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strojnjashka [21]

4 years ago

7

The amplitude, or magnitude, of a sinusoidal source is the maximum value of the source. What is the amplitude of the voltage sou

rce described as v(t)=50cos(2000t−45∘) mVv(t)=50cos⁡(2000t−45∘) mV ?

1 answer:

liraira [26]4 years ago

7 0

Answer:

<em>A = 0.05 V</em>

Explanation:

<u>Sinusoidal Functions</u>

A sinusoid or sinusoidal function is a sine or cosine which general equation is

$F(x)=A.sin(wt-\phi)$

Or also

$F(x)=A.cos(wt-\phi)$

Where A is the amplitude or maximum value, w is the angular frequency, t is the time and $\phi$ is the phase shift.

Comparing the given expression with the general formula

$v(t)=50cos(2000t-45^o) mV$

We can establish that A=50 mV = 0.05 V

$A = 0.05\ V$

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State the laws of vibration of a stringed Instrument

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If the length and linear density are constant, the frequency is directly proportional to the square root of the tension.

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3 years ago

The mean diameters of Mars and Earth are 6.9 ✕ 103 km and 1.3 ✕ 104 km, respectively. The mass of Mars is 0.11 times Earth's mas

Roman55 [17]

Answer:

(a) Ratio of mean density is 0.735

(b) Value of g on mars 0.920 $m,/sec^2$

(c) Escape velocity on earth is $3.563\times 10^4m/sec$

Explanation:

We have given radius of mars $R_{mars}=6.9\times 10^3km=6.9\times 10^6m$ and radius of earth $R_{E}=1.3\times 10^4km=1.3\times 10^7m$

Mass of earth $M_E=5.972\times 10^{24}kg$

So mass of mars $M_m=5.972\times\times 0.11 \times 10^{24}=0.657\times 10^{24}kg$

Volume of mars $V=\frac{4}{3}\pi R^3=\frac{4}{3}\times 3.14\times (6.9\times 10^6)^3=1375.357\times 10^{18}m^3$

So density of mars $d_{mars}=\frac{mass}{volume}=\frac{0.657\times 10^{24}}{1375.357\times 10^{18}}=477.69kg/m^3$

Volume of earth $V=\frac{4}{3}\pi R^3=\frac{4}{3}\times 3.14\times (1.3\times 10^7)^3=9.198\times 10^{21}m^3$

So density of earth $d_{E}=\frac{mass}{volume}=\frac{5.972\times 10^{24}}{9.198\times 10^{21}}=649.271kg/m^3$

(A) So the ratio of mean density $\frac{d_{mars}}{d_E}=\frac{477.69}{649.27}=0.735$

(B) Value of g on mars

g is given by $g=\frac{GM}{R^2}=\frac{6.67\times 10^{-11}\times0.657\times 10^{24}}{(6.9\times 10^6)^2}=0.920m/sec^2$

(c) Escape velocity is given by

$v=\sqrt{\frac{2GM}{R}}=\sqrt{\frac{2\times 6.67\times 10^{-11}\times 0.657\times 10^{24}}{6.9\times 10^6}}=3.563\times 10^4m/sec$

5 0

4 years ago

Read 2 more answers

Why are a 12 ounce hard seltzer and 1.5 ounces of liquor both standard drinks? alchohol edu

Mars2501 [29]

12 ounce hard seltzer and 1.5 ounces of liquor are standard drinks because they contain the <u>same amount</u><u> </u><u>of </u><u>pure alcohol</u>

<h3>What is standard drink?</h3>

This is a term used to refer to the measure of alcoholic content of drinks such that the drink should have 14 grams of pure alcohol or 0.6 fluid ounces of pure alcohol. This concept of standard drink is applicable in the United States of America.

The equivalence of a standard drink is 5 percent alcohol as seen in regular beer, 12 percent as seen in wines, and 40 percent as seen in distilled spirits. This measurement in taken by the percentage of the total volume of the beverage. hence the quantity of the drink or beverage may be different as the percentage is what determines the amount of alcohol present

From standard drink chart both drinks has same amount of pure alcohol so they are said to be standard drinks

Read more on standard drinks here: brainly.com/question/17645986

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6 0

2 years ago

A car is traveling 35 mph on a smooth surface. If a balanced force is applied to the car, what happens?

zepelin [54]

Answer:

Here is the answer.

Explanation:

Balanced forces- they are those forces that produce 0 resultant forces.

therefore, on applying a balanced force on the object, it wouldn't result in any change, as resultant force would be 0.

7 0

3 years ago

A horizontal spring is lying on a frictionless surface. One end of the spring is attached to a wall, and the other end is connec

juin [17]

Answer:

Explanation:

Given that:

angular frequency = 11.3 rad/s

Spring constant (k) = $= \omega^2 \times m$

k = (11.3)² m

k = 127.7 m

where;

$x_1$ = 0.065 m

$x_2$ = 0.048 m

According to the conservation of energies;

$E_1=E_2$

∴

$\Big(\dfrac{1}{2} \Big) kx_1^2 =\Big(\dfrac{1}{2} \Big) mv_2^2 + \Big(\dfrac{1}{2} \Big) kx_2^2$

$kx_1^2 = mv_2^2 + kx_2^2$

$(127.7 \ m) \times 0.065^2 = v_2^2 + (127.7 \ m) \times 0.048^2$

$0.5395325 = v_2^2 +0.2942208 \\ \\ 0.5395325 - 0.2942208 = v_2^2 \\ \\ v_2^2 = 0.2453117 \\ \\ v_2 = \sqrt{0.2453117} \\ \\ \mathbf{ v_2 \simeq0.50 \ m/s}$

4 0

3 years ago