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3 years ago

Why does the background radiation in the universe have a temperature of 3ºC above absolute zero?

1 answer:

5 0

The background radiation is associated with a temperature of about 2.73 Kelvins (not Celsius). There is no particular reason for it to be at this specific value. This temperature is a snapshot of an ongoing process taken at our point in time. This temperature continues to change from where it used to be - many orders of magnitude higher in the early stages of our universe model - to today's measured value. The expanding universe view implies this temperature continues to decline as the universe expands, as you pointed out. It is just happening on an extremely slow time scale that we consider this value to be a constant.

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Both a fire blanket and the spray from a fire extinguisher put out a fire by pressing down on the fire and pushing any oxygen aw

Sladkaya [172]

Answer:

Water can carry electricity

Explanation:

While pure water does not conduct electricity, ionized water does. It is common that water is ionized than is pure, unless it is purified. The dissolved ions in the water are what makes the water conduct electricity. Handling electric equipment with wet hands may lead to electrocution.

7 0

4 years ago

Read 2 more answers

If the time of the impact in a collision is extended by 4 times ,how much does the force of impact change

Rama09 [41]

We know, I = F.Δt
As Δt is increased to 4 times, then, F would decrease to 4 times, in order to keep that impulse constant.

In short, Your force will change to 1/4th of it's initial value

Hope this helps!

7 0

3 years ago

Jumping up before the elevator hits. After the cable snaps and the safety system fails, an elevator cab free-falls from a height

noname [10]

Answer:

a) I = 2279.5 N s , b) F = 3.80 10⁵ N, c) I = 3125.5 N s and d) F = 5.21 10⁵ N

Explanation:

The impulse is equal to the variation in the amount of movement.

I =∫ F dt = Δp

I = m $v_{f}$ - m v₀

Let's calculate the final speed using kinematics, as the cable breaks the initial speed is zero

$v_{f}$ ² = V₀² - 2g y

$v_{f}$ ² = 0 - 2 9.8 30.0

$v_{f}$ = √588

$v_{f}$ = 24.25 m/s

a) We calculate the impulse

I = 94 24.25 - 0

I = 2279.5 N s

b) Let's join the other expression of the impulse to calculate the average force

I = F t

F = I / t

F = 2279.5 / 6 10⁻³

F = 3.80 10⁵ N

just before the crash the passenger jumps up with v = 8 m / s, let's take the moments of interest just when the elevator arrives with a speed of 24.25m/s down and as an end point the jump up to vf = 8 m / n

c) I = m $v_{f}$ - m v₀

I = 94 8 - 94 (-24.25)

I = 3125.5 N s

d) F = I / t

F = 3125.5 / 6 10⁻³

F = 5.21 10⁵ N

7 0

3 years ago

Worldâs tallest building. suppose that you drop a marble from the top of the burj khalifa building in dubai, which is about 830

-Dominant- [34]

<span>a) 13 seconds b) 130 m/s The formula for the distance an object moves while under constant acceleration is d = 1/2AT^2. So let's define d as 830 m, A as 9.8m/s^2, and solve for T 830 m = 1/2 9.8 m/s^2 T^2 830 m = 4.9 m/s^2 T^2 Divide both sides by 4.9 m/s^2 169.3878 s^2 = T^2 Take the square root of both sides 13.01491 s = T Since we only have 2 significant figures, round the result to 13 seconds which is the answer to the first part of the question. To find out how fast the marble is moving, just multiply T and A together 13 s * 9.8 m/s^2 = 127.4 m/s Since we only have 2 significant figures, round the result to 130 m/s.</span>

7 0

4 years ago

. Two astronauts are 1.90 m apart in their spaceship. One speaks to the other. The conversation is transmitted to earth via elec

melomori [17]

Answer:

d= 1650 km.

Explanation:

Assuming that sound waves travel along a straight line at a constant speed of 346 m/s, we can find the time needed for the sound to travel the distance of 1.9 m between the astronauts, just applying the definition of average velocity, as follows:

$t = \frac{\Delta x}{v} =\frac{1.9m}{346m/s} = 5.5 msec (1)$

The electromagnetic waves travel in free space at the same speed of light in vacuum ( since light is a electromagnetic wave indeed), i.e., 3*10⁸ m/s.
Applying the same formula than in (1) we can solve for the distance d, as follows:

$d = v*t = 3e8 m/s*5.5e-3s = 1650 km (2)$

3 0

3 years ago