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Kazeer [188]
3 years ago
13

3.

Physics
1 answer:
weeeeeb [17]3 years ago
8 0

Answer:

I.72m/s²

II.8m/s²

Explanation:

acceleration equal velocity² divided by length

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7 0
3 years ago
Read 2 more answers
Un proyectil de masa m (kg) se dispara con una velocidad v (m/s) contra un bloque de masa 4m inicialmente en reposo. Tras la col
Maru [420]

Answer:

  x = \frac{v^2}{10 \mu g}

Explanation:

Let's start the exercise with the definition of a system formed by the projectile and the block, in this case the forces during the collision are internal and the moment is conserved,

initial instant. Just before the crash

         p₀ = m v + 4m 0

final instant. Right after the crash, before the block began to move

        p_{f} = (m + 4m) v_{f}

how the moment is preserved

        p₀ = p_{f}

        m v = 5m v_{f}

       v_{f} = v/5

knowing the speed of the system (projectile + block) we can use the relationship between work and energy

       W = ΔK

starting point. Just when the projectile + block system starts to move

       Em₀ = K = ½ m v_{f}²

final point. When the system is stopped

       Em_{f} = 0

The work of the friction force is

       W = - fr x

the negative sign is because the friction force opposes the motion, let's use Newton's second law to find the friction force

Y Axis  

      N- W = 0

      N = W = mg

The expression for the friction force is

      fr = μ N

     

substituting

     fr = μ mg

      W = - μ mg x

using the energy duty ratio

      - μ mg x = 0 - ½ m v_{f}^2

        x =   \frac{v_{f}^{2} }{2  \mu  g}

we substitute speed

       x = \frac{v^2}{10 \mu g}

7 0
3 years ago
A ball is throw horizontally from the top of a building 29.3 m high. The ball strikes the ground at a point 99.4 m from the base
Aleonysh [2.5K]

We can solve this via the kinematic equation:

h=v_{y}t+\frac{1}{2}gt^2

Where the vertical velocity is zero and so:

h=0 \times t +\frac{1}{2}gt^2\\\\h=\frac{1}{2}gt^2\\\\t=\sqrt{\frac{2h}{g}}

Since the height of the building is 29.3 meters then:

t= \sqrt{\frac{2(29.3)}{9.8}}\\\\t \approx 2.45s

The ball was in motion for approximately 2.45 seconds

6 0
3 years ago
The gravity on Mars is 3.69 meters/second². If you had a pendulum on Mars that was 1.8 meters long, what would the period be? 3.
Tomtit [17]

The solution would be like this for this specific problem:

T = 2 * pi * sqrt (Length / g)

 

T = 2  * 31.4 * sqrt (1.8m / 3.69 m/second2)

T = 4.386142257432951112677107108824

<span>So if you had a pendulum on Mars that was 1.8 meters long, the period would be 4.4.</span>

4 0
4 years ago
A current carrying wire wrapped around an iron ore is called a(n) ____.
Sladkaya [172]
I believe the correct answer from the choices listed above is the last option. A current carrying wire wrapped around an iron ore is called an electromagnet. They are <span>usually consist of a large number of closely spaced turns of wire that create the magnetic field. Hope this answers the question.</span>
7 0
4 years ago
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