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zheka24 [161]
3 years ago
12

A comet is traveling through space with a speed of 3.40 ✕ 104 m/s when it collides with an asteroid that was at rest. The comet

and the asteroid stick together during the collision process. The mass of the comet is 1.05 ✕ 1014 kg, and the mass of the asteroid is 7.05 ✕ 1020 kg.
a. What is the speed of the center of mass of the asteroid-comet system before the collision?
b. What is the speed of the system's center of mass after the collision?
Physics
1 answer:
Dvinal [7]3 years ago
6 0

Answer:

a) vcm = 5.06*10⁻³ m/s

b) vcm = 5.06*10⁻³ m/s

Explanation:

v₁ = 3.40*10⁴ m/s

v₂ = 0 m/s

m₁ = 1.05*10¹⁴ Kg

m₂ = 7.05*10²⁰ Kg

vcm initial = ?

vcm final = ?

a)  Before the collision

vcm = (m₁*v₁ + m₂*v₂) / (m₁ + m₂)

vcm = (1.05*10¹⁴ Kg*3.40*10⁴ m/s + 7.05*10²⁰ Kg*0 m/s) / (1.05*10¹⁴ Kg + 7.05*10²⁰ Kg)

vcm = 5.06*10⁻³ m/s

a)  After the collision

m₁*v₁ + m₂*v₂ = (m₁ + m₂)*vcm

⇒  vcm = (m₁*v₁ + m₂*v₂) / (m₁ + m₂)

vcm = (1.05*10¹⁴ Kg*3.40*10⁴ m/s + 7.05*10²⁰ Kg*0 m/s) / (1.05*10¹⁴ Kg + 7.05*10²⁰ Kg)

vcm = 5.06*10⁻³ m/s

The speed of the system's center of mass is the same value.

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Answer:

a)  v = 16.57 m / s, b)  a = 19.6 m / s², d)    N = 1.76 10³ N,     N / W = 3

Explanation:

This exercise looks interesting, but I think you have some problem with the writing, the questions seem a bit disconnected from the initial text.

Let's answer the questions.

a) For this part we can use energy considerations.

Starting point. The upper part of the trajectory indicates that the arm is horizontally

          Em₀ = U = m g h

in this case h = r

Final point. For lower of the trajectory

          Em_f = K = ½ m v²

as they indicate that there is no friction

         Em₀ = em_f

         mgh = ½ m v²

         v = \sqrt{2gh}

let's calculate

        v = \sqrt{2 \ 9.8 \ 14.0}

         v = 16.57 m / s

b) the centripetal acceleration has the formula

           a = v² / r

           a = 16.57² / 14.0

           a = 19.6 m / s²

c) see attached where the diagram is

where N is the normal and w the weight

d) let's use Newton's second law

               N-W = m a

               N - mg = m ar

               N = m (g + a)

let's calculate

               N = 60.0 (9.8 + 19.6)

               N = 1.76 10³ N

the relationship with weight is

              N / W = 1.76 10³/( 60 9.8)

              N / W = 3

normal is three times greater than body weight

e) the answer is reasonable since by Newton's first law the body must continue in a straight line, therefore to change its trajectory a force must be applied to deflect it

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Answer:

Stars create new elements in their cores by squeezing elements together in a process called nuclear fusion.

Explanation:

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A 30000 grams boy is riding a merry-go-round with a radius of 600 cm. What is the centripetal force and acceleration on the boy
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Answer:

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Explanation:

so 30000g= 30kg

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7 0
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Why is the answer C?
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Explanation:

We want to find the statement that is proven by the fact that the balls reach the same height.

A isn't supported by the evidence.  Balls can reach the same height without having the same initial speed.

B isn't supported by the evidence.  Balls can reach the same height without having the same launch angle.

C is supported.  Projectiles spend the same amount of time going up as they do coming down, so if two projectiles reach the same height, then they must spend the same amount of time in the air.

D isn't supported by the evidence.  Balls thrown at the same speed and complementary angles have the same range but different heights.

E isn't supported by the evidence.  The mass of the ball doesn't affect the height.

7 0
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