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emmainna [20.7K]
3 years ago
6

Before the big bang, the universe was much ____ and _____ than it is now.

Physics
2 answers:
kherson [118]3 years ago
7 0
My answer is that Before the big bang, the universe was much cooler and denser. There are three circumstances before the big bang theory, and it shows that the flatness problem and horizon problem affect the universe to be cooler and denser. In flatness problem, it states there that the universe seems to have enough energy density in the form of matter and in horizon problem it affects the coolness of the universe. 
Dahasolnce [82]3 years ago
5 0
The answer to the question is that before the big bang, the universe was much hotter and more dense than it is now. Letter B.
It is because after the big bag occurred, the universe became cooler and less dense.
a. - does not correspond in the answer because the universe became less dense after the big bang.
c - the universe became cool and less dense after the big bang so being cool and less dense does not correspond to the question.
d - cooler does not answer the question because it only became cooler after the big bang.
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Two charged objects are 1 meter apart. Calculate the magnitude of the electric force between them if the two charges are +1.0 μC
Solnce55 [7]

Answer:

0.00899 N

Explanation:

The magnitude of the electrostatic force between two charges is given by the equation:

F=k\frac{q_1 q_2}{r^2}

where:

k=8.99\cdot 10^9 Nm^{-2}C^{-2} is the Coulomb's constant

q_1, q_2 are the charges

r is the distance between the two charges

And the force is:

- Repulsive if the two charges have same sign

- Attractive if the two charges have opposite sign

In this problem we have:

q_1=1.0\mu C = 1.0\cdot 10^{6}C (charge of object 1)

q_2=1.0\mu C = 1.0\cdot 10^{6}C (charge of object 2)

r = 1 m (separation between the objects)

So, the electric force is

F=(8.99\cdot 10^9)\frac{(1.0\cdot 10^{-6})^2}{1^2}=0.00899 N

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2 years ago
A violin string vibrates at 260 Hz when unfingered. At what frequency will it vibrate if it is fingered one fourth of the way do
Advocard [28]

Answer:

346.66 Hz

Explanation:

l_1 = Length of string which is unfingered = l

l_2 = Length of string which is vibrate when fingered = l-\dfrac{1}{4}l=\dfrac{3}{4}l

f_1 = Unfingered frequency = 260 Hz

f_2 = Fingered frequency

Frequency is inversely proportional to length

f=\dfrac{1}{l}

So,

\dfrac{f_1}{f_2}=\dfrac{l_2}{l_1}\\\Rightarrow \dfrac{f_1}{f_2}=\dfrac{\dfrac{3}{4}l}{l}\\\Rightarrow \dfrac{f_1}{f_2}=\dfrac{3}{4}\\\Rightarrow f_2=\dfrac{4}{3}f_1\\\Rightarrow f_2=\dfrac{4}{3}260\\\Rightarrow f_2=346.66\ Hz

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