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s2008m [1.1K]
3 years ago
12

Fe + Br2 --> FeBr3

Physics
1 answer:
eimsori [14]3 years ago
6 0

Explanation:

  • 1
  • 2
  • 1
  • 1
  • 3
  • 4
  • No
  • 2Fe+3Br2-->2FeBr3
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Jamal is testing the effect of four different fertilizers on a particular plant. For his experiment, Jamal obtained twenty ident
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hockey puck slides across the ice with an initial velocity of 7.2 m/s. It has a deceleration of 1.1 m/s2 and is traveling toward
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For this use the formula:

d = Vo * t - (at^2) / 2

Clearing t:

t = d/(v + 0.5*a)

Replacing:

t = 5 m / (7.2 m/s + 0.5 * (-1.1 m/s²)

Resolving:

t = 5 m / (7.2 m/s + (-0.55 m/s²)

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Weathering of rocks can occur in many ways. In the western United States, strong winds can erode huge rock formations by blowing
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Answer: physical or mechanical weathering

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The Astronomer Carl Sagan is famous. Why?
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Point A of the circular disk is at the angular position θ = 0 at time t = 0. The disk has angular velocity ω0 = 0.17 rad/s at t
statuscvo [17]

Given that,

Angular velocity = 0.17 rad/s

Angular acceleration = 1.3 rad/s²

Time = 1.7 s

We need to calculate the angular velocity

Using angular equation of motion

\omega=\omega_{0}+\alpha t

Put the value in the equation

\omega=0.17+1.3\times1.7

\omega=2.38(k)\ m/s

We need to calculate the angular displacement

Using angular equation of motion

\theta=\theta_{0}+\omega_{0}t+\dfrac{\alpha t^2}{2}

Put the value in the equation

\theta=0+0.17\times1.7+\dfrac{1.3\times1.7^2}{2}

\theta=2.1675\times\dfrac{180}{\pi}

\theta= 124.18^{\circ}

We need to calculate the velocity at point A

Using equation of motion

v_{A}=v_{0}+\omega\times r

Put the value into the formula

v_{A}=0+2.38(k) \times0.2(\cos(124.18)i+\sin(124.18)j))

v_{A}=0.476\cos(124.18)j+0.476\sin(124.18)i

v_{A}=(-0.267j-0.393i)\ m/s

We need to calculate the acceleration at point A

Using equation of motion

a_{A}=a_{0}+\alpha\times r+\omega\times(\omega\times r)

Put the value in the equation

a_{A}=0+1.3(k)\times0.2(\cos(124.18)i+\sin(124.18)j)+2.38\times2.38\times0.2(\cos(124.18)i+\sin(124.18)j)

a_{A}=0.26\cos(124.18)i+0.26\sin(124.18)j+(2.38)^2\times0.2(\cos(124.18)i+\sin(124.18)j)

a_{A}=-0.146j-0.215i−0.636i+0.937j

a_{A}=0.791j-0.851i

a_{A}=-0.851i+0.791j\ m/s^2

Hence, (a). The velocity at point A is (-0.267j-0.393i)\ m/s

(b). The acceleration at point A is (-0.851i+0.791j)\ m/s^2

3 0
4 years ago
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