the object.
m = Mass of the object in kg
a = Acceleration of the object in
Now, we substitute m = 20 kg and a = in Newton's second law,
Since, the exerted force on the bag is less than 250 N, the groceries will stay in the bag.
Answer:
the stopping distance is greater than the free length of the track, the vehicle leaves the track before it can brake
Explanation:
This problem can be solved using the kinematics relations, let's start by finding the final velocity of the acceleration period
v² = v₀² + 2 a₁ x
indicate that the initial velocity is zero
v² = 2 a₁ x
let's calculate
v =
v = 143.666 m / s
now for the second interval let's find the distance it takes to stop
v₂² = v² - 2 a₂ x₂
in this part the final velocity is zero (v₂ = 0)
0 = v² - 2 a₂ x₂
x₂ = v² / 2a₂
let's calculate
x₂ =
x₂ = 573 m
as the stopping distance is greater than the free length of the track, the vehicle leaves the track before it can brake