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1 year ago

Suppose you have a 100 mhz cpu and a a/d converter with an 8 khz sampling rate. Samples of the a/d converter are 32-bits

Physics

1 answer:

seropon [69]1 year ago

3 0

The sampling rate that's given regarding the CPU will produce a rate of 256kps.

<h3>How to calculate the rate? </h3>

The following can be gotten from the question:

Sampling rate = 8KHx

Number of bits = 32

The data rate will be:

= 32 × 8kbps

= 256kps

In conclusion, the sampling rate that's given regarding the CPU will produce a rate of 256kps.

Learn more about rates on:

brainly.com/question/2375289

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State the formula for calculating power

Alex

Where power P is in watts, voltage V is in volts and current I is in amperes (DC).Power Formula 2 – Mechanical power equation: Power P = E ⁄ t where power P is in watts, Power P = work / time (W ⁄ t). Energy E is in joules, and time t is in seconds.

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2 years ago

Which statement describes something that should always be done at the end of a calculation?

Reika [66]

Answer:

C) Check that the numerical answer is reasonable and the units are what you expected.

Explanation:

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3 years ago

Read 2 more answers

g a small smetal sphere, carrying a net charge is held stationarry. what is the speed are 0.4 m apart

Veseljchak [2.6K]

Answer:

The speed of q₂ is $4\sqrt{10}\ m/s$

Explanation:

Given that,

Distance = 0.4 m apart

Suppose, A small metal sphere, carrying a net charge q₁ = −2μC, is held in a stationary position by insulating supports. A second small metal sphere, with a net charge of q₂ = −8μC and mass 1.50g, is projected toward q₁. When the two spheres are 0.800m apart, q₂ is moving toward q₁ with speed 20m/s.

We need to calculate the speed of q₂

Using conservation of energy

$E_{i}=E_{f}$

$\dfrac{1}{2}mv_{i}^2+\dfrac{kq_{1}q_{2}}{r_{i}}=\dfrac{kq_{1}q_{2}}{r_{f}}+\dfrac{1}{2}mv_{f}^2$

$\dfrac{1}{2}m(v_{i}^2-v_{f}^2)=kq_{1}q_{2}(\dfrac{1}{r_{f}}-\dfrac{1}{r_{i}})$

Put the value into the formula

$\dfrac{1}{2}\times1.5\times10^{-3}(20^2-v_{f}^2)=9\times10^{9}\times-2\times10^{-6}\times-8\times10^{-6}(\dfrac{1}{(0.4)}-\dfrac{1}{(0.8)})$

$0.00075(400-v_{f}^2)=0.18
$

$400-v_{f}^2=\dfrac{0.18}{0.00075}$

$-v_{f}^2=240-400$

$v_{f}^2=160$

$v_{f}=4\sqrt{10}\ m/s$

Hence, The speed of q₂ is $4\sqrt{10}\ m/s$

7 0

2 years ago

Assume that the electric field E is equal to zero at a given point. Does it mean that the electric potential V must also be equa

lyudmila [28]

Answer:

No, this doesn't mean the electric potential equals zero.

Explanation:

In electrostatics, the electric field $\vec{E}$ is related to the gradient of the electric potential V with :

$\vec{E} (\vec{r}) = - \vec{\nabla} V (\vec{r})$

This means that for constant electric potential the electric field must be zero:

$V(\vec{r}) = k$

$\vec{E} (\vec{r}) = - \vec{\nabla} V (\vec{r}) = - \vec{\nabla} k$

$\vec{E} (\vec{r}) = - (\frac{\partial}{\partial x} , \frac{\partial}{\partial y } , \frac{\partial}{\partial z}) k$

$\vec{E} (\vec{r}) = - (\frac{\partial k}{\partial x} , \frac{\partial k}{\partial y } , \frac{\partial k}{\partial z})$

$\vec{E} (\vec{r}) = - (0,0,0)$

This is not the only case in which we would find an zero electric field, as, any scalar field with gradient zero will give an zero electric field. For example:

$V(\vec{r})= (x+2)^2 (y+4)^3 (z+5)^4$

give an electric field of zero at point (0,0,0)

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3 years ago

The number of planets, besides the earth, that are visible to the unaided eye is

sergejj [24]

The answer should be 5 it’s easy

3 0

2 years ago