Given the Hubble's constant, the approximate age of the universe is 5.88 × 10⁹ Years.
Given the data in the question;
Hubble's constant; 
Age of the universe; 
We know that, the reciprocal of the Hubble's constant ( 
) gives an estimate of the age of the universe ( 
). It is expressed as:
Now,
Hubble's constant;
We know that;
so
![1\ Million\ light\ years = [9.46 * 10^{15}m] * 10^6 = 9.46 * 10^{21}m](https://tex.z-dn.net/?f=1%5C%20Million%5C%20light%5C%20years%20%3D%20%5B9.46%20%2A%2010%5E%7B15%7Dm%5D%20%2A%2010%5E6%20%3D%209.46%20%2A%2010%5E%7B21%7Dm)
Therefore;
Now, we input this Hubble's constant value into our equation;
Therefore, given the Hubble's constant, the approximate age of the universe is 5.88 × 10⁹ Years.
Learn more: brainly.com/question/14019680
<u>Answer;</u>
<em>D. The object’s weight changes, but its mass stays the same.</em>
<u>Explanation;</u>
- Mass is the amount of matter in a object, which is measured in kilograms. Mass of an object is measured using a beam balance. It is important to note that the mass of an object or a body remains constant, and does not vary from one place to another. For instance the mass of a person on the moon will be the same as when the person is on the earth surface.
- Weight on the other hand is the measurement of gravitational pull of an object. weight is measured using a spring balance and measured in Newtons. Weight varies from one place to another depending on the gravitational pull of a given surface.
<h3>
Answer: 22.5 m/s</h3>
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Work Shown:
acceleration = ( finalVelocity - initialVelocity )/(change in time)
1.5 = (60 - x)/(25)
1.5*25 = 60-x
37.5 = 60-x
x = 60-37.5
x = 22.5
The initial velocity is 22.5 m/s
Answer:
E. two times the original diameter
Explanation:
Resistance of a wire is:
R = ρ L/A
where ρ is the resistivity of the material, L is the length, and A is the cross-sectional area.
For a round wire with diameter d:
R = ρ L / (¼ π d²)
The two wires must have the same resistance, so:
ρ₁ L₁ / (¼ π d₁²) = ρ₂ L₂ / (¼ π d₂²)
The wires are made of the same material, so ρ₁ = ρ₂:
L₁ / (¼ π d₁²) = L₂ / (¼ π d₂²)
The new length is four times the old, so 4 L₁ = L₂:
L₁ / (¼ π d₁²) = 4 L₁ / (¼ π d₂²)
1 / (¼ π d₁²) = 4 / (¼ π d₂²)
Solving:
1 / (d₁²) = 4 / (d₂²)
(d₂²) / (d₁²) = 4
(d₂ / d₁)² = 4
d₂ / d₁ = 2
So the new wire must have a diameter twice as large as the old wire.
