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3 years ago

5

Consider two waves defined by the wave functions y1(x,t)=0.50msin(2π3.00mx+2π4.00st) and y2(x,t)=0.50msin(2π6.00mx−2π4.00st). Wh

at are the similarities and differences between the two waves

1 answer:

guapka [62]3 years ago

4 0

Answer:

They two waves has the same amplitude and frequency but different wavelengths.

Explanation: comparing the wave equation above with the general wave equation

y(x,t) = Asin(2Πft + 2Πx/¶)

Let ¶ be the wavelength

A is the amplitude

f is the frequency

t is the time

They two waves has the same amplitude and frequency but different wavelengths.

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Two blocks are connected by a massless rope that passes over a 1 kg pulley with a radius of 12 cm. The rope moves over the pulle

tangare [24]

Answer:

$F=1.159$

Explanation:

From the question we are told that:

Mass of pulley $M=1kg$

Radius $r=12cm$

Mass of block A $M_a=2.1kg$

Mass of block B $m_b=4.1kg$

Spring constant $\mu= 358 J/m2$

Generally the equation for Torque is mathematically given by

Since $\sumF=ma$

At mass A

$T_2-f_3=2.1a$

At mass B

$4.8-T_1=4.1a$

At Pulley

$R(T_1-T_2)=\frac{1*1*R^2}{2}\frac{a}{R}$

$R(T_1-T_2)=0.55a$

Therefore the equation for total force F

At mass A+At mass B+At Pulley

$(T_2-f_3+4.8-T_1+R(T_1-T_2)=2.1a+4.1a+0.55a$

$(T_2-f_3+4.8-T_1+R(T_1-T_2)=2.7a+4.8a+0.55a$

$-f_3+4.1=6.75a$

$-f_3=6.75a+4.8$

Since From above equation

$M_{eff}=6.7kg$

Therefore

$T=2\pi \sqrt{{\frac{M_{eff}}{k}}$

$T=2\pi \sqrt{{\frac{6.75}{\mu}}$

$T=0.862s$

Generally the equation for frequency is mathematically given by

$F=\frac{1}{T} \\F=\frac{1}{0.862}$

$F=1.159$

3 0

3 years ago

If a CFOC was launched and travels 65 meters and is in the air for 3 seconds, what is the launch velocity and angle?

labwork [276]

Answer:

Lauch velocity (u) = 26.15 m/s

Lauch Angle (θ) = 35°

Explanation:

From the question given above, the following data were obtained:

Range (R) = 65 m

Time of flight (T) = 3 s

Acceleration due to gravity (g) = 10 m/s²

Lauch velocity (u) =?

Lauch Angle (θ) =?

R = u²Sin2θ /g

65 = u² × Sin2θ /10

Recall:

Sin2θ = 2SinθCosθ

65 = u² × 2SinθCosθ / 10

65 = u² × SinθCosθ / 5

Cross multiply

65 × 5 = u² × SinθCosθ

325 = u² × SinθCosθ .....(1)

T = 2uSinθ / g

3 = 2uSinθ / 10

3 = uSinθ / 5

Cross multiply

3 × 5 = uSinθ

15 = u × Sinθ

Divide both side by Sinθ

u = 15 / Sinθ....... (2)

Substitute the value of u in equation (2) into equation (1)

325 = u² × SinθCosθ

u = 15 / Sinθ

325 = (15 / Sinθ)² × SinθCosθ

325 = 225 / Sin²θ × SinθCosθ

325 = 225 × SinθCosθ / Sin²θ

325 = 225 × Cosθ / Sinθ

Cross multiply

325 × Sineθ = 225 × Cosθ

Divide both side by Cosθ

325 × Sineθ / Cosθ = 225

Divide both side by 325

Sineθ / Cosθ = 225 / 325

Sineθ / Cosθ = 0.6923

Recall:

Sineθ / Cosθ = Tanθ

Tanθ = 0.6923

Take the inverse of Tan

θ = Tan¯¹ 0.6923

θ = 35°

Substitute the value of θ into equation (2) to obtain the value of u.

u = 15 / Sinθ

θ = 35°

u = 15 / Sin 35

u = 15 / 0.5736

u = 26.15 m/s

Summary:

Lauch velocity (u) = 26.15 m/s

Lauch Angle (θ) = 35°

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