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kicyunya [14]
3 years ago
13

PLEASE HELPP I NEED TO PASS THIS !!

Physics
1 answer:
goldfiish [28.3K]3 years ago
3 0
Option D is correct.
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A particle is traveling in the positive direction along an x axis, at a constant 5 m/s. Which of the following
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The particle will accelerate 5m/s every second until it reaches a maximum of whatever your graph/diagram goes to, I'm in physical science and this is somewhat similar to what I am doing now but I'm not sure if that was what your looking for.
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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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3241004551 [841]
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3 0
3 years ago
I NEED HELP ASAP PLEASE!!!<br> (19 pts!)
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They go in the boxes in this order:
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3 years ago
Find the change in velocity of a 369 g hockey puck subject to the force shown below.
Elena-2011 [213]

Answer:

The change in velocity is 15.83 [m/s]

Explanation:

Using the Newton's second law we have:

ΣF = m*a

The force in the graph is 185 N, therefore:

185=0.369*a\\Where\\a=acceleration made it by the force [m/s^2]

a=501.35[m/s^2]

Now using the following kinematic equation:

V^{2}=Vi^{2} + 2*a*(x-xi) \\where\\V=final velocity [m/s]\\Vi= initial velocity [m/s] = 0 the hockey disk is in rest when receives the hit.\\ x = Final position [m] = 0.4 m\\xi = initial position [m] = 0.15m\\

Now replacing the values:

V^{2}=0 + 2*501.35*(0.4-0.15)\\ \\V= 15.83[m/s]

3 0
3 years ago
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