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Marina86 [1]
3 years ago
5

The floor of a railroad flatcar is loaded with loose crates having a coefficient of static friction of 0.32 with the floor. If t

he train is initially moving at a speed of 54 km/h, in how short a distance can the train be stopped at constant acceleration without causing the crates to slide over the floor?
Physics
1 answer:
coldgirl [10]3 years ago
7 0

Answer:

The shortest braking distance is 35.8 m

Explanation:

To solve this problem we must use Newton's second law applied to the boxes, on the vertical axis we have the norm up and the weight vertically down

On the horizontal axis we fear the force of friction (fr) that opposes the movement and acceleration of the train, write the equation for each axis

    Y axis

     N- W = 0

     N = W = mg

  X axis

     -Fr = m a

     -μ N = m a

     -μ mg = ma

     a = μ g

     a  = - 0.32 9.8

     a =  - 3.14 m/s²

We calculate the distance using the kinematics equations

    Vf² = Vo² + 2 a x

     x = (Vf² - Vo²) / 2 a

When the train stops the speed is zero (Vf = 0)

 Vo = 54 km/h (1000m/1km) (1 h/3600s)= 15 m/s

     x = ( 0 - 15²) / 2 (-3.14)

     x=  35.8 m

The shortest braking distance is  35.8 m

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A 70.0 kg ice hockey goalie, originally at rest, has a 0.110 kg hockey puck slapped at him at a velocity of 31.5 m/s. Suppose th
NISA [10]

Answer

given,

mass of the goalie(m₁) = 70 kg

mass of the puck (m₂)= 0.11 kg

velocity of the puck = 31.5 m/s

elastic collision

v_1=\dfrac{m_2-m_1}{m_1+m_2}v_1+\dfrac{2m_2}{m_1+m_2}v_2

v_{pf}=\dfrac{0.11-70}{0.11+70}31.5+\dfrac{2m_2}{m_1+m_2}\times (0)

v_{pf}=-31.4\ m/s

v'_2 = \dfrac{2m_1v_1}{m_1+m_2}-\dfrac{(m_2-m_1)v_2}{m_2+m_1}

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v_{gf} = \dfrac{2\times 0.11\times 31.5}{0.11+70}

v_{gf} = 0.0988\ m/s

4 0
3 years ago
A person carries a plank of wood 2.00 m long with one hand pushing down on it at one end with a force f1 and the other hand hold
slega [8]

Answer:


F₁ = 4,120.2 N


F₂ = 3,924N


Explanation:



1) Balance of angular momentum around the end where F₁ is applied.


F₂ × 0.5m - F₁ × 0 = mass × g × 1m


⇒ F2 × 0.5 m= 20 kg × 9.81 m/s² × 1 m = 1,962 N×m


F₂ = 196.2 Nm / 0.5m = 3,924 N


2) Balance of forces


F₁ - F₂ = mg


F₁ = F₂ + mg = 3,924N + 20kg (9.81 m/s²) = 4,120.2 N

4 0
3 years ago
A typical nuclear fission power plant produces about 1.00 GW of electrical power. Assume the plant has an overall efficiency of
mamaluj [8]

Answer:

mass consumed by 235U each day = 2 kg

Explanation:

electrical power produced = 1 GW = 1 × 10⁹ × (6.24151 × 10¹⁸ ) eV

                                            = 6.24151× 10²¹ MeV/s

thermal energy =  0.420 * 250 = 105 MeV

\dfrac{1 GW}{150 MeV}= \dfrac{6.24151\times 10^{21}}{105}

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                                      =  5.13 × 10²⁴ fission/day

mu = 235.04393 ×  1.660× 10 ⁻²⁷ = 390.1729× 10⁻²⁷ Kg

M = mu ×5.13 × 10²⁴

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M   =  2 kg(approx.)

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3 0
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Alisiya [41]

Answer:

The induced current direction as viewed is clockwise

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Therefore, as the magnet approaches the coil with the south pole,  the coil produces current equivalent to the upward movement of the south pole of a permanent magnet through it which according to Flemings Right Hand Rule is clockwise

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The direction of the induced current in the loop (as viewed from above, looking down the magnet) is clockwise

5 0
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