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3 years ago

What are the units for the spring constant, k?

Physics

2 answers:

vitfil [10]3 years ago

5 0

Answer:

C. newtons/meter

Explanation:

To find the units of spring constant you can use the hook's law, which says,

F = k × x where

F = force (unit Newton)

k = spring constant

x = expansion or contraction ( unit m)

So by this equation you can get that

k = F / X

So you get the units as N/m

Tanya [424]3 years ago

4 0

The units for the spring constant, k, are Newtons per meter (N/m).)

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Assuming typical speeds of 8.5 km/s and 5.5 km/s for P and S waves, respectively, how far away did the earthquake occur if a par

taurus [48]

Answer:

The earthquake occurred at a distance of 1122 km

Explanation:

Given;

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speed of the S wave, v₂ = 5.5 km/s

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let the time taken by the wave with greater speed = t₁

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v₁t₁ = v₂t₂

v₁t₁ = v₂(t₁ + 1.2 min)

v₁t₁ = v₂(t₁ + 72 s)

v₁t₁ = v₂t₁ + 72v₂

v₁t₁ - v₂t₁ = 72v₂

t₁(v₁ - v₂) = 72v₂

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3 years ago

Calculate the ratio of the drag force on a passenger jet flying with a speed of 1200 km/h at an altitude of 10 km to the drag fo

Sonbull [250]

Answer:

$2.267$

Explanation:

Drag force is given by

$F=\dfrac{1}{2}\rho Av^2C$

C = Drag coefficient is constant

A = Area is constant

$v_1$ = Velocity of the passenger jet = 1200 km/h = $\dfrac{1200}{3.6}\ \text{m/s}$

$v_2$ = Velocity of the prop plane = $\dfrac{1}{4}v_1$

$\rho_1$ = Density of the air where the jet was flying = $0.38\ \text{kg/m}^3$

$\rho_2$ = Density of the air where the prop plane was flying = $0.67\ \text{kg/m}^3$

$F\propto \rho v^2$

$\dfrac{F_1}{F_2}=\dfrac{\rho_1 v_1^2}{\rho_2 v_2^2}\\\Rightarrow \dfrac{F_1}{F_2}=\dfrac{0.38 v_1^2}{0.67 (\dfrac{1}{4}v_1^2)}\\\Rightarrow \dfrac{F_1}{F_2}=2.267$

The ratio of the drag forces is $2.267$ .

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3 years ago

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3 years ago