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Tomtit [17]
3 years ago
7

A section of uniform pipe is bent into an upright U shape and partially filled with water, which can then oscillate back and for

th in simple harmonic motion. The inner radius of the pipe is r = 0.011 m. The radius of curvature of the curved part of the U is R = 0.15 m. When the water is not oscillating, the depth of the water in the straight sections is d = 0.21 m.
Write an expression for the mass of water in the tube, in terms of the defined quantities and the density of water, rho. Use the approximation r«R.

Physics
1 answer:
nadezda [96]3 years ago
7 0

Answer:

The expression is shown in the picture below

Explanation:

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A 16-cm-long straight line connects the center of a turntable to its edge. The turntable rotates counter-clockwise at 45 rpm. A
Bond [772]

Answer:

\mathbf{V_x = 3.25 \ cm/s}

\mathbf{V_y = 1.29\ cm/s}

Explanation:

Given that:

The radius of the table r = 16 cm  = 0.16 m

The angular velocity = 45 rpm

= 45 \times \dfrac{1}{60}(2 \pi)

= 4.71 rad/s

However, the relative velocity of the bug with turntable is:

v = 3.5 cm/s = 0.035 m/s

Thus, the time taken to reach the bug to the end is:

t = \dfrac{r}{v}

t = \dfrac{0.16}{0.035}

t = 4.571s

So the angle made by the radius r  with the horizontal during the time the bug gets to the end is:

\theta = \omega t

\theta = 4.712 \times 4.571

\theta = 21.54^0

Now, the velocity components of the bug with respect to the table is:

V_x = Vcos \theta

V_x = 0.035 \times cos (21.54^0)

V_x = 0.0325 \ m/s

\text {V_x = 3.25 \ cm/s}\mathbf{V_x = 3.25 \ cm/s}

Also, for the vertical component of the velocity V_y

V_y = V sin \theta

V_y = 0.035 \times sin (21.54^0)

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2 years ago
What is the acceleration along the ground of a 10 kg wagon when it is pulled with a force of 44 N at an angle of 35° above the
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The acceleration of the wagon along the ground is 3.6 m/s².

To solve the problem above, we need to use the formula of acceleration as related to force and mass.

Acceleration: This can be defined as the rate of change of velocity.

⇒ Formula:

  • Fcos∅ = ma................. Equation 1

⇒ Where:

  • F = Force
  • ∅ = angle above the horizontal
  • m = mass of the wagon
  • a = acceleration of the wagon

⇒ make a the subject of equation 1

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From the question,

⇒ Given:

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  • m = 10 kg

⇒ Substitute these values into equation 2

  • a = 44(cos35°)/10
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Hence, The acceleration of the wagon along the ground is 3.6 m/s²

Learn more about acceleration here: brainly.com/question/9408577

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