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2 years ago

7

A guitar string 0.750 m long has a fundamental frequency of 440 hz. The guitarist wants to accent the 5th harmonic by picking at

the first loop from the end of the string for this harmonic. How far from the end of the string (cm) should she pick?

1 answer:

weeeeeb [17]2 years ago

7 0

Answer: a b c or d

Explanation

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Air at the poles tends to flow close to the surface toward the equator. What can you conclude about the characteristics of this

LenaWriter [7]

Answer:

That the polar air has has more pressure than the air at the equator.

Explanation:

5 0

3 years ago

A golf club hits a stationary 0.05kg golf ball with and average force of 5.0 x 10^3 newtons accelerating the ball at 44 meters p

maxonik [38]

Answer: The magnitude of impulse imparted to the ball by the golf club is 2.2 N seconds

Explanation:

Force applied on the golf ball = $5.0\times 10^3 N$

Mass of the ball = 0.05 kg

Velocity with which ball is accelerating = 44 m/s

Time period over which forece applied = t

$f=ma=\frac{m\times v}{t}$

$t=\frac{0.05 kg\times 44m/s}{5.0\times 10^3 N}=4.4\times 10^{-4} seconds$

$Impulse=(force)\times (time)=f\times t = 5.0\times 10^3\times 4.4\times 10^{-4} seconds=2.2$ Newton seconds

The magnitude of impulse imparted to the ball by the golf club is 2.2 N seconds

7 0

3 years ago

A cell phone weighing 80 grams is flying through the air at 15 m/s. What is its kinetic energy

Contact [7]

Answer:

The kinetic energy of the cell phone is 9J

Explanation:

The kinetic energy is the energy possessed by a body by virtue of motion.

The kinetic energy is expressed as

KE= 1/2m(v)²

Given data

Mass of cell phone m= 80g--to kg=80/1000= 0.08kg

Velocity of cell phone v= 15m/s

Substituting our given data we have

KE= 1/2*0.08(15)²

KE= (0.08*225)/2

KE=18/2

KE= 9J

8 0

3 years ago

What will change the velocity of a periodic wave?

Gwar [14]

The one that will change the velocity of a periodic wave is :
B. Changing the medium of the wave
Waves is always determined by the properties of the medium, which means that changing the medium will change the velocity of the wave

hope this helps

7 0

2 years ago

Read 2 more answers

Cesium-137 undergoes beta decay and has a half-life of 30.0 years. How many beta particles are emitted by a 14.0-g sample of ces

Mandarinka [93]

Answer: $0.81\times 10^{16}$ beta particles

Explanation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass = 14.0 g

Molar mass = 137 g/mol

$\text{Number of moles of cesium}=\frac{14.0g}{137g/mol}=0.102moles$

According to avogadro's law, 1 mole of every substance weighs equal to its molecular mass and contains avogadro's number $6.023\times 10^{23}$ of particles.

1 mole of cesium contains atoms = $6.023\times 10^{23}$

0.102 moles of cesium contains atoms = $\frac{6.023\times 10^{23}}{1}\times 0.102=0.614\times 10^{23}$

The relation of atoms with time for radioactivbe decay is:

$N_t=N_0\times \frac{1}{2}^{\frac{t}{t_{\frac{1}{2}}}}$

Where $N_t$ =atoms left undecayed

$N_0$ = initial atoms

t = time taken for decay = 3 minutes

${t_{\frac{1}{2}}}$ = half life = 30.0 years = $1.577\times 10^7$ minutes

The fraction that decays : $1-(\frac{1}{2})^{\frac{3}{1.577\times 10^7}}=1.32\times 10^{-7}$

Amount of particles that decay is = $0.614\times 10^{23}\times 1.32\times 10^{-7}=0.81\times 10^{16}$

Thus $0.81\times 10^{16}$ beta particles are emitted by a 14.0-g sample of cesium-137 in three minutes.

7 0

3 years ago