What is the outcome of a star that runs out of hydrogen?

You find yourself in a place that is unimaginably hot and dense. A rapidly changing gravitational field randomly warps space and

Marta_Voda [28]

You find yourself in a place that is unimaginably hot and dense. A rapidly changing gravitational field randomly warps space and time. Gripped by these huge fluctuations, you notice that there is but a single, unified force governing the universe, you are in the early universe before the Planck time.

<h3>What is Planck time?</h3>

The Planck time is approximately 10^-44 seconds. The smallest time interval, or "zeptosecond," that has so far been measured is 10^-21 seconds. A photon traveling at the speed of light would need one Planck time to traverse a distance of one Planck length.

<h3>What is Planck length?</h3>

Planck units are a set of measuring units used only in particle physics and physical cosmology. They are defined in terms of four universal physical constants in such a way that when expressed in terms of these units, these physical constants have the numerical value 1. These units are a system of natural units because its definition is based on characteristics of nature, more especially the characteristics of free space, rather than a selection of prototype object, as was the case with Max Planck's original 1899 proposal. They are pertinent to the study of unifying theories like quantum gravity.

To learn more about Plank time:

brainly.com/question/23791066

#SPJ4

8 0

1 year ago

What is the pressure drop due to the Bernoulli Effect as water goes into a 3.00-cm-diameter nozzle from a 9.00-cm-diameter fire

Marianna [84]

Answer:

The pressure and maximum height are $1.58\times10^{6}\ N/m^2$ and 161.22 m respectively.

Explanation:

Given that,

Diameter = 3.00 cm

Exit diameter = 9.00 cm

Flow = 40.0 L/s²

We need to calculate the pressure

Using Bernoulli effect

$P_{1}+\dfrac{1}{2}\rho v_{1}^2+\rho g h_{1}=P_{2}+\dfrac{1}{2}\rho v_{2}^2+\rho g h_{2}$

When two point are at same height so ,

$P_{1}+\dfrac{1}{2}\rho v_{1}^2=P_{2}+\dfrac{1}{2}\rho v_{2}^2$ ....(I)

Firstly we need to calculate the velocity

Using continuity equation

For input velocity,

$Q=A_{1}v_{1}$

$v_{1}=\dfrac{Q}{A_{1}}$

$v_{1}=\dfrac{40.0\times10^{-3}}{\pi\times(1.5\times10^{-2})^2}$

$v_{1}=56.58\ m/s$

For output velocity,

$v_{2}=\dfrac{40.0\times10^{-3}}{\pi\times(4.5\times10^{-2})^2}$

$v_{2}=6.28\ m/s$

Put the value into the formula

$P_{1}-P_{2}=\dfrac{1}{2}\rho(v_{1}^2-v_{2}^2)$

$\Delta P=\dfrac{1}{2}\times1000\times(56.58^2-6.28^2)$

$\Delta P=1.58\times10^{6}\ N/m^2$

(b). We need to calculate the maximum height

Using formula of height

$\Delta P=\rho g h$

Put the value into the formula

$1.58\times10^{6}=1000\times9.8\times h$

$h=\dfrac{1.58\times10^{6}}{1000\times9.8}$

$h=161.22\ m$

Hence, The pressure and maximum height are $1.58\times10^{6}\ N/m^2$ and 161.22 m respectively.

8 0

3 years ago

A building is being knocked down with a wrecking ball, which is a big metal sphere that swings on a 12-m-long cable. You are (un

leva [86]

Because you know that gravity is in m/s^2 so, period will be measured in seconds. You know the cable is 12m long and gravity is 9.81 solve for T (period) 2π12sqrt(9.81)=6.94922

8 0

3 years ago

Read 2 more answers

A 37 cm long solenoid, 1.8 cm in diameter, is to produce a 0.50 T magnetic field at its center. If the maximum current is 4.4 A,

taurus [48]

Answer:

33,458.71 turns

Explanation:

Given: L = 37 cm = 0.37 m, B= 0.50 T, I = 4.4 A, n= number of turn per meter

μ₀ = Permeability of free space = 4 π × 10 ⁻⁷

Solution:

We have B = μ₀ × n × I

⇒ n = B/ (μ₀ × I)

n = 0.50 T / ( 4 π × 10 ⁻⁷ × 4.4 A)

n = 90,428.94 turn/m

No. of turn through 0.37 m long solenoid = 90,428.94 turn/m × 0.37

= 33,458.71 turns

8 0

2 years ago

Read 2 more answers

PLEASE ANSWER FAST In which of the following situations is the greatest amount of work accomplished? 1. A boy lifts a 2-newton b

GuDViN [60]

Explanation:

Work done is given by the product of force and displacement.

Case 1,

1. A boy lifts a 2-newton box 0.8 meters.

W = 2 N × 0.8 m = 1.6 J

2. A boy lifts a 5-newton box 0.8 meters.

W = 5 N × 0.8 m = 4 J

3. A boy lifts a 8-newton box 0.2 meters.

W = 8 N × 0.2 m = 1.6 J

4. A boy lifts a 10-newton box 0.2 meters.

W = 10 N × 0.2 m = 2 J

Out of the four options, in option (2) ''A boy lifts a 5-newton box 0.8 meters'', the work done is 4 J. Hence, the greatest work done is 4 J.

3 0

3 years ago