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TEA [102]
2 years ago
9

.A car skid for 290 m. assuming that the car skidded to a stop with a constant

Physics
1 answer:
snow_tiger [21]2 years ago
8 0

Answer:

The speed of the car before it began to skid is 47.56 m/s.

Explanation:

We can use kinematics to solve this problem.

We are given three known variables:

  • Δx = 290 m
  • a = -3.90 m/s²
  • v = 0 m/s (final velocity is 0 m/s because the car skids to a stop).

We can use this kinematic equation to <u><em>solve</em></u><em> for the initial velocity</em>, v₀.

  • <u>v² = v₀² + 2aΔx</u>

Substitute the known variables into the equation.

  • (0)² = v₀² + 2(-3.9)(290)
  • 0 = v₀² - 2262
  • 2262 = v₀²
  • <u>v₀ = 47.56 m/s</u>

The speed of the car before it began to skid is 47.46 m/s.

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Answer:

Each thruster has to applied a force of 294.5N in tangential direction

Explanation:

Mass of the ring ,M =2.1×105kg

Mass of the ship ,m = 3.5×104kg

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Let force applied by each thruster be F

Time taken to reach  gravity ,t = 3hrs = 3600× 3 =10800sec

The movement of ring make the object kept at the edge feel a force of centrifuge in outward direction.

Centrifugal force = weight of the object on earth

Assume the ring is moving with angular speed ω

Centripetalforce of the object kept at ring

m₁R ω²=m₁g  (m₁=mas of object)

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⇒ω = √g/R

The ring start from 0 angular speed with constant angular acceleration

Let the constant angular acceleration be ∝

∝ = ω  / t

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∝ = \frac{1}{t } \sqrt{\frac{g}{R} } }

Consider Torque on the ring and ship system

T = FR + FR = 2FR

Moment of inertia of ring ship system

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= MR² + mr² + mr² = MR² + 2mr²

angular acceleration of the ring ship system

∝ = \frac{2FR}{MR^2 + 2mr^2}

Now we have ,

∝ = \frac{1}{t} \sqrt{\frac{g}{R} }  , ∝ = \frac{2FR}{MR^2+2mr^2}

equating both values

We have,

F = \frac{MR^2+2mr^2}{2Rt} \sqrt{\frac{g}{R} }

where,

m = 2.1 × 10⁵kg, R = 86m, m = 3.5 × 10₄kg,

r = 31m, g = 9.8m/s² , t = 10800sec

F = \frac{(2.1 \times 10^5 \times86^2)+(2\times3.5v10^4\times31^2) }{2\times86\times10800} \times\sqrt{\frac{9.8}{86} } \\\\F = 294.47N\\\approx294.5N

Each thruster has to applied a force of 294.5N in tangential direction

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