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3 years ago

12

a vibrating source with a frequency of 20.0Hz produces water waves that have a wavelength of 3.0 cm. Calculate the speed of the

waves

1 answer:

Shtirlitz [24]3 years ago

7 0

Answer:

60m/s

Explanation:

v=frequency ×wave length

v signifies the speed

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A food processor draws 8.47 A of current when connected to a potential difference of 110 V.

Andreyy89

Answer:

27.95[kW*min]

Explanation:

We must remember that the power can be determined by the product of the current by the voltage.

$P=V*I$

where:

P = power [W]

V = voltage [volt]

I = amperage [Amp]

Now replacing:

$P=110*8.47\\P=931.7[W]$

Now the energy consumed can be obtained mediate the multiplication of the power by the amount of time in operation, we must obtain an amount in Kw per hour [kW-min]

$Energy = 931.7[kW]*30[days]*10[\frac{min}{1day} ]=279510[W*min]or 27.95[kW*min]$

3 0

3 years ago

A mass is hung from a spring and set in motion so that it oscillates continually up and down. The velocity v of the weight at ti

otez555 [7]

To solve the problem it is necessary to identify the equation in the manner given above.

This equation corresponds to the displacement of a body under the principle of simple harmonic movement.

Where,

$\xi = Acos(\omega t +\phi)$

PART A) Our equation corresponds to

$y = -5cos(4\pi t)$

Therefore the value of omega is equivalent to that of

$\omega = 4\pi$

From the definition we know that the period as a function of angular velocity is equivalent to

$T = \frac{2\pi}{\omega}$

$T = \frac{2\pi}{4\pi}$

$T = \frac{1}{2}$

This same point is the equivalent of the maximum point of the speed that the body can reach, since the internal expression of the $cos\theta$ Is equivalent to . So the maximum speed that the body can reach is,

$y = -5cos(4\pi t)$

$y = -5cos(4\pi (1/2))$

$y = -5*(-1)$

$y = 5$

Therefore the maximum felocity will be 5ft / s

PART B) The period of graph is the time taken to reach from one maximum point to next point maximum point, then

$t = \frac{T}{2} = \frac{1}{2}*\frac{1}{2}$

$t = \frac{1}{4}s$

5 0

3 years ago

A 2.00 kg block hangs from a spring balance calibrated in Newtons that is attached to the ceiling of an elevator.(a) What does t

tresset_1 [31]

Answer:

Part a)

$Reading = 2.00 kg$

Part b)

$Reading = 2.00 kg$

Part c)

$Reading = 4.04 kg$

Part d)

from t = 0 to t = 4.9 s

so the reading of the scale will be same as that of weight of the block

Then its speed will reduce to zero in next 3.2 s

from t = 4.9 to t = 8.1 s

The reading of the scale will be less than the actual mass

Explanation:

Part a)

When elevator is ascending with constant speed then we will have

$F_{net} = 0$

$T - mg = 0$

$T = mg$

So it will read same as that of the mass

$Reading = 2.00 kg$

Part b)

When elevator is decending with constant speed then we will have

$F_{net} = 0$

$T - mg = 0$

$T = mg$

So it will read same as that of the mass

$Reading = 2.00 kg$

Part c)

When elevator is ascending with constant speed 39 m/s and acceleration 10 m/s/s then we will have

$F_{net} = ma$

$T - mg = ma$

$T = mg + ma$

Reading is given as

$Reading = \frac{mg + ma}{g}$

$Reading = 2.00\frac{9.81 + 10}{9.81}$

$Reading = 4.04 kg$

Part d)

Here the speed of the elevator is constant initially

from t = 0 to t = 4.9 s

so the reading of the scale will be same as that of weight of the block

Then its speed will reduce to zero in next 3.2 s

from t = 4.9 to t = 8.1 s

The reading of the scale will be less than the actual mass

3 0

3 years ago

Precautions to be taken for the displacement method

finlep [7]

Volume by Displacement. The displacement method (submersion, or dunking method) can be used to accurately measure the volume of the human body and other oddly shaped objects by measuring the volume of fluid displaced when the object is submerged.

The following precautions should be taken very observantly:-

The line of sight must be perpendicular to measuring scale to avoid parallax error. Formation of bubbles inside the cylinder should be completely avoided. Any bubbles within leads to wrong measurements.

8 0

3 years ago

A race car moving along a circular track has a centripetal acceleration of 15.4 m/s? If the car has

Helen [10]

Answer:

r = 58.44 [m]

Explanation:

To solve this problem we must use the following equation that relates the centripetal acceleration with the tangential velocity and the radius of rotation.

a = v²/r

where:

a = centripetal acceleration = 15.4 [m/s²]

v = tangential speed = 30 [m/s]

r = radius or distance [m]

r = v²/a

r = 30²/15.4

r = 58.44 [m]

3 0

3 years ago