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What beat frequencies result if a piano hammer hits three strings that emit frequencies of 392.0, 587.3, and 146.8 hz? (enter yo

FinnZ [79.3K]

128.1-127.8= 0.3Hz
129.1-128.1= 1.0Hz 
129.1-127.8= 1.3Hz

3 0

3 years ago

Electrical systems are governed by Ohm’s law, which states that V = IR, where V is the voltage, I is the current, and R is the r

ella [17]

Answer:

$\frac{dR(t)}{dt}=0.06\Omega$

Explanation:

Since $R(t)=\frac{V}{I(t)}$ , we calculate the resistance rate by deriving this formula with respect to time:

$\frac{dR(t)}{dt}=\frac{d}{dt}(\frac{V}{I(t)})=V\frac{d}{dt}(\frac{1}{I(t)})$

Deriving what is left (remember that $(\frac{1}{f(x)})'=-\frac{1}{f(x)^2}f'(x)$ ):

$\frac{d}{dt}(\frac{1}{I(t)})=-\frac{1}{I(t)^2}\frac{dI(t)}{dt}$

So we have:

$\frac{dR(t)}{dt}=-\frac{V}{I(t)^2}\frac{dI(t)}{dt}$

Which for our values is (the rate of I(t) is decreasing so we put a negative sign):

$\frac{dR(t)}{dt}=-\frac{24V}{(56A)^2}(-8A/s)=0.06\Omega$

8 0

3 years ago

A golf ball is struck with a five iron on level ground. it lands 100.0 m away 4.60 s later. what was the magnitude and direction

finlep [7]

consider the motion in x-direction

$v_{ox}$ = initial velocity in x-direction = ?

X = horizontal distance traveled = 100 m

$a_{x}$ = acceleration along x-direction = 0 m/s²

t = time of travel = 4.60 sec

Using the equation

X = $v_{ox}$ t + (0.5) $a_{x}$ t²

100 = $v_{ox}$ (4.60)

$v_{ox}$ = 21.7 m/s

consider the motion along y-direction

$v_{oy}$ = initial velocity in y-direction = ?

Y = vertical displacement = 0 m

$a_{y}$ = acceleration along x-direction = - 9.8 m/s²

t = time of travel = 4.60 sec

Using the equation

Y = $v_{oy}$ t + (0.5) $a_{y}$ t²

0 = $v_{oy}$ (4.60) + (0.5) (- 9.8) (4.60)²

$v_{oy}$ = 22.54 m/s

initial velocity is given as

$v_{o}$ = sqrt(( $v_{ox}$ )² + ( $v_{oy}$ )²)

$v_{o}$ = sqrt((21.7)² + (22.54)²) = 31.3 m/s

direction: θ = tan⁻¹(22.54/21.7) = 46.12 deg

6 0

3 years ago

A ball is thrown into the air by a baby alien on a planet in the system of Alpha Centauri with a velocity of 27 ft/s. Its height

Ahat [919]

Answer:

$V_{3.01}=-93.2m/s$

$V_{3.005}=-93.1m/s$

$V_{3.002}=-93.04m/s$

$V_{3.001}=-93.02m/s$

$V_{3}=-93m/s$

Explanation:

To calculate average velocity we need the position for both instants t0 and t1.

Now we will proceed to calculate all the positions we need:

$Y_{3}=-99m/s$

$Y_{3.01}=-99.932m/s$

$Y_{3.005}=-99.4655m/s$

$Y_{3.002}=-99.18608m/s$

$Y_{3.001}=-99.09302m/s$

Replacing these values into the formula for average velocity:

$V_{3-3.01}=\frac{Y_{3.01}-Y_{3}}{3.01-3}=-93.2m/s$

$V_{3-3.005}=\frac{Y_{3.005}-Y_{3}}{3.005-3}=-93.1m/s$

$V_{3-3.002}=\frac{Y_{3.002}-Y_{3}}{3.005-3}=-93.04m/s$

$V_{3-3.001}=\frac{Y_{3.001}-Y_{3}}{3.001-3}=-93.02m/s$

To know the actual velocity, we derive the position and we get:

$V=27-40t = -93m/s$

5 0

3 years ago

A sling is used to give a stone an initial velocity of 20 at an angle of 30 above the horizontal. The stone travels through the

Luba_88 [7]

Answer:

Option E is correct.

There must be a horizontal wind opposite the direction of the stone's motion, because ignoring air resistance when calculating the horizontal range would yield a value greater than 32 m.

Explanation:

Normally, ignoring air resistance, for projectile motion, the range (horizontal distance teavelled) of the motion is given as

R = (u² sin 2θ)/g

where

u = initial velocity of the projectile = 20 m/s

θ = angle above the horizontal at which the projectile was launched = 30°

g = acceleration due to gravity = 9.8 m/s²

R = (30² sin 60°) ÷ 9.8

R = 78.53 m

So, Normally, the stone should travel a horizontal distance of 78.53 m. So, travelling a horizontal distance of 32 m (less than half of what the range should be without air resistance) means that, the motion of the stone was impeded, hence, option E is correct.

There must be a horizontal wind opposite the direction of the stone's motion, because ignoring air resistance when calculating the horizontal range would yield a value greater than 32 m.

Hope this Helps!!!

7 0

3 years ago