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3 years ago

What is the restoring force of a spring stretched 0.35 meters with a spring contact of 55 newtons?

Physics

2 answers:

Vesna [10]3 years ago

6 0

F=-kx = -19,25 answer D

Bogdan [553]3 years ago

6 0

Answer:

F=19.25 N

Explanation:

The restoring force for a spring is given by:

$\vec{F}=-k \vec{x}$

Where k is the spring constant and $\vec{x}$ is the spring deformation.

The minus sign indicates that force is opposed to the deformation direction.

According to the problem:

$k=55 \;N/m$ and $x=0.35 \;m$

Thus, in magnitude, the restoring force is:

$F=55\frac{N}{m} *0.35m=19.25\; N$

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The curvature of the helix r(t)equals(a cosine t )iplus(a sine t )jplusbt k (a,bgreater than or equals0) is kappaequalsStartF

4vir4ik [10]

Answer:

$\kappa = \frac{1}{2 b}$

Explanation:

The equation for kappa ( κ) is

$\kappa = \frac{a}{a^2 + b^2}$

we can find the maximum of kappa for a given value of b using derivation.

As b is fixed, we can use kappa as a function of a

$\kappa (a) = \frac{a}{a^2 + b^2}$

Now, the conditions to find a maximum at $a_0$ are:

$\frac{d \kappa(a)}{da} \left | _{a=a_0} = 0$

$\frac{d^2\kappa(a)}{da^2} \left | _{a=a_0} < 0$

Taking the first derivative:

$\frac{d}{da} \kappa = \frac{d}{da} (\frac{a}{a^2 + b^2})$

$\frac{d}{da} \kappa = \frac{1}{a^2 + b^2} \frac{d}{da}(a)+ a * \frac{d}{da} (\frac{1}{a^2 + b^2} )$

$\frac{d}{da} \kappa = \frac{1}{a^2 + b^2} * 1 + a * (-1) (\frac{1}{(a^2 + b^2)^2} ) \frac{d}{da} (a^2+b^2)$

$\frac{d}{da} \kappa = \frac{1}{a^2 + b^2} * 1 - a (\frac{1}{(a^2 + b^2)^2} ) (2* a)$

$\frac{d}{da} \kappa = \frac{1}{a^2 + b^2} * 1 - 2 a^2 (\frac{1}{(a^2 + b^2)^2} )$

$\frac{d}{da} \kappa = \frac{a^2+b^2}{(a^2 + b^2)^2} - 2 a^2 (\frac{1}{(a^2 + b^2)^2} )$

$\frac{d}{da} \kappa = \frac{1}{(a^2 + b^2)^2} (a^2+b^2 - 2 a^2)$

$\frac{d}{da} \kappa = \frac{b^2 - a^2}{(a^2 + b^2)^2}$

This clearly will be zero when

$a^2 = b^2$

as both are greater (or equal) than zero, this implies

$a=b$

The second derivative is

$\frac{d^2}{da^2} \kappa = \frac{d}{da} (\frac{b^2 - a^2}{(a^2 + b^2)^2} )$

$\frac{d^2}{da^2} \kappa = \frac{1}{(a^2 + b^2)^2} \frac{d}{da} ( b^2 - a^2 ) + (b^2 - a^2) \frac{d}{da} ( \frac{1}{(a^2 + b^2)^2} )$

$\frac{d^2}{da^2} \kappa = \frac{1}{(a^2 + b^2)^2} ( -2 a ) + (b^2 - a^2) (-2) ( \frac{1}{(a^2 + b^2)^3} ) (2a)$

$\frac{d^2}{da^2} \kappa = \frac{-2 a}{(a^2 + b^2)^2} + (b^2 - a^2) (-2) ( \frac{1}{(a^2 + b^2)^3} ) (2a)$

We dcan skip solving the equation noting that, if a=b, then

$b^2 - a^2 = 0$

at this point, this give us only the first term

$\frac{d^2}{da^2} \kappa = \frac{- 2 a}{(a^2 + a^2)^2}$

if a is greater than zero, this means that the second derivative is negative, and the point is a minimum

the value of kappa is

$\kappa = \frac{b}{b^2 + b^2}$

$\kappa = \frac{b}{2* b^2}$

$\kappa = \frac{1}{2 b}$

3 0

3 years ago

The intensity of a sound wave at a fixed distance from a speaker vibrating at 1.00 kHz is 0.750 W/m2. (a) Determine the intensit

sveticcg [70]

Answer:

a) I = 3.63 W / m² , b) I = 0.750 W / m²

Explanation:

The intensity of a sound wave is given by the relation

I = P / A = ½ ρ v (2π f $s_{max}$ )²

I = (½ ρ v 4π² s_{max}²) f²

a) with the initial condition let's call the intensity Io

cte = (½ ρ v 4π² s_{max}²)

I₀ = cte s² f₀²

I₀ = cte 10 6

If frequency is increase f = 2.20 10³ Hz

I = constant (2.20 10³) 2

I = cte 4.84 10⁶

let's find the relationship of the two quantities

I / Io = 4.84

I = 4.84 Io

I = 4.84 0.750

I = 3.63 W / m²

b) in this case the frequency is reduced to f = 0.250 10³ Hz and the displacement s = 4 s or

I = cte (f s)²

I = constant (0.250 10³ 4)²

I = cte 1 10⁶

the relationship

I / Io = 1

I = Io

I = 0.750 W / m²

6 0

2 years ago

When refrigeration can’t work how does the molecules work to keep it cool

Simora [160]

The compressor constricts the refrigerant vapor, raising its pressure, and pushes it into the coils on the outside of the refrigerator. 2. When the hot gas in the coils meets the cooler air temperature of the kitchen, it becomes a liquid. ... The refrigerant absorbs the heat inside the fridge, cooling down the air.

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3 years ago

40. The emotions of this drug vary from euphoria to detachment to panic.

statuscvo [17]

Answer:

Marijuana

Explanation:

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2 years ago

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tigry1 [53]

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3 years ago

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