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3 years ago

The big bang theory has finally answered one of the biggest questions of science—the origin of the universe.

Physics

1 answer:

ss7ja [257]3 years ago

6 0

Answer:

False

Explanation:

According to the big bang theory, matter was an infinitely small and very high density point which at one point exploded and expanded in all directions, creating what we know as our Universe, which also includes space and time . This happened about 13.8 billion years ago. Theoretical physicists have managed to reconstruct this chronology of events from 1/100 of a second after the Big Bang. After the explosion, while the Universe expanded, it cooled sufficiently and the first subatomic particles were formed: Electrons, Positrons, Mesons, Barions, Neutrinos, Photons among others. Today more than 90 particles are known. This theory solves many unknowns and is very well received by the scientific community, however there is still much to solve, for example, one of the great unsolved scientific problems in the expanding Universe model is whether the Universe is open or closed.

An attempt to solve this problem is to determine if the average density of matter in the Universe is greater than the critical value in Friedmann's model. The mass of a galaxy can be measured by observing the movement of its stars; multiplying the mass of each galaxy by the number of galaxies, it is seen that the density is only 5 to 10% of the critical value.

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A 50 kg skydiver is falling downwards and accelerating 6 m/s2 down. What is the net force on the skydiver?

Montano1993 [528]

Net Force = (mass) x (acceleration) (Newton #2)

Net Force = (50 kg) x (6 m/s² down)

Net Force = (50 * 6) (kg-m/s² down)

<em>Net Force = 300 Newtons down</em>

6 0

3 years ago

Two point charges, a +45nC charge X and a +12nC charge Y are separated by a distance of 0.5m.

Gnoma [55]

A) Calculate the resultant electric field strength at the midpoint between the charges.

Qx is the charge at X and Qy is the charge at Y.

E at midpoint = k×Qx/0.25² - k×Qy/0.25²

k = 9×10⁹Nm²C⁻², Qx = 45nC, Qy = 12nC

E = 4752N/C

Well done.

B) Calculate the distance from X at which the electric field strength is zero.

Let D be some point between X and Y for which the net E field is 0.

Let d be the distance from X to D.

Set up the following equation:

E at D = k×Qx/d² - k×Qy/(0.5-d)² = 0

Do some algebra to solve for d:

k×Qx/d² = k×Qy/(0.5-d)²

Qx/d² = Qy/(0.5-d)²

Qx(0.5-d)² = Qyd²

(0.5-d)√Qx = d√Qy

0.5√Qx-d√Qx = d√Qy

d(√Qx+√Qy) = 0.5√Qx

d = (0.5√Qx)/(√Qx+√Qy)

Plug in Qx = 45nC, Qy = 12nC

d ≈ 330mm

C) Calculate the magnitude of the electric field strength at the point P on the diagram below.

First determine the angles of the triangle. The sides of the triangle are 0.3m, 0.4m, and 0.5m, so this is a right triangle where the angle between the 0.3m and 0.4m sides is 90°

∠Y = tan⁻¹(0.4/0.3) = 53.13°

∠X = 90-∠Y = 36.87°

Determine the horizontal component of E at P:

Ex = E from Qx × cos(∠X) - E from Qy × cos(∠Y)

Ex = k×Qx/0.4²×cos(36.87°) - k×Qy/0.3²×cos(53.13°)

Ex = 1305N/C

Determine the vertical component of E at P:

Ey = E from Qx × sin(∠X) - E from Qy × sin(∠Y)

Ey = k×Qx/0.4²×sin(36.87°) - k×Qy/0.3²×sin(53.13°)

Ey = 2479N/C

Use the Pythagorean theorem to determine the magnitude of E at P:

E = √(Ex²+Ey²)

E ≈ 2802N/C

4 0

3 years ago

12) Water flows through a horizontal pipe of cross-sectional area 10.0 cm2 at a pressure of 0.250 atm with a flow rate is 1.00 L

masha68 [24]

Answer:

The pressure after passing the valve is 23,8 [Kpa] ( 0,234 atm) and the pressure drop is about 1,53 [Kpa]

Explanation:

We need to use the formula of bernoulli, in the attached image we can see the fluid throw the pipe, we also can calculate the velocity inside the pipe using the flow rate and the cross sectional area.

For this case, we don't use the elevation difference and therefore those terms can be cancelled.

When the area has reduced the velocity of the fluid is increased but there is a drop pressure through the valve.

5 0

3 years ago

The voltage ???? in a simple electrical circuit is slowly decreasing as the battery wears out. The resistance ???? is slowly inc

zimovet [89]

Answer:

The change in current at $R =456 \Omega$ is $\frac{dI}{dt} = 7.032 * 10^{-5} A/s$

Explanation:

From the question we are told that

The resistance is $R = 465 \Omega$

The current is $I = 0.09A$

The change in voltage with respect to time is $\frac{dV}{dt} = - 0.03 V/s$

The change in resistance with time is $\frac{dR}{dt} = 0.03 \Omega /s$

According to ohm's law

$V = IR$

differentiating with respect to time using chain rule

$\frac{dV}{dt} = I \frac{dR}{dt} + R * \frac{dI}{dt}$

substituting value at R = 456

$-0.0327 = 0.09 * 0.03 + 456* \frac{dI}{dt}$

$\frac{dI}{dt} = 7.032 * 10^{-5} A/s$

6 0

3 years ago

Two atoms that are isotopes of one another must have the same number of what?

fredd [130]

Answer:

Explanation:

Two atoms which are isotopes of one another must have a different number of neutrons.

Isotopes are defined as atoms of the same element which have the same numbers of protons i.e. atomic number remains the same, but has different numbers of neutrons. It is observed that they have same chemical properties due to the same electronic configuration but physical properties differs.

3 0

3 years ago