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RUDIKE [14]
3 years ago
8

When you slosh the water back and forth in a tub at just the right frequency, the water alternately rises and falls at each end,

remaining relatively calm at the center. Suppose the frequency to produce such a standing wave in a 75 cm -wide tub is 0.90 Hz .
What is the speed of the water wave?
Physics
1 answer:
Natali5045456 [20]3 years ago
5 0
Slot in wwatwe to be determined for use
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Consider a bicycle wheel to be a ring of radius 30 cm and mass 1.5 kg. Neglect the mass of the axle and sprocket. If a force of
vredina [299]

Answer:

The angular speed after 6s  is \omega = 1466.67s^{-1}.

Explanation:

The equation

I\alpha  = Fd

relates the moment of inertia I of a rigid body, and its angular acceleration \alpha, with the force applied F at a distance d from the axis of rotation.

In our case, the force applied is F = 22N, at a distance d = 6cm =0.06m, to a ring with the moment of inertia of I =mr^2; therefore, the angular acceleration is

$\alpha =\frac{Fd}{I} $

$\alpha =\frac{22N*0.06m}{(1.5kg)*(0.06)^2} $

\alpha  = 244.44\: s^{-2}

Therefore, the angular speed \omega which is

\omega  = \alpha t

after 6 seconds is

\omega = 244.44$\: s^{-2}* 6s

\boxed{\omega = 1466.67s^{-1}}

7 0
3 years ago
Which of the following is a strength training option?
Vilka [71]

Answer:

D. All of the above

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5 0
2 years ago
Read 2 more answers
5 milligrams into quintal​
Greeley [361]

Answer:

divide the mass value by 1e+8

4 0
3 years ago
What is the mass of a falling rock if it produces a force of 50 N?
Rasek [7]
××F=m \times a×

50N is your force and the acceleration is -9.8m/s^2 due to gravity. 

So, you just plug that in. 

50 N=m\times-9.8m/s^2\\
\frac{50}{-9.8}=m\\
m=-5.102

BUT you know that mass cannot be negative, so you just disregard the negative sign and the mass of the rock is 5.102 grams.
8 0
3 years ago
A nonconducting spherical shell, with an inner radius of 4 cm and an outer radius of 6 cm, has charge spread nonuniformly throug
Vaselesa [24]

Answer:

1.57 * 10^{3} Q

Explanation:

The volume charge density is defined by ρ = \frac{Q}{V} (Equation A), where Q is the charge and V, the volume.

The units in the S.I. are \frac{Coulombs}{m^{3} }, so we have to express the radius in meters:

inner radius = 4 cm * \frac{1 m}{100 cm} = 0.04m

outer radius = 6 cm * \frac{1m}{100cm}  = 0.06m

Now, we know that the volume of the sphere is calculated by the formula:

V = \frac{4}{3}\pi r^{3}, and as we have an spherical shell, the volume is calculated by the difference between the outher and inner spheres:

V = \frac{4}{3}\pi (r_{o} ^{3} - r_{i} ^{3}), where r_{o} is the outer radius and r_{i} is the inner radius.

Replacing the volume formula in the Equation A:

ρ = \frac{Q}{\frac{4}{3}\pi(r^{3} _{o}-r_{i} ^{3})}

ρ = \frac{3Q}{4\pi (r_{o} ^{3}-r_{i} ^{3} ) }

Replacing the values of the outer and inner radius whe have:

ρ = \frac{3Q}{4\pi (1.52 * 10^{-4})}

ρ = 1.57 * 10^{3} Q

4 0
3 years ago
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