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Stells [14]
2 years ago
6

URGANT!!!!!!! NEED ANSWERS TODAY!!!!

Physics
1 answer:
natima [27]2 years ago
5 0
2H2 + O2 = 2H2O + Energy
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3m/s square

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A thin block of soft wood with a mass of 0.072 kg rests on a horizontal frictionless surface. A bullet with a mass of 4.67 g is
kozerog [31]

Answer:

v’= 279.66 m / s

Explanation:

We work this exercise using the conservation of the moment. For this we define the system formed by the two blocks, therefore the forces during the collision are internal of the action and reaction type.

Initial instant. Before the crash

        p₀ = m v₀ + 0

Final moment. After the crash

        p_f = m v + M v ’

how the tidal wave is preserved

       p₀ = p_f

       m v₀ = m v + M v ’

       v = \frac{m v_o - Mv'}{m}

let's calculate

       v ’= \frac{0.00467 \ 619 - 0.072 \ 22}{0.004676}

       v ’= \frac{2.89- 1.584}{ 0.00467}

       v ’= 279.66 m / s

4 0
2 years ago
A historical society is testing an old cannon. They
sveta [45]

Answer:

26.2 m/s

Explanation:

We can find the speed of the cannonball just by analyzing its vertical motion. In fact, the initial vertical velocity is

u_y = u sin \theta (1)

where u is the initial speed and \theta = 45.0^{\circ} is the angle of projection.

We can therefore use the following suvat equation for the vertical motion of the ball:

v_y = u_y + at

where v_y is the vertical velocity at time t, and a=g=-9.8 m/s^2 is the acceleration of gravity. The time of flight is 3.78 s, so we know that the ball reaches its maximum height at half this time:

t=\frac{3.78}{2}=1.89 s

And at the maximum height, the vertical velocity is zero:

v_y=0

Substituting these values, we find the initial vertical velocity:

u_y = v_y - at = 0-(-9.8)(1.89)=18.5 m/s

And using eq.(1) we now find the initial speed:

u=\frac{u_y}{sin \theta}=\frac{18.5}{sin 45.0^{\circ}}=26.2 m/s

4 0
2 years ago
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