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ASHA 777 [7]
3 years ago
7

If Hubble's constant is 73 km/s/Mpc then the age of the universe is 13.4 billion years, assuming a uniform expansion. Suppose it

were discovered that Hubble's constant is actually larger than 73 km/s/Mpc. What effect would this have on the calculated age of the universe?
Physics
1 answer:
ki77a [65]3 years ago
3 0

Answer:

It would decrease the calculated age of the universe

Explanation:

Since the age of the universe is the reciprocal of Hubble's constant, it's therefore if Hubble's constant is increased the age decreases but if the Hubble's constant is decreased, the age of universe increases. Therefore, the age of universe and Hubble's constant are inversely proportional. Conclusively, any attempt to increase Hubble's constant would imply the calculated age of the universe decreases.

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Help pleasseeee URGENT
Zina [86]

Answer:

The speed of the 8-ball is 2.125 m/s after the collision.

Explanation:

<u>Law Of Conservation Of Linear Momentum</u>

The total momentum of a system of masses is conserved unless an external force is applied. The momentum of a body with mass m and velocity v is calculated as follows:

P=mv

If we have a system of masses, then the total momentum is the sum of all the individual momentums:

P=m_1v_1+m_2v_2+...+m_nv_n

When a collision occurs, the velocities change to v' and the final momentum is:

P'=m_1v'_1+m_2v'_2+...+m_nv'_n

In a system of two masses, the law of conservation of linear momentum is simplified to:

m_1v_1+m_2v_2=m_1v'_1+m_2v'_2

The m1=0.16 Kg 8-ball is initially at rest v1=0. It is hit by an m2=0.17 Kg cue ball that was moving at v2=2 m/s.

After the collision, the cue ball comes to rest v2'=0. It's required to find the final speed v1' after the collision.

The above equation is solved for v1':

\displaystyle v'_1=\frac{m_1v_1+m_2v_2-m_2v'_2}{m_1}

\displaystyle v'_1=\frac{0.16*0+0.17*2-0.17*0}{0.16}

\displaystyle v'_1=\frac{0.34}{0.16}

v'_1=2.125\ m/s

The speed of the 8-ball is 2.125 m/s after the collision.

8 0
3 years ago
When a potential difference of 13 V is placed across a resistor, the current in the resistor is 1.4
vitfil [10]
Just divide the two numbers with each other.
I mean 13/1.4=9.2857...
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4) The object that changes its position relative to a fixed point with time is called ..<br>​
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5 0
3 years ago
Which statements are true concerning Newton's law of gravitation? The gravitational force is related to the mass of each object.
OLEGan [10]

Answer:

The gravitational force is related to the mass of each object.

The gravitational force is an attractive force.

Explanation:

Gravitational force is a long range force of attraction between any two masses.

Mathematically given as :

F=G.\frac{m_1.m_2}{r^2}

where:

m_1 & m_2 are the masses

r= distance between the center of mass of the two objects.

G= gravitational constant = 6.67\times 10^{-11} m^3.kg^{-1}.s^{-2}

From the above relation of eq. (1) it is clear that,

Gravitational force is inversely proportional to the square of the distance and directly proportional to the masses.

The mass of an object is independent of its size due to the fact that density may vary for different objects.

The force of gravity varies with height as:

\frac{g}{g_x} =(\frac{r_x}{r} )^2

where:

g=9.8\,m.s^{-2}

g_x= gravity at height r_x of the center of mass of the object from the center of mass of the earth.

and we know that force:

F=m\times g

where: m= mass of the object.

5 0
3 years ago
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