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3 years ago

What is the relationship between two variables if the product of the variables is constant?

Physics

1 answer:

OverLord2011 [107]3 years ago

6 0

Answer:

If the product of two variables (x ⋅ y) is equal to a constant (k = x ⋅ y), then the two are said to be inversely proportional to each other with the proportionality constant k

Explanation:

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A 2000 kg car moves along a horizontal road at speed vo

cluponka [151]

Answer:

The shortest possible stopping distance of the car is 175.319 meters.

Explanation:

In this case we see that driver use the brakes to stop the car by means of kinetic friction force. Deceleration of the car is directly proportional to kinetic friction coefficient and can be determined by Second Newton's Law:

$\Sigma F_{x} = -\mu_{k}\cdot N = m \cdot a$ (Eq. 1)

$\Sigma F_{y} = N-m\cdot g = 0$ (Eq. 2)

After quick handling, we get that deceleration experimented by the car is equal to:

$a = -\mu_{k}\cdot g$ (Eq. 3)

Where:

$a$ - Deceleration of the car, measured in meters per square second.

$\mu_{k}$ - Kinetic coefficient of friction, dimensionless.

$g$ - Gravitational acceleration, measured in meters per square second.

If we know that $\mu_{k} = 0.0735$ and $g = 9.807\,\frac{m}{s^{2}}$ , then deceleration of the car is:

$a = -(0.0735)\cdot (9.807\,\frac{m}{s^{2}} )$

$a = -0.721\,\frac{m}{s^{2}}$

The stopping distance of the car ( $\Delta s$ ), measured in meters, is determined from the following kinematic expression:

$\Delta s = \frac{v^{2}-v_{o}^{2}}{2\cdot a}$ (Eq. 4)

Where:

$v_{o}$ - Initial speed of the car, measured in meters per second.

$v$ - Final speed of the car, measured in meters per second.

If we know that $v_{o} = 15.9\,\frac{m}{s}$ , $v = 0\,\frac{m}{s}$ and $a = -0.721\,\frac{m}{s^{2}}$ , stopping distance of the car is:

$\Delta s = \frac{\left(0\,\frac{m}{s} \right)^{2}-\left(15.9\,\frac{m}{s} \right)^{2}}{2\cdot \left(-0.721\,\frac{m}{s^{2}} \right)}$

$\Delta s = 175.319\,m$

The shortest possible stopping distance of the car is 175.319 meters.

8 0

4 years ago

A satellite orbiting the earth is directly over a point on the equator at 12:00 midnight every four days. It is not over that po

Mrrafil [7]

Answer:

106417026.88435 m

Explanation:

T = Time period of the satellite = 4 days

m = Mass of the Earth = 5.972 × 10²⁴ kg

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

Time period is given by

$T=2\pi\sqrt{\dfrac{r^3}{GM}}\\\Rightarrow r=\left(\dfrac{T^2GM}{4\pi ^2}\right)^{\dfrac{1}{3}}\\\Rightarrow r=\left(\dfrac{(4\times 24\times 60\times 60)^2\times 6.67\times 10^{-11}\times 5.972\times 10^{24}}{4\pi ^2}\right)^{\dfrac{1}{3}}\\\Rightarrow r=106417026.88435\ m$