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sveta [45]
3 years ago
13

A train departs from its station at a constant acceleration of 5 m/s. What is the speed of the train at the end of 20s?

Physics
2 answers:
shutvik [7]3 years ago
7 0

Answer:

Velocity, v = 100 m/s

Explanation:

It is given that,

Initial speed of the train, u = 0

Acceleration of the train, a=5\ m/s^2

Let v is the velocity of the train at the end of 20 seconds, t = 20 s

Using the equation of kinematics as :

v = u + a × t

v = a × t

v = 5 × 20

v = 100 m/s

So, the velocity of the train after 20 s is 100 meters. Hence, this is the required solution.

morpeh [17]3 years ago
4 0
Acceleration = (final velocity-initial velocity)/time
5 = (v-0)/20
v = 100m/s
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Bryce, a mouse lover, keeps his four pet mice in a roomy cage, where they spend much of their spare time (when they are not slee
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Answer:

I₁ = (7.78 i ^ - 6.71 j ^) 10⁻³ J s ,  I₂ = (-12.5 i ^ -14.6 j ^) 10⁻³ J s ,  I₃ = (19.1i ^ + 18.6 j ^) 10⁻³ J s  and I₄ = (-9.14i ^ + 7.24 j ^) 10⁻³ J s

Explanation:

The impulse is equal to the variation of the moment, to apply this relationship to our case, we will assume that initially the mouse was at rest

    I = Δp = m v_{f} -m v₀

    I = m (v_{f}  -v₀)

Bold indicates vector quantities, let's calculate the momentum of each mouse in for the x and y axes

We recommend bringing all units to the SI system

Mouse 1.

It has a mass of 22.3 g = 22.3 10⁻³ kg, a final velocity of (v = 0.349 i ^ - 0.301 j ^) m / s with an initial velocity of zero

    Iₓ = m (v_{fx}  - v₀ₓ)

    Iₓ = 22.3 10⁻³ (0.349 -0)

    Iₓ = 7.78 10⁻³ J s

   I_{y} = m (v_{fy}  -v_{oy} )

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Mouse 2

Mass 17.9 g = 17.9 10⁻³ kg

Speed ​​(-0.699 i ^ - 0.815 j ^) m / s

    Iₓ = m (v_{fx}  - v₀ₓ)

    Iₓ = 17.9 10⁻³ (-0.699 -0)

    Iₓ = -12.5 10⁻³ J s

    I_{y} = 17.9 10⁻³ (-0.815 - 0)

    I_{y} = -14.6 10⁻³ J s

   I₂ = (-12.5 i ^ -14.6 j ^) 10⁻³ J s

Mouse 3

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Speed ​​(0.745i ^ + 0.975 j ^) m / s

    Iₓ = 19.1 10⁻³ (0.745 -0)

    Iₓ = 14.2 10⁻³ J s

    I_{y} = 19.1 10⁻³(0.975 -0)

    I_{y} = 18.6 10⁻³ J s

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Mouse 4

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    Iₓ = 10.1 10⁻³ (-0.905 -0)

    Iₓ = -9.14 10⁻³ J s

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    I_{y} = 7.24 10⁻³ J s

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Answer:

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