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marin [14]
3 years ago
15

An airplane traveling 919 m above the ocean at 505 m/s is to drop a case of twinkies to the victims below. How much time before

being directly overhead should the box be dropped? What is the horizontal distance between the plane and the victims when the box is dropped?
Physics
1 answer:
Elena L [17]3 years ago
6 0

Answer:

a. 13.7 s b. 6913.5 m

Explanation:

a. How much time before being directly overhead should the box be dropped?

Since the box falls under gravity we use the equation

y = ut - 1/2gt² where y = height of plane above ocean = 919 m, u = initial vertical velocity of airplane = 0 m/s, g = acceleration due to gravity = -9.8 m/s² and t = time it takes the airplane to be directly overhead.

So,

y = ut - 1/2gt²

y = 0 × t - 1/2gt²

y = 0 - 1/2gt²

y = - 1/2gt²

t² = -2y/g

t = √(-2y/g)

So, t = √(-2 × 919 m/-9.8 m/s²)

t = √(-1838 m/-9.8 m/s²)

t = √(187.551 m²/s²)

t = 13.69 s

t ≅ 13.7 s

So, the box should be dropped 13.69 s before being directly overhead.

b. What is the horizontal distance between the plane and the victims when the box is dropped?

The horizontal distance x between plane and victims, x = speed of plane × time it takes for box to drop = 505 m/s × 13.69 s = 6913.45 m ≅ 6913.5 m

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serg [7]

Answer:

b= 2.14 m

Explanation:

Given that

Weight of the board ,wt = 40 N

Wight of the first children , wt₁=500 N

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The distance of the 500 N child from center ,a= 1.5 m

lets take distance of the 350 N child from center = b m

Now by taking the moment about the center of the board

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b= 2.14 m

Therefore the distance of the 350 N weight child from the center is 2.14 m.

5 0
3 years ago
Tendons are strong elastic fibers that attach muscles to bones. To a reasonable approximation, they obey Hooke's law. In laborat
meriva

Answer:

k = 200 N/m

Explanation:

given,

mass of the object  = 251 g

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using the formula

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where

k is the force constant of the tendon

F = m g

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2.4598 = k x 0.0123

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k = 200 N/m

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3 years ago
A 55-kg block, starting from rest, is pushed a distance of 5.0 m across a floor by a horizontal force Fp whose magnitude is 140
max2010maxim [7]

Answer:

The answer is "151.25 J and -547.64 J".

Explanation:

u = 0\\\\v = 2.35\  \frac{m}{sec}\\\\d = 5.0 \ m\\\\

Using formula:

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The above work is converted into thermal energy.

Now,

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Answer:

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