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snow_tiger [21]
3 years ago
5

Como se encuentra el agua a una temperatura menor de 0ºC

Physics
1 answer:
Rudiy273 years ago
6 0

Answer:

yes

Explanation:

Yes, water can stay liquid below zero degrees Celsius. There are a few ways in which this can happen.

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“A place where things are baked”

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2 years ago
What’s the name of the “super continent” that broke up about 225 million years ago?
Salsk061 [2.6K]

pangeea.............


6 0
3 years ago
Read 2 more answers
A horizontal force of 200 N is applied to move a 55 kg television set across a 10 m level surface. What is the work done by the
vivado [14]

Answer:

The work done is "2000 J".

Explanation:

The given values are:

Force,

F = 200 N

Mass,

m = 55 kg

Displacement,

d = 10 m

Now,

The work done will be:

⇒  Work \ done= Force\times displacement

On substituting the given values, we get

⇒                     =200\times 10

⇒                     =2000 \ J

3 0
3 years ago
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
bija089 [108]

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

so

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

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

Learn more: brainly.com/question/14019680

6 0
2 years ago
A 0.5 kg rock is dropped from a height of 1.0 m above the ground. Approximately how much kinetic energy will be stored in the ro
irina1246 [14]

Answer:

2.45 J

Explanation:

The following data were obtained from the question:

Mass (m) = 0.5 kg

Height (h) = 1 m

Kinetic energy (KE) =?

Next, we shall determine the velocity of the rock after it has fallen half way. This can be obtained as follow:

Initial velocity (u) = 0 m/s

Acceleration due to gravity (g) = 9.8 m/s²

Height (h) = 1/2 = 0.5 m

Final velocity (v) =?

v² = u² + 2gh

v² = 0² + (2 × 9.8 × 0.5)

v² = 9.8

Take the square root of both side

v = √9.8

v = 3.13 m/s

Finally, we shall determine the kinetic energy of the rock after it has fallen half way. This can be obtained as follow:

Mass (m) = 0.5 kg

Velocity (v) = 3.13 m/s

Kinetic energy (KE) =?

KE = ½mv²

KE = ½ × 0.5 × 3.13²

KE = 0.25 × 9.8

KE = 2.45 J

Therefore, the kinetic energy of the rock after it has fallen half way is 2.45 J

8 0
2 years ago
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