What is the force of an object with a mass of 30 kg that is free falling?

Physics

1 answer:

disa [49]3 years ago

4 0

Answer:

F = 294.3 [N]

Explanation:

To solve this problem we must use Newton's second law which tells us that force is equal to the product of mass by acceleration. It is this particular case the acceleration is due to the gravitational acceleration since the body is in free fall.

Therefore we have:

F = m*g

where:

F = force [N]

m = mass = 30 [kg]

g = gravity acceleration = 9.81 [m/s^2]

F = 30*9.81

F = 294.3 [N]

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Suppose that Hubble's constant were H0 = 51 km/s/Mly (which is not its actual value). What would the approximate age of the univ

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Given the Hubble's constant, the approximate age of the universe is 5.88 × 10⁹ Years.

Given the data in the question;

Hubble's constant; $H_0 = 51km/s/Mly$

Age of the universe; $t = \ ?$

We know that, the reciprocal of the Hubble's constant ( $H_0$ ) gives an estimate of the age of the universe ( $t$ ). It is expressed as:

$Age\ of\ Universe; t = \frac{1}{H_0}$

Now,

Hubble's constant; $H_0 = 51km/s/Mly$

We know that;

$1\ light\ years = 9.46*10^{15}m$

$1\ Million\ light\ years = [9.46 * 10^{15}m] * 10^6 = 9.46 * 10^{21}m$

Therefore;

$H_0 = 51\frac{km}{\frac{s}{Mly} } = 51000\frac{m}{s\ *\ Mly} \\\\H_0 = 51000\frac{m}{s\ *\ (9.46*10^{21}m)} \\\\H_0 = 5.39 *10^{-18}s^{-1}\\$

Now, we input this Hubble's constant value into our equation;

$Age\ of\ Universe; t = \frac{1}{H_0}\\\\t = \frac{1}{ 5.39 *10^{-18}s^{-1}} \\\\t = 1.855 * 10^{17}s\\\\We\ convert\ to\ years\\\\t = \frac{ 1.855 * 10^{17}}{60*60*24*365}yrs \\\\t = \frac{ 1.855 * 10^{17}}{31536000}yrs\\\\t = 5.88 *10^9 years$