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kenny6666 [7]
3 years ago
5

A rescue pilot drops a survival kit while her plane is flying at an altitude of 1.5 km with a forward velocity of 100.0 m/s. If

air friction is disregarded, how far in advance of the starving explorer’s drop zone should she release the package?
Physics
2 answers:
n200080 [17]3 years ago
7 0
Given: Initial Velocity Vi = 100 m/s; Altitude y = 1.5 Km; Time t = ?

The aircraft is moving horizontally on x direction

Formula to follow: x = VicosΘt   but Time t is unknown. 

Solving for Time "t" formula will be y = VisineΘt + 1/2 gt² 

Since sineΘt = 0 therefore  y = 1/2gt² Solve for t

t = √2y/g  t = √2(1500 m)/9.8m/s²)   t = 31.24 s

Now solving for distance

x = VicosΘt  x = (100 m/s) (31.24 s) x =  3,124 m or 3.124 Km



 
tresset_1 [31]3 years ago
3 0

1750 meters.  
First, determine how long it takes for the kit to hit the ground. Distance over constant acceleration is: 
d = 1/2 A T^2
 where
 d = distance
 A = acceleration
 T = time 
 Solving for T, gives
 d = 1/2 A T^2
 2d = A T^2
 2d/A = T^2
 sqrt(2d/A) = T 
 Substitute the known values and calculate.
 sqrt(2d/A) = T
 sqrt(2* 1500m / 9.8 m/s^2) = T
 sqrt(3000m / 9.8 m/s^2) = T
 sqrt(306.122449 s^2) = T
 17.49635531 s = T 
 Rounding to 4 significant figures gives 17.50 seconds. Since it will take
17.50 seconds for the kit to hit the ground, the kit needs to be dropped 17.50
seconds before the plane goes overhead. So just simply multiply by the velocity. 
17.50 s * 100 m/s = 1750 m
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A thin beam of light of wavelength 625 nm goes through a thin slit and falls on a screen 3.00 m past the slit. You observe that
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a = 2.275 10⁻⁴ m

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The first dark minimum occurs for m = 1

         a = λ  / sin θ

The angle can be found by trigonometry,

       tan θ = y / x

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Let's reduce the magnitudes to the SI system

        y = 8.24 mm = 8.24 10⁻³ m

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A 0.032-kg bullet is fired vertically at 230 m/s into a 0.15-kg baseball that is initially at rest. How high does the combined b
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Answer:

heigth=83.44m

Explanation:

Given data

Baseball mass m₁=0.15 kg

initial speed v₁=0

Bullet mass m₂=0.032 kg

final speed v₂=230 m/s

To find

height h=?

Solution

From conservation of momentum we know that

m_{1}v_{1}+m_{2}v_{2}=(m_{1}+m_{2})v\\ (0.032kg)(230m/s)+(0.15kg)(0m/s)=(0.15kg+0.032kg)v\\7.36+0=0.182v\\v=7.36/0.182\\v=40.44m/s

Now from the conservation of mechanical energy

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