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2 years ago

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What would be your estimate of the age of the universe if you measured a value for Hubble's constant of H0 = 30 km/s/Mly ? You c

an assume that the expansion rate has remained unchanged during the history of the universe.

1 answer:

Maksim231197 [3]2 years ago

6 0

Answer:

The age of the universe would be 9.9 billion years

Explanation:

We can calculate an estimate for the age of the Universe from Hubble's Law. Let's suppose the distance between two galaxies is D and the apparent velocity with which they are separating from each other is v. At some point, the galaxies were touching, and we can consider that time the moment of the Big Bang.

Thus, the time it has taken for the galaxies to reach their current separations is:

$\displaystyle{t=D/v}$

and from Hubble's Law:

$v =H_0D$

Therefore:

$\displaystyle{t=D/v=D/(H_0\times D)=1/H_0}$

With the given value for the Hubble's constant we have:

$H_0=(30\ km/s/Mly) \times (1 Mly/ 9.461 \times 10^{18} km) = 3.17\times 10^{-18}\ 1/s$

and thus,

$t=1/H_0 = 1/(3.17\times 10^{-18} 1/s) = 0.315 \times 10^{18}\ s \approx 9988584474.8858\ years \approx 9.9\ billion\ years$

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A force of 3600 N is exerted on a piston that has an area of 0.030 m2. What force is exerted on a second piston that has an area

Airida [17]

For Pascal's law, the pressure is transmitted with equal intensity to every part of the fluid:
$p_1 = p_2$
which becomes
$\frac{F_1}{A_1}= \frac{F_2}{A_2}$
where
$F_1=3600 N$ is the force on the first piston
$A_1=0.030 m^2$ is the area of the first piston
$F_2$ is the force on the second piston
$A_2=0.015 m^2$ is the area of the second piston

If we rearrange the equation and we use these data, we can find the intensity of the force on the second piston:
$F_2=F_1 \frac{A_2}{A_1}=(3600 N) \frac{0.015 m^2}{0.030 m^2}= 1800 N$

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3 years ago

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Kalea throws a baseball directly upward at time t = 0 at an initial speed of 13.7 m/s. How high h does the ball rise above its r

Trava [24]

Answer:

h = 9.57 seconds

Explanation:

It is given that,

Initial speed of Kalea, u = 13.7 m/s

At maximum height, v = 0

Let t is the time taken by the ball to reach its maximum point. It cane be calculated as :

$v=u-gt$

$u=gt$

$t=\dfrac{u}{g}$

$t=\dfrac{13.7}{9.8}$

t = 1.39 s

Let h is the height reached by the ball above its release point. It can be calculated using second equation of motion as :

$h=ut+\dfrac{1}{2}at^2$

Here, a = -g

$h=ut-\dfrac{1}{2}gt^2$

$h=13.7\times 1.39-\dfrac{1}{2}\times 9.8\times (1.39)^2$

h = 9.57 meters

So, the height attained by the ball above its release point is 9.57 meters. Hence, this is the required solution.

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3 years ago

An 75-kg hiker climbs to the summit of Mount Mitchell in western North Carolina. During one 2.0-h period, the climber's vertical

Ostrovityanka [42]

Answer:

The change in gravitational potential energy of the climber-Earth system is $\Delta PE = 396900 \ J$

Explanation:

From the question we are told that

The mass of the hiker is $m = 75 \ kg$

The time taken is $T = 2 \ hr = 2 * 3600 = 7200 \ s$

The vertical elevation after time T is $H = 540 \ m$

The change in gravitational potential is mathematically represented as

$\Delta PE = mgH$

here g is the acceleration due to gravity with value $g = 9.8 \ m/s^2$

substituting values

$\Delta PE = 75 * 9.8 * 540$

$\Delta PE = 396900 \ J$

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3 years ago

Which of the following is a unit of rotational speed?

gladu [14]

C. Rotations per second
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2 years ago

When a 5 kg rock is dropped from a height of 6 m on Planet X, it loses 24 J of GPE. What is the acceleration due to gravity on P

yan [13]

Answer:

g = 1.25m/s²

Explanation:

Given the following data;

Mass = 5kg

Height = 6m

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To find the acceleration due to gravity;

Potential energy can be defined as an energy possessed by an object or body due to its position.

Mathematically, potential energy is given by the formula;

$P.E = mgh$

Where,

P.E represents potential energy measured in Joules.

m represents the mass of an object.

g represents acceleration due to gravity measured in meters per seconds square.

h represents the height measured in meters.

GPE = mgh

Substituting into the equation, we have;

24 = 5*6*g

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g = 30/24

g = 1.25m/s²

Therefore, the acceleration due to gravity on Planet X is 1.25m/s².

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