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natita [175]
4 years ago
11

using hooke's law F spring = k Δ x find the force needed to stretch a spring 2 cm if it has elastic constant of 3 N/cm. A. 2/3 N

B. 3/2 N C. 6 N D. 3 N
Physics
1 answer:
Assoli18 [71]4 years ago
4 0
Data:
<span>Hooke represented mathematically his theory with the equation:
 
F = K * x
 
On what:
F (elastic force) = ?
K (elastic constant) = 3 N/cm
x (deformation or elongation of the elastic medium) = 2cm

Solving:

</span>F = K * x
F = 3 \frac{N}{\diagup\!\!\!\!\!\!cm}  * 2\diagup\!\!\!\!\!\!cm
\boxed{\boxed{F = 6N}}

Answer:
<span>C. 6 N</span>
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Answer:

Tension in the cable is T = 16653.32 N

Explanation:

Give data:

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Velocity V = 4.3 m/s

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Density \rho \ of\  water= 1000 kg/m3

 Drag force FD is given as

F_{D} = \fracP{1}{2} \rho v^{2} C_{D} A

        = 0.5\times 1000\times 4.32\times  1.2\times 1.3

Drag force = 14422.2 N acting opposite to the motion

As cable made angle  of 30 degree with horizontal  thus horizontal component is take into action to calculate drag force

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3 years ago
A hot air balloon is traveling vertically upward at a constant speed of 4.5 m/s. When it is 28 m above the ground, a package is
ella [17]

Answer:

1.97 seconds

Explanation:

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u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.8 m/s²

s=ut+\frac{1}{2}at^2\\\Rightarrow 21=4.5t+\frac{1}{2}\times -9.8\times t^2\\\Rightarrow 21=-4.5t-4.9t^2\\\Rightarrow 4.9t^2+4.5t-28=0\\\Rightarrow 49t^2+45t-280=0

Solving the above equation we get

t=\frac{-45+\sqrt{45^2-4\cdot \:49\left(-280\right)}}{2\cdot \:49}, \frac{-45-\sqrt{45^2-4\cdot \:49\left(-280\right)}}{2\cdot \:49}\\\Rightarrow t=1.97, -2.89

So, the time the package was in the air is 1.97 seconds

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