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Nookie1986 [14]
3 years ago
7

A food packet is dropped from a helicopter during a flood-relief operation from a height of 750 meters. Assuming no drag (air fr

iction), what will the velocity of the packet be when it reaches the ground? Also, at what height will the packet have half this velocity?
Physics
1 answer:
Leviafan [203]3 years ago
3 0

1. Velocity at which the packet reaches the ground: 121.2 m/s

The motion of the packet is a uniformly accelerated motion, with constant acceleration a=g=9.8 m/s^2 directed downward, initial vertical position d=750 m, and initial vertical velocity v_0 = 0. We can use the following SUVAT equation to find the final velocity of the packet after travelling for d=750 m:

v_f^2 -v_i^2 =2ad

substituting, we find

v_f^2 = 2ad\\v_f = \sqrt{2ad}=\sqrt{2(9.8 m/s^2)(750 m)}=121.2 m/s

2. height at which the packet has half this velocity: 562.6 m

We need to find the heigth at which the packet has a velocity of

v_f=\frac{121.2 m/s}{2}=60.6 m/s

In order to do that, we use again the same SUVAT equation substituting v_f with this value, so that we find the new distance d that the packet travelled from the helicopter to reach this velocity:

v_f^2-v_i^2=2ad\\d=\frac{v_f^2}{2a}=\frac{(60.6 m/s)^2}{2(9.8 m/s^2)}=187.4 m

Which means that the heigth of the packet was

h=750 m-187.4 m=562.6 m

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Explanation:

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The motion of a car on a position-time graph is represented with a horizontal line. What does this indicate about the car's moti
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A house brick has a volume of 1900 cm³ and a weight in air of 80N.What is its apparent weight in water?The density of water is 1
Shtirlitz [24]

Answer:

61 N

Explanation:

We'll begin by calculating the mass of the brick when placed in water. This can be obtained as follow:

Volume of brick = 1900 cm³

Density of water = 1 g/cm³

Mass of brick in water =…?

Density = mass / volume

1 = mass of brick in water / 1900

Cross multiply

Mass of brick in water = 1 × 1900

Mass of brick in water = 1900 g

Next, we shall convert 1900 g to Kg.

1000 g = 1 Kg

Therefore,

1900 g = 1900 g × 1 Kg / 1000 g

1900 g = 1.9 Kg

Next, we shall determine the weight in water. This can be obtained as follow:

Mass (m) = 1.9 Kg

Acceleration due to gravity (g) = 10 m/s²

Weight (W) =?

W = m × g

W = 1.9 × 10

W = 19 N

Thus, the weight of the brick in water is 19 N.

Finally, we shall determine the apparent weight of the brick in water. This can be obtained as follow:

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3 years ago
Two small nonconducting spheres have a total charge of 93.0 μC . Part A
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Answer:

charge on each

Q1 = 2.06 ×10^{-5}  C

Q2 = 7.23 × 10^{-5} C

when force were attractive

Q1 = 1.07 × 10^{-4} C

Q2 = -1.39 × 10^{-5} C

Explanation:

given data

total charge = 93.0 μC

apart distance r  = 1.14 m

force exerted  F = 10.3 N

to find out

What is the charge on each and What if the force were attractive

solution

we know that force is repulsive mean both sphere have same charge

so total charge on two non conducting sphere is

Q1 + Q2 = 93.0 μC  = 93 ×10^{-6} C

and

According to Coulomb's law force between two sphere is

Force F = \frac{K Q1 Q2 }{r^2}      .........1

Q1Q2 = \frac{F*r^2}{k}

here F is force and r is apart distance and k is 9 × 10^{9} N-m²/C² put all value we get

Q1Q2 = \frac{ 10.3*1.14^2}{9*10^9}

Q1Q2 = 1.49 ×  10^{-9} C²

and

we have  Q2 = 93 ×10^{-6} C - Q1

put here value

Q1²  - 93 ×10^{-6}  Q1 + 1.49 ×  10^{-9} = 0

solve we get

Q1 = 2.06 ×10^{-5}  C

and

Q1Q2 = 1.49 ×  10^{-9}

2.06 ×10^{-5}  Q2 = 1.49 ×  10^{-9}

Q2 = 7.23 × 10^{-5} C

and

if force is attractive we get here

Q1Q2 = - 1.49 ×  10^{-9} C²

then

Q1²  - 93 ×10^{-6}  Q1 - 1.49 ×  10^{-9} = 0

we get here

Q1 = 1.07 × 10^{-4} C

and

Q1Q2 = - 1.49 ×  10^{-9}

2.06 ×10^{-5}  Q2 = - 1.49 ×  10^{-9}

Q2 = -1.39 × 10^{-5} C

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