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Masja [62]
3 years ago
9

Based on what you’ve learned from the raisin cake analogy, which two properties of distant galaxies do astronomers have to measu

re to show that we live in an expanding universe?
Physics
1 answer:
Kisachek [45]3 years ago
4 0
Here is the answer to the given question above. Based on the Raisin Cake Analogy, the two properties of distant galaxies that astronomers have to measure to show<span> that we live in an expanding universe are DISTANCE and SPEED. Hope this answers your question. Have a great day!</span>
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One observer stand on a train moving at a constant speed, and one observer stands at rest on the ground. The person on the train
ANTONII [103]

Answer:

b) Equal to c

Explanation:

According to relativity, the speed of light in free space is constant in all inertial reference frame.

3 0
3 years ago
please help! easy science! second pic is the question and possible answers (A B C or D) its one question! brainliest promised!
kirza4 [7]

Answer:

Please do not take my word for this at all, but this is what I found, "When the pendulum swings back down, the potential energy is converted back into kinetic energy. At all times, the sum of potential and kinetic energy is constant." So I think the answer is B also you are anime fan too lol :DD I love hinata

Explanation:

4 0
3 years ago
What is the acceleration of this object? The object's mass is 60 kg.
andre [41]
The acceleration is 3.3 m/s2
8 0
3 years ago
Read 2 more answers
A basketball is held over head at a height of 2.4 m. The ball is lobbed to a teammate at 8 m/s at an angle of 40'. If the ball i
cupoosta [38]

Explanation:

since both the teammates are of the same height, their height won't matter. Because now the basketball won't cover any vertical distance.

We have to calculate its range the horizontal distance covered by it when tossed from one teammate to the other.

range can be calculated by the formula :-

\boxed{\mathfrak{range =  \frac{  u  {}^{2}   \sin 2\theta }{g} }}

u is the velocity during its take off and \theta is the angle at which its thrown

Given that

  • u = 8m/ s
  • \theta = 40°

calculating range using the above formula

= \frac{ {8}^{2} \sin2(40)  }{10}

=  \frac{64 \times  \sin(80) }{10}

value of sin 80 = 0. 985

=  \frac{64 \times 0.985}{10}

=  \frac{63.027}{10}

= 6.3027

Hence,

\mathfrak { \blue{the \: teammate \: is \:  \red{\underline{6.3027 \: meters} }\: away } }

7 0
3 years ago
Two charges, Q1 and Q2, are separated by 6·cm. The repulsive force between them is 25·N. In each case below, find the force betw
Misha Larkins [42]

Answer:

a) 5 N b) 225 N c) 5 N

Explanation:

a) Per Coulomb's Law the repulsive force between 2 equal sign charges, is directly proportional to the product of the charges, and inversely proportional to the square of  the distance between them, acting along  the  line that joins the charges, as follows:

F₁₂ = K Q₁ Q₂ / r₁₂²

So, if we make Q1 = Q1/5, the net effect will be to reduce the force in the same factor, i.e. F₁₂ = 25 N / 5 = 5 N

b) If we reduce the distance, from r, to r/3, as the  factor is squared, the net effect will be to increase the force in a factor equal to 3² = 9.

So, we will have F₁₂ = 9. 25 N = 225 N

c) If we make Q2 = 5Q2, the force would be increased 5 times, but if at the same , we increase the distance 5 times, as the factor is squared, the net factor will be 5/25 = 1/5, so we will have:

F₁₂ = 25 N .1/5 = 5 N

3 0
3 years ago
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