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pychu [463]
3 years ago
11

If an object left earth which direction would it travel

Physics
1 answer:
IceJOKER [234]3 years ago
4 0

Answer:

In a straight line

Explanation:

Why, because of the power of inerta

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Science cannot be used to answer questions about Moral judgements
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How many cubic feet are in a 55-gallon drum?
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1 gallon = 231 cubic inches
1 cubic foot = 1728 cubic inches

                       (55 gal) x (231 in³/gal) x (1 ft³/1728 in³)

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                   =         7.352 cubic feet       (rounded)  
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3 years ago
The gravitational force between two objects is 2400 N. What will be the gravitational force between the objects if the mass of o
motikmotik

Answer:

4800N

Explanation:

Lets assume,

Mass of first object = m₁

Mass of second object = m₂

Distance between the two objects = r

Thus the force between the two objects will be

F = \frac{G\times m_{1}\times m_{2}}{r^{2}}

where, G = Universal gravitational constant

Given, F = 2400N

New mass of second object = 2m₂

Now, the force will be

F_{2} = \frac{G\times m_{1}\times 2m_{2}}{r^{2}}

F_{2}= 2\frac{G\times m_{1}\times m_{2}}{r^{2}}

F_{2}= 2F

F_{2}= 2\times2400

Thus, F₂ = 4800N

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A rifle bullet with mass 8.00g strikes and embeds itself in a block with mass 0.992kg that rests on a frictionless, horizontal s
mafiozo [28]

Answer:

block velocity   v = 0.09186 = 9.18 10⁻² m/s  and speed bollet   v₀ = 11.5 m / s

Explanation:

We will solve this problem using the concepts of the moment, let's try a system formed by the two bodies, the bullet and the block; In this system all scaffolds during the crash are internal, consequently, the moment is preserved.

Let's write the moment in two moments before the crash and after the crash, let's call the mass of the bullet (m) and the mass of the Block (M)

Before the crash

     p₀ = m v₀ + 0

After the crash

   p_{f} = (m + M) v

    p₀ = p_{f}

    m v₀ = (m + M) v                    (1)

Now let's lock after the two bodies are joined, in this case the mechanical energy is conserved, write it in two moments after the crash and when you have the maximum compression of the spring

Initial

    Em₀ = K = ½ m v2

Final

    E m_{f}= Ke = ½ k x2

   Emo = E m_{f}

   ½ m v² = ½ k x²

   v² = k/m  x²

Let's look for the spring constant (k), with Hook's law

   F = -k x

   k = -F / x

   k = - 0.75 / -0.25

   k = 3 N / m

Let's calculate the speed

  v = √(k/m)   x

  v = √ (3/8.00)   0.15

  v = 0.09186 = 9.18 10⁻² m/s

This is the spped of  the  block  plus bullet rsystem right after the crash

We substitute calculate in equation  (1)

   m v₀ = (m + M) v

  v₀ = v (m + M) / m

  v₀ = 0.09186 (0.008 + 0.992) /0.008

  v₀ = 11.5 m / s

6 0
3 years ago
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