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Andre45 [30]
3 years ago
8

Newton’s first law of motion applies to what?

Physics
1 answer:
lapo4ka [179]3 years ago
3 0

Vehicles, Rockets etc.

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An information signal consists of a 25 Hz and a 75 Hz sine waves summed together. It is sampled at a frequency of 500 Hz, the hi
denpristay [2]

Answer:

Explanation:

An information contains

25Hz and 75Hz sine wave

Sample frequency is 500Hz

The analogy signal are generally

y(t) = Asin(2πx/λ - wt), w=2πf

y1(t)=Asin(2πx/λ - wt)

y1(t)=Asin(2πx/λ - 2π•25t)

y1(t)=Asin(2πx/λ - 50πt)

Similarly

y2(t)=Asin(2πx/λ - 150πt)

Using Nyquist theorem

Nyquist Theorem states that in order to adequately reproduce a signal it should be periodically sampled at a rate that is 2 times the highest frequency you wish to record.

From sampling

f(nyquist)=f(sample)/2

f(nyquist)=500/2

f(nyquist)=250Hz

From signal

The highest frequency is 150Hz

F(nyquist) = 2×F(highest)

f(nyquist)= 2×150

f(nyquist)= 300Hz

Sample per frequency Ns is given as

Ns=F(sample)/F(highest signal)

Ns=500/150

Ns=3.33sample/period

This is above nyquist rate of 2sample/period

So signal below 300Hz reproduced without aliasing.

The highest resulting frequency is 300Hz

6 0
3 years ago
An object moving with a constant
jeka57 [31]

Answer:

Acceleration:

{ \tt{a =  \frac{v - u}{t} }} \\ { \tt{a =  \frac{20 - 10}{5} }} \\ { \tt{a = 2 \: m {s}^{ - 2} }}

From third equation:

{ \bf{ {v}^{2}  =  {u}^{2}  + 2as}} \\ { \tt{s =  \frac{ {20}^{2}  -  {10}^{2} }{2 \times 2} }} \\   = { \tt{s = 75 \: m}}

5 0
2 years ago
Read 2 more answers
A block is at rest on a plank whose angle can be varied. The angle is gradually increased from 0 deg. At 31.8°, the block starts
galina1969 [7]

Answer:

\mu_s = 0.62

\mu_k = 0.415

The motion of the block is downwards with acceleration 1.7 m/s^2.

Explanation:

First, we will calculate the acceleration using the kinematics equations. We will denote the direction along the incline as x-direction.

x - x_0 = v_0t + \frac{1}{2}at^2\\3.4 = 0 + \frac{1}{2}a(2)^2\\a = 1.7~m/s^2

Newton’s Second Law can be used to find the net force applied on the block in the -x-direction.

F = ma\\F = 1.7m

Now, let’s investigate the free-body diagram of the block.

Along the x-direction, there are two forces: The x-component of the block’s weight and the kinetic friction force. Therefore,

F = mg\sin(\theta) - \mu_k mg\cos(\theta)\\1.7m = mg\sin(31.8) - \mu_k mg\cos(31.8)\\1.7 = (9.8)\sin(\theta) - mu_k(9.8)\cos(\theta)\\mu_k = 0.415

As for the static friction, we will consider the angle 31.8, but just before the block starts the move.

mg\sin(31.8) = \mu_s mg\cos(31.8)\\\mu_s = tan(31.8) = 0.62

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3 years ago
How does your education contribute to comunity development
Angelina_Jolie [31]
Because everybody in community needs to be smart & have some type of knowledge
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3 years ago
If you let him have Czechoslovakia, he is going to.​
valentinak56 [21]

I - ok. Czechoslovakia is officially taken.

4 0
2 years ago
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