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andrey2020 [161]
3 years ago
11

Did the bigbang violate the law of conservation of energy and matter?

Physics
1 answer:
Pie3 years ago
3 0
Under general relativity, there is no 'before the Big Bang'. The problem is that time is itself a part of the universe and is affected by matter and energy. Because of the huge densities just after the Big Bang, time itself is warped in such a way that it cannot go back before that event. It is somewhat like asking what is north of the north pole.

The conservation of matter and energy states that the total amount of mass and energy at one time is the same at any other time. Notice how time is a crucial part of this statement. To even talk about conservation laws, you have to have time.

The upshot is that the Big Bang did not break the conservation laws because time itself is part of the universe and started at the Big Bang and because the conservation laws need to have time in their statements.
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Four students are each pushing on one side of a box. Mara is pushing to the west, Cindi is pushing to the east, Chris is pushing
jenyasd209 [6]

Answer:

cindi

Explanation:

cindi's work done is larger than all the other students combined

5 0
2 years ago
1.- Se desea elevar un cuerpo de 1500kg utilizando una elevadora hidráulica de plato grande
kvv77 [185]

Answer:

181.48 N

Explanation:

Calculate the area :

Area = pi * r² ;

pi = 3.14 ; r1 = 90cm /100 = 0.9m ; r2 = 10/100 = 0.1m

Area 1, A1 = 3.14 * 0.1² = 0.0314 m²

Area 2, A2 = 3.14 * 0.9² = 2.5434 m²

Force, F = mass * acceleration due to gravity

F2 = 1500 * 9.8 = 14700 N

Force 1 / Area 1 = Force 2 / Area 2

Force 1 = (Force 2 / Area 2), * Area 1

Force 1 = (14700 / 2.5434) * 0.0314

Force = 5779.6650 * 0.0314

= 181.48 N

5 0
3 years ago
A sample of helium behaves as an ideal gas as energy is
alisha [4.7K]

Answer:

0.0321 g

Explanation:

Let helium specific heat c_h = 5.193 J/g K

Assuming no energy is lost in the process, by the law of energy conservation we can state that the 20J work done is from the heat transfer to heat it up from 273K to 393K, which is a difference of ΔT = 393 - 273 = 120 K. We have the following heat transfer equation:

E_h = m_hc_h \Delta T = 20 J

where m_h is the mass of helium, which we are looking for:

m_h = \frac{20}{c_h \Delta T} = \frac{20}{5.193 * 120} \approx 0.0321 g

4 0
3 years ago
The 8 kg block is then released and accelerates to the right, toward the 2 kg block. The surface is rough and the coefficient of
natita [175]

Answer:

3.258 m/s

Explanation:

k = Spring constant = 263 N/m (Assumed, as it is not given)

x = Displacement of spring = 0.7 m (Assumed, as it is not given)

\mu = Coefficient of friction = 0.4

Energy stored in spring is given by

U=\dfrac{1}{2}kx^2\\\Rightarrow U=\dfrac{1}{2}\times 263\times 0.7^2\\\Rightarrow U=64.435\ J

As the energy in the system is conserved we have

\dfrac{1}{2}mv^2=U-\mu mgx\\\Rightarrow v=\sqrt{2\dfrac{U-\mu mgx}{m}}\\\Rightarrow v=\sqrt{2\dfrac{64.435-0.4\times 8\times 9.81\times 0.7}{8}}\\\Rightarrow v=3.258\ m/s

The speed of the 8 kg block just before collision is 3.258 m/s

7 0
3 years ago
The blades in a blender rotate at a rate of 6100 rpm. When the motor is turned off during operation, the blades slow to rest in
MissTica

Answer:

<em>155.80rad/s</em>

Explanation:

Using the equation of motion to find the angular acceleration:

\omega_f = \omega_i + \alpha t

\omega_f is the final angular velocity in rad/s

\omega_i  is the initial angular velocity in rad/s

\alpha is the angular acceleration

t is the time taken

Given the following

\omega_f = 6100rpm

Time = 4.1secs

Convert the angular velocity to rad/s

1rpm = 0.10472rad/s

6100rpm = x

x = 6100 * 0.10472

x  = 638.792rad/s

\omega_f = 638.792rad/s\\

Get the angular acceleration:

Recall that:

\omega_f = \omega_i + \alpha t

638.792 = 0 + ∝(4.1)

4.1∝ = 638.792

∝ = 638.792/4.1

∝ = 155.80rad/s

<em>Hence the angular acceleration as the blades slow down is 155.80rad/s</em>

5 0
2 years ago
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