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Zielflug [23.3K]
4 years ago
14

How does noise affect signals? what happens if the level of noise becomes too high relative to the strength of the signal.

Physics
1 answer:
Nat2105 [25]4 years ago
6 0

Noise could be defined as electromagnetic fields that affect analog signals that are constantly changing. This process does not occur in a similar way with digital signals, which have fixed electrostatic pulses (For this reason they are able to withstand 'noise' because the power of these signals are much stronger than the power coming from noise).

That phenomenon does not happen with the analog signals which have a variable intensity and become vulnerable to any electronic noise interference.

When very high electromagnetic fields are generated, the waves of the analog signal cannot be perceived which causes problems in the transmitted signal (making it unintelligible to the receiver)

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How far (in meters) will you travel in 3 minutes running at a rate of 6 m/s?
trasher [3.6K]
The formula for distance is speed times time.

So 3 times 6 would be 18 m/s
7 0
3 years ago
REAL ANSWERS ONLY PLS
just olya [345]

Answer:

The statement of the student is correct.

Since B attained a higher velocity in a short amount of time, that is it accelerated faster(having a larger slope).

Slope = dy/dx

That is, <u>Velocity</u>

Time

which is acceleration.

That's my guess.

Hope it's right.

7 0
3 years ago
How much heat is needed to change the temperature of 3 grams of gold (c = 0.129 ) from 21°C to 363°C? The answer is expressed to
madam [21]

The correct answer to the question is : B) 132 Joules.

EXPLANATION :

As per the question, the mass of the gold m = 3 gram.

The initial temperature of the gold T = 21 degree celsius.

The final temperature of the gold  T' = 363 degree celsius.

Hence, the change in temperature  dT = T' - T

                                                          = 363 - 21 degree celsius

                                                          = 342 degree celsius.

The specific heat of the gold is given as c = 0.129\ J/g^0C

We are asked to calculate the heat required ( dQ ) to raise the temperature of gold.

The heat required for this is calculated as -

                                   dQ = mcDT

                                         = 3 × 342 × 0.129 J

                                         = 132.354 J

                                         ≈ 132 J

Hence, the correct answer is 132 J.

8 0
3 years ago
Read 2 more answers
1 point
7nadin3 [17]

Answer:

Gravity

Explanation:

It is gravity because the sun's gravity pulls all the planets towards itself and keeps the planets in the sun's orbit

8 0
3 years ago
Consider the collision of a 1500-kg car traveling east at 20.0 m/s (44.7 mph) with a 2000-kg truck traveling north at 25 m/s (55
masha68 [24]

Answer:

1. X_{cm}=-8.57m

2. Y_{cm}=-14.29m

3. V_{cm} = 16.66m/s

4. α = 59.05°

5. V_{cm} = 16.66m/s

6. α = 59.05°

Explanation:

The position of the center of mass 1s before the collision is:

X_{cm}=\frac{X_c*mc+X_t*m_t}{m_c+m_t}

where

X_c=-20m;  m_c=1500kg;  

X_t=0m;   m_t=2000kg;

Replacing these values:

X_{cm}=-8.57m

Y_{cm}=\frac{Y_c*mc+Y_t*m_t}{m_c+m_t}

where

Y_c=0m;  m_c=1500kg;  

Y_t=-25m;   m_t=2000kg;

Replacing these values:

Y_{cm}=-14.29m

The velocity of their center of mass is:

V_{cm-x}=\frac{V_{c-x}*mc+V_{t-x}*m_t}{m_c+m_t}

where

V_{c-x}=20m/s;  m_c=1500kg;  

V_{t-x}=0m/s;   m_t=2000kg;

Replacing these values:

V_{cm-x}=8.57m/s

V_{cm-y}=\frac{V_{c-y}*mc+V_{t-y}*m_t}{m_c+m_t}

where

V_{c-y}=0m;  m_c=1500kg;  

V_{t-y}=-25m;   m_t=2000kg;

Replacing these values:

V_{cm-y}=-14.29m

So, the magnitude of the velocity is:

V_{cm}=\sqrt{V_{cm-x}^2+V_{cm-y}^2}

V_{cm}=16.66m/s

The angle of the velocity is:

\alpha =atan(V_{cm-y}/V_{cm-x})

\alpha=59.05\°

Since on any collision, the velocity of the center of mass is preserved, then the velocity after the collision is the same as the previously calculated value of 16.66m/s at 59.0° due north of east

4 0
3 years ago
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