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irina [24]
3 years ago
13

How do I figure work done using W=Fd when I have a mass of 61.09kg, weight of 599.79N, d=153cm, and t=4.52s?

Physics
1 answer:
Sergio039 [100]3 years ago
4 0

You have the correct formula.  Work = F d .
You use it by gathering up 'F' and 'd' and throwing away everything else.

F = 599.79 N
d = 153 cm

That's all you need to calculate 'W' with your formula.  Just because they
also gave you the mass and the time doesn't mean you MUST use them.

My guess is that there are more parts to this problem where you may need
the mass and the time.  For example, if they ask you for the power, then
you'll need to divide the work by the time.


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B

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6 0
3 years ago
The sampling rate of an ADC is 8.1 kHz. What will be an appropriate cut-off frequency (break frequency) for the anti-aliasing fi
KiRa [710]

Answer:

4000 Hz

Explanation:

An anti-alias filter is usually added in front of the ADC to limit a certain range of input frequencies in order to avoid aliasing. This filter is usually a low pass filter that passes low frequencies but attenuates the high frequencies.

The Nyquist sampling criteria states that the sampling rate should be at least twice the maximum frequency component of the desired signal.

Sampling rate = 2(max input frequency)

From the relation we can find out the cut-off frequency for the anti-aliasing filter.

max input frequency = sampling rate/2

max input frequency = 8100/2 = 4050 Hz

Therefore, 4000 Hz would be an appropriate cut-off frequency for the anti-aliasing filter.

3 0
3 years ago
Unpolarized light is incident upon two polarization filters that do not have their transmission axes aligned. If 21% of the ligh
Debora [2.8K]

Answer:

49.6°

Explanation:

I_0 = Unpolarized light

I_2 = Light after passing though second filter = 0.21I_0

Polarized light passing through first filter

I_1=\frac{I_0}{2}

Polarized light passing through second filter

I_2=\frac{I_0}{2}cos^2\theta\\\Rightarrow 0.21I_0=\frac{I_0}{2}cos^2\theta\\\Rightarrow cos^2\theta=\frac{0.21I_0}{\frac{I_0}{2}}\\\Rightarrow cos\theta=\sqrt{\frac{0.21I_0}{\frac{I_0}{2}}}\\\Rightarrow \theta=cos^{-1}\sqrt{\frac{0.21I_0}{\frac{I_0}{2}}}\\\Rightarrow \theta=cos^{-1}\sqrt{0.21\times 2}\\\Rightarrow \theta=cos^{-1}\sqrt{0.42}\\\Rightarrow \theta=49.6^{\circ}

The angle between the two filters is 49.6°

5 0
4 years ago
Uniform solid sphere of mass M and radius R rotates with an angular speed w about an axis through is center A uniform solid cyli
Maru [420]

Answer:

e. 2ωs / √5

Explanation:

The rotational kinetic energy of any rigid body, like an extension of the translational kinetic energy, is defined as follows:

Krot = 1/2 *I * ω²

For a solid sphere of mass M and radius R, the moment of inertia regarding any axis through its center, is as follows:

I =2/5 M*R²⇒ Krot(sp) = 1/2 (2/5 M*R²)*ωs² (1)

For a solid cylinder, rotating through an axis running through the central axis of the cylinder, the moment of inertia can be calculated as follows:

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As both rotational kinetic energies must be equal each other, we can equate (1) and (2), as follows:

1/2 (2/5 M*R²)*ωs² = 1/2 (1/2*M*R²)*ωc²

Simplifying common terms, and solving for ωc, we have:

ωc = 2*ωs /√5

4 0
3 years ago
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