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3 years ago

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The greater the amplitude, the greater the wave's:

1 answer:

dolphi86 [110]3 years ago

4 0

The greater the waves will be mass

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A solid ball is released from rest and slides down a hillside that slopes downward at 65.0" from the horizontal

PilotLPTM [1.2K]

Setting reference frame so that the x axis is along the incline and y is perpendicular to the incline
<span>X: mgsin65 - F = mAx </span>
<span>Y: N - mgcos65 = 0 (N is the normal force on the incline) N = mgcos65 (which we knew) </span>
<span>Moment about center of mass: </span>
<span>Fr = Iα </span>
<span>Now Ax = rα </span>
<span>and F = umgcos65 </span>
<span>mgsin65 - umgcos65 = mrα -------------> gsin65 - ugcos65 = rα (this is the X equation m's cancel) </span>
<span>umgcos65(r) = 0.4mr^2(α) -----------> ugcos65(r) = 0.4r(rα) (This is the moment equation m's cancel) </span>
<span>ugcos65(r) = 0.4r(gsin65 - ugcos65) ( moment equation subbing in X equation for rα) </span>
<span>ugcos65 = 0.4(gsin65 - ugcos65) </span>
<span>1.4ugcos65 = 0.4gsin65 </span>
<span>1.4ucos65 = 0.4sin65 </span>
<span>u = 0.4sin65/1.4cos65 </span>
<span>u = 0.613 </span>

3 0

3 years ago

Consider the power dissipated in a series R–L–C circuit with R=280Ω, L=100mH, C=0.800μF, V=50V, and ω=10500rad/s. The current an

ki77a [65]

Answer:

0.28802

2.57162 W

14.28 W

53.55 W

6.07142 W

Explanation:

R = 280Ω

L = 100 mH

C = 0.800 μF

V = 50 V

ω = 10500rad/s

For RLC circuit impedance is given by

$Z=\sqrt{R^2+(X_L-X_C)^2}\\\Rightarrow Z=\sqrt{R^2+(\omega L-\dfrac{1}{\omega C})^2}\\\Rightarrow Z=\sqrt{280^2+(10500\times 100\times 10^{-3}-\dfrac{1}{10500\times 0.8\times 10^{-6}})^2}\\\Rightarrow Z=972.1483\ \Omega$

Power factor is given by

$F=\dfrac{R}{Z}\\\Rightarrow F=\dfrac{280}{972.1483}\\\Rightarrow F=0.28802$

The power factor is 0.28802

The average power to the circuit is given by

$P=\dfrac{V^2}{Z}\\\Rightarrow P=\dfrac{50^2}{972.1483}\\\Rightarrow P=2.57162\ W$

The average power to the circuit is 2.57162 W

Power to resistor

$P_R=IR\\\Rightarrow P_R=5.1\times 10^{-2}\times 280\\\Rightarrow P_R=14.28\ W$

Power to resistor is 14.28 W

Power to inductor

$P_L=IX_L\\\Rightarrow P_L=5.1\times 10^{-2}\times 10500\times 100\times 10^{-3}\\\Rightarrow P_L=53.55\ W$

Power to the inductor is 53.55 W

Power to the capacitor

$P_C=IX_C\\\Rightarrow P_C=5.1\times 10^{-2}\times \dfrac{1}{10500\times 0.8\times 10^{-6}}\\\Rightarrow P_C=6.07142\ W$

The power to the capacitor is 6.07142 W

8 0

3 years ago

A pulley system lifts a 500-lb. block 2.0 feet with an effort of only 50 lb. If the 50 lb. moves 30 feet, calculate the efficien

tatiyna

The answer to this is 67%

4 0

3 years ago

Read 2 more answers

What phytophotodermatitis caused by?

Paraphin [41]

Phytophotodermatitis happens when certain plant chemicals cause the skin to become inflamed following exposure to sunlight.

3 0

3 years ago

Read 2 more answers

Can i get help for the parts for the two questions? MATHPHYSSSS

Anestetic [448]

Explanation:

003 (part 1 of 2)

Pressure is force divided by area.

P = F / A

P = (117 kg × 9.8 m/s²) / (2 × (0.05 m)²)

P = 229,320 Pa

003 (part 2 of 2)

There are approximately 6895 Pa in 1 psi.

P = 229,320 Pa × (1 psi / 6895 Pa)

P = 33.3 psi

004 (part 1 of 2)

Since the collisions are elastic, the angle of reflection is the same as the angle of incidence (it bounces off at the same angle).

Impulse = change in momentum

F Δt = m Δv

F (36 s) = (300 × 0.003 kg) (5.2 sin 57° m/s − (-5.2 sin 57° m/s))

F = 0.218 N

004 (part 2 of 2)

Pressure is force over area.

P = F / A

P = 0.218 N / 0.712 m²

P = 0.306 N/m²

7 0

3 years ago