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Flura [38]
3 years ago
11

Why do skydivers use parachutes ???

Physics
2 answers:
Ksivusya [100]3 years ago
8 0
The parachutes slow them down as they fall, therefore making it safer.

hope that helped
Ierofanga [76]3 years ago
7 0
The parachute help go down slowly. it catches the air and slows it down

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Sindrei [870]

Answer: _____= beautiful, yet annoying frustrating death

Explanation:

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A small but measurable current of 5.8 × 10-10 A exists in a copper wire whose diameter is 3.0 mm. The number of charge carriers
swat32

Answer:

a) The current density ,J = 2.05×10^-5

b) The drift velocity Vd= 1.51×10^-15

Explanation:

The equation for the current density and drift velocity is given by:

J = i/A = (ne)×Vd

Where i= current

A = Are

Vd = drift velocity

e = charge ,q= 1.602 ×10^-19C

n = volume

Given: i = 5.8×10^-10A

Raduis,r = 3mm= 3.0×10^-3m

n = 8.49×10^28m^3

a) Current density, J =( 5.8×10^-10)/[3.142(3.0×10^-3)^2]

J = (5.8×10^-10) /(2.83×10^-5)

J = 2.05 ×10^-5

b) Drift velocity, Vd = J/ (ne)

Vd = (2.05×10^-5)/ (8.49×10^28)(1.602×10^-19)

Vd = (2.05×10^-5)/(1.36 ×10^10)

Vd = 1.51× 10^-5

8 0
3 years ago
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(answer fast and correctly and i will give you brainiest and 5 star's. )
oee [108]

Answer:

Energy cannot be created or destroyed.

Explanation:

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3 years ago
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If it requires 2.0 J of work to stretch a particular spring by 2.0 cm from its equilibrium length, how much more work will be re
valina [46]

Answer:

16 J

Explanation:

It is given that,

Work done, W = 2 J

A spring is stretched by 2.0 cm from its equilibrium length

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Let k is the spring constant of the spring. When W = 2J, and x = 2 cm, then energy required to stretch the spring is :

U=\dfrac{1}{2}kx^2\\\\k=\dfrac{2U}{x^2}\\\\k=\dfrac{2(2)}{(0.02)^2}\\\\k=10000\ N/m

The energy required to stretch the spring from 2 cm to additional 4 cm i.e. 2+4= 6 cm.

W=\dfrac{1}{2}k(x_2^2-x_1^2)\\\\=\dfrac{1}{2}\times 10000\times ((0.06)^2-(0.02)^2)\\\\W=16\ J

So, the required work done is 16 J.

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3 years ago
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Answer:

1 kg at 1 m/s

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