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raketka [301]
2 years ago
8

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.
Physics
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 mass m is traveling at an initial speed of 25.0 m/s. It is brought to rest in a distance of 62.5 m by a net force of 15.0 N.
harkovskaia [24]

Answer:

m = 3 kg

The mass m is 3 kg

Explanation:

From the equations of motion;

s = 0.5(u+v)t

Making t thr subject of formula;

t = 2s/(u+v)

t = time taken

s = distance travelled during deceleration = 62.5 m

u = initial speed = 25 m/s

v = final velocity = 0

Substituting the given values;

t = (2×62.5)/(25+0)

t = 5

Since, t = 5 the acceleration during this period is;

acceleration a = ∆v/t = (v-u)/t

a = (25)/5

a = 5 m/s^2

Force F = mass × acceleration

F = ma

Making m the subject of formula;

m = F/a

net force F = 15.0N

Substituting the values

m = 15/5

m = 3 kg

The mass m is 3 kg

7 0
3 years ago
if a bowling ball hits a wall a force of 6 N, the wall exerts a force of how much back. on the bowling ball
Ray Of Light [21]
There has to be 6 N for a reactant


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How far from the castle wall does the launched rock hit the ground?
lina2011 [118]
King Arthur's knights use a catapult to launch a rock from their vantage point on top of the castle wall, 14 m above the moat. The rock is launched at a speed of 27 m/s and an angle of 32degrees above the horizontal.
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3 years ago
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A(n) 1100 kg car is parked on a 4◦ incline. The acceleration of gravity is 9.8 m/s 2 . Find the force of friction keeping the ca
igomit [66]

Answer:

Force of friction, f = 751.97 N

Explanation:

it is given that,

Mass of the car, m = 1100 kg

It is parked on a 4° incline. We need to find the force of friction keeping the car from sliding down the incline.

From the attached figure, it is clear that the normal and its weight is acting on the car. f is the force of friction such that it balances the x component of its weight i.e.

f=mg\ sin\theta

f=1100\ kg\times 9.8\ m/s^2\ sin(4)

f = 751.97 N

So, the force of friction on the car is 751.97 N. Hence, this is the required solution.

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