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Triss [41]
3 years ago
7

Is density exponentially proportional to volume or inversely?

Physics
1 answer:
djverab [1.8K]3 years ago
8 0
An exponential relationship takes the form y = k(a^x), where a and b are constants. Often the value of a turns out to be Euler's number, e = 2.718281828..., because it has the special property that d/dx(e^x) = e^x. Exponential growth might model how the population of a bacterium which divides every 5 seconds increases, and exponential decay might model how the mass of a radioactive isotope sample decreases with time.

Inverse proportion takes the form

An inverse proportion takes the form y = k/x, for a constant k. This kind of relationship might model how the time taken to complete a job varies as the number of workers varies, for example.

We know the formula for density is mass over volume (d = m/v). We therefore see that when mass is considered an arbitrary constant, substitute y = d, k = m and x = v, and we have our inverse proportion.

Intuitively this makes sense. Double the volume for the same mass and the density halves. Quarter the volume for the same mass and the density quadruples.

I hope this helps you :)
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A baseball is batted from a height of 1.09 m with a speed of
kobusy [5.1K]

(a) The horizontal and vertical components of the ball’s initial velocity is 37.8 m/s and 12.14 m/s respectively.

(b) The maximum height above the ground reached by the ball is 8.6 m.

(c) The distance off course the ball would be carried is 0.38 m.

(d) The ball's velocity after 2.0 seconds if there is no crosswind is 38.53 m/s.

<h3>Horizontal and vertical components of the ball's velocity</h3>

Vx = Vcosθ

Vx = 39.7 x cos(17.8)

Vx = 37.8 m/s

Vy = Vsin(θ)

Vy = 39.7 x sin(17.8)

Vy = 12.14 m/s

<h3>Maximum height reached by the ball</h3>

H = \frac{v^2 sin^2(\theta)}{2g} \\\\H = \frac{(39.7)^2 \times (sin17.8)^2}{2(9.8)} \\\\H = 7.51 \ m

Maximum height above ground = 7.51 + 1.09 = 8.6 m

<h3>Distance off course after 2 second </h3>

Upward speed of the ball after 2 seconds, V = V₀y - gt

Vy = 12.14 - (2x 9.8)

Vy = - 7.46 m/s

Horizontal velocity will be constant = 37.8 m/s

Resultant speed of the ball after 2 seconds = √(Vy² + Vx²)

V = \sqrt{(-7.46)^2 + (37.8)^2} \\\\V = 38.53 \ m/s

<h3>Resultant speed of the ball and crosswind</h3>

V = \sqrt{38.52^2 + 4^2} \\\\V = 38.72 \ m/s

<h3>Distance off course the ball would be carried</h3>

d = Δvt = (38.72 - 38.53) x 2

d = 0.38 m

The ball's velocity after 2.0 seconds if there is no crosswind is 38.53 m/s.

Learn more about projectiles here: brainly.com/question/11049671

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