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vesna_86 [32]
3 years ago
12

A river flows at a velocity of 3 km/h relative to the riverbank. A boat moves downstream at a velocity of 15 km/h relative to th

e river. What is the velocity of the boat relative to the riverbank?
Physics
1 answer:
Alexus [3.1K]3 years ago
3 0
15+3=18km/hour
Think about it like this. The boat is going 15 faster than the river, and the river is going 3 faster than the bank, so the boat is going 18 faster than the river bank
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Interactive Solution 8.29 offers a model for this problem. The drive propeller of a ship starts from rest and accelerates at 2.3
MAXImum [283]

Answer:

Δθ = 15747.37 rad.

Explanation:

  • The total angular displacement is the sum of three partial displacements: one while accelerating from rest to a certain angular speed, a second one rotating at this same angular speed, and a third one while decelerating to a final angular speed.
  • Applying the definition of angular acceleration, we can find the final angular speed for this first part as follows:

       \omega_{f1} = \alpha * \Delta t = 2.38*e-3rad/s2*2.04e3s = 4.9 rad/sec (1)

  • Since the angular acceleration is constant, and the propeller starts from rest, we can use the following kinematic equation in order to find the first angular displacement θ₁:

       \omega_{f1}^{2} = 2* \alpha *\Delta\theta (2)

  • Solving for Δθ in (2):

       \theta_{1} = \frac{\omega_{f1}^{2}}{2*\alpha } = \frac{(4.9rad/sec)^{2}}{2*2.38*e-3rad/sec2} = 5044.12 rad (3)

  • The second displacement θ₂, (since along it the propeller rotates at a constant angular speed equal to (1), can be found just applying the definition of average angular velocity, as follows:

       \theta_{2} =\omega_{f1} * \Delta_{t2} = 4.9 rad/s * 1.48*e3 s = 7252 rad (4)

  • Finally we can find the third displacement θ₃, applying the same kinematic equation as in (2), taking into account that the angular initial speed is not zero anymore:

       \omega_{f2}^{2} - \omega_{o2}^{2} = 2* \alpha *\Delta\theta (5)

  • Replacing by the givens (α, ωf₂) and ω₀₂ from (1) we can solve for Δθ as follows:

      \theta_{3} = \frac{(\omega_{f2})^{2}- (\omega_{f1}) ^{2} }{2*\alpha } = \frac{(2.42rad/s^{2}) -(4.9rad/sec)^{2}}{2*(-2.63*e-3rad/sec2)} = 3451.25 rad (6)

  • The total angular displacement is just the sum of (3), (4) and (6):
  • Δθ = θ₁ + θ₂ + θ₃ = 5044.12 rad + 7252 rad + 3451.25 rad
  • ⇒ Δθ = 15747.37 rad.
4 0
2 years ago
a steel sphere and brass ring have diameter 25cm and 24.9cm at 15°C.If the sphere and the ring are heated together.what is the t
Oksi-84 [34.3K]

Answer:

Explanation:

Due to heat energy , metal expands . Formula for linear expansion is as follows .

L = l ( 1 + α Δt )

where L is expanded length , l is original length , α is coefficient of linear expansion and Δt is increase in temperature .

To pass the sphere through the ring , the diameter of both ring and sphere should be same after heating . Let after increase of temperature Δt , their diameter becomes same as L  . The linear coefficient of brass and steel are

20 x 10⁻⁶ and 12 x 10⁻⁶ respectively .

For steel sphere ,

L = 25 ( 1 + 12 x 10⁻⁶ Δt )

For brass ring

L = 24.9 ( 1 + 20 x 10⁻⁶ Δt )

25 ( 1 + 12 x 10⁻⁶ Δt ) = 24.9 ( 1 + 20 x 10⁻⁶ Δt )

1.004( 1 + 12 x 10⁻⁶ Δt ) =  ( 1 + 20 x 10⁻⁶ Δt )

1.004 + 12.0482 x 10⁻⁶ Δt  =   1 + 20 x 10⁻⁶ Δt

.004 = 7.9518 x 10⁻⁶ Δt

Δt  = 4000 / 7.9518

= 503⁰C.

final temp = 503 + 15 = 518⁰C  .

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