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Sonbull [250]
3 years ago
12

What do astonomers use to calculate the age of the universe

Physics
1 answer:
makvit [3.9K]3 years ago
3 0

Answer: Stars

Explanation: Some stars have remained intact since the Big Bang. These stars are shown in something called the HR diagram with portrays the brightness and largeness of the star. Also, white dwarfs and black dwarfs can be traced back which may correspond with the initiation of the universe.

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Could someone explain to me how to got the answer B, thank you very much​
Mnenie [13.5K]

Answer:

since -6 lasted for 5 seconds, multiplying both would result in -30

3 lasted for 10 seconds, so multiplying both would give +30

average = ( 30 + (-30) ) / 2

30 -30 is already equal to zero, so the answer should be 0

4 0
3 years ago
Blocks A (mass 3.50 kg) and B (mass 6.50 kg) move on a frictionless, horizontal surface. Initially, block B is at rest and block
Vedmedyk [2.9K]

Answer:

(a) V (A) =  0.7 m/s,

(b) V (A) =  0.7 m/s,

(c) V (B) =  0.7 m/s

(d) u= - 0.60 m/s

(e) v = 0.75 m/s

Explanation:

Given:

M(A) =3.50 Kg, M(B)=6.50 Kg, V(A) = 2.00 m/s, V(B) = 0 m/s

Sol:

a)  law of conservation of momentum

M(a) x V(A) + M(B) x V(B) = ( M(a) + M(B) ) V      (let V is Common Velocity of Both block)

so 3.50 Kg x 2.00 m/s + 6.50 Kg x 0 m/s = (3.50 Kg + 6.50 Kg ) V

after solving V =  0.7 m/s

After the collision the velocities of the both block will be as the the spring is compressed maximum.

V (A) =  0.7 m/s

b)  V(A) =  0.7 m/s ( Part (a) and Part (a) are repeated )

c) as stated above the in the Part (a)

V(B) =  0.7 m/s

d) When the both blocks moved apart after the collision:

Let u=velocity of block A after the collision.

and v = velocity of block B after the collision.

then conservation of momentum

M(a) x V(A) + M(B) x V(B) = M(a) x v + M(B) x u

⇒ 3.50 Kg x 2.00 m/s + 6.50 Kg x 0 m/s =  3.50 Kg x u + 6.50 Kg x v

⇒ 2.00 m/s = u + 1.86 v -----eqn (1)  ( dividing both side by 3.50 Kg)

For elastic collision  

the velocity relative approach = velocity relative separation

so 2.00 m/s = v-u  ----- eqn (2)

⇒v = u + 2.00 m/s

putting this value in eqn (1) we get

2.00 m/s = u + 1.86 (v + 2.00 m/s)

u= - 0.60 m/s

e) putting v= 2.00 m/s in eqn (1)

2.00 m/s = - 2.32 m/s + 1.86 v

v = 0.75 m/s

5 0
3 years ago
8. A 30-kg box is sliding down a frictionless plane that is sloped at 24º. Assuming the object starts at rest,
Katena32 [7]

The net force on the box parallel to the plane is

∑ F[para] = mg sin(24°) = ma

where mg is the weight of the box, so mg sin(24°) is the magnitude of the component of its weight acting parallel to the surface, and a is the box's acceleration.

Solve for a :

g sin(24°) = a ≈ 3.99 m/s²

The box starts at rest, so after 7.0 s it attains a speed of

(3.99 m/s²) (7.0 s) ≈ 28 m/s

6 0
2 years ago
The arctic tundra can be found in _______. a. Greenland b. Canada c. The United States d. all of the above
dangina [55]
The artic tundra can be found in D. All of the above.
7 0
3 years ago
Read 2 more answers
How many of the identified objects are not accelerating? a race car going around a circular track at 150 mph a sky diver falling
Nonamiya [84]
A race car, sky diver.
3 0
3 years ago
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