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Margarita [4]
2 years ago
14

A stretched string has a mass per unit length of 5.00 g/cm and a tension of 10.0 N. A sinusoidal wave on this string has an ampl

itude of 0.12 mm and a frequency of 100 Hz and is traveling in the negative direction of an x axis. If the wave equation is of the form y(x, t) = ym sin(kx ± ωt), what are (a) ym, (b) k, (c) ω, and (d) the correct choice of sign in front of ω?
Physics
1 answer:
8_murik_8 [283]2 years ago
4 0

Answer:

0.12 mm ; 140.50 rad/m ; 628.32 rad/sec ; +

Explanation:

Given the wave equation of the form :

y(x, t) = ym sin(kx ± ωt)

Mas per unit length (u) = 5 g/cm = (5÷1000)kg / 0.01m) = 0.005kg/0.01m = 0.5kg/m

Tension, T = 10 N

Amplitude, A = 0.12 mm

Frequency, F = 100 Hz

Comparing with the general wave equation :

y = Asin(kx ± ωt)

A = amplitude = ym = 0.12 mm

2.) k = 2π / λ

Recall :

v = fλ

v = sqrt(T/u) = sqrt(10/0.5) = sqrt(20) = 4.472

λ = v/ f = 4.472 / 100 = 0.04472

Hence,

k = (2 * π) / 0.04472

k = 140.50 rad/m

3.) Angular frequency, ω

ω = 2πf = 2 * 3.14 * 100 = 628.32 rad/sec

4.) sign is +ve

Direction of wave propagation as given is in the negative x axis

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6 0
1 year ago
Two point charges are 10.0cm apart and have charges of 2.0uC and -2.0uC, respectively. What is the magnitude of the electric fie
elena-14-01-66 [18.8K]
The electric field generated by a point charge is given by:
E= k_e \frac{Q}{r^2}
where
k_e = 8.99 \cdot 10^9 Nm^2 C^{-2} is the Coulomb's constant
Q is the charge
r is the distance from the charge

We want to know the net electric field at the midpoint between the two charges, so at a distance of r=5.0 cm=0.05 m from each of them. 

Let's calculate first the electric field generated by the positive charge at that point:
E_1=k_e  \frac{Q_1}{r^2}=(8.99 \cdot 10^9 Nm^2C^{-2}) \frac{(2.0 \cdot 10^{-6} C)}{(0.05 m)^2} =+7.19 \cdot 10^6 N/C
where the positive sign means its direction is away from the charge.

while the electric field generated by the negative charge is:
E_2=k_e \frac{Q_1}{r^2}=(8.99 \cdot 10^9 Nm^2C^{-2}) \frac{(-2.0 \cdot 10^{-6} C)}{(0.05 m)^2} =-7.19 \cdot 10^6 N/C
where the negative sign means its direction is toward the charge.

If we assume that the positive charge is on the left and the negative charge is on the right, we see that E1 is directed to the right, and E2 is directed to the right as well. This means that the net electric field at the midpoint between the two charges is just the sum of the two fields:
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3 years ago
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emmainna [20.7K]

Given

The projectile is in air for a time of t=8 sec

To find

The time it takes to reach the highest point

Explanation

A projectile moves up to the highest point and then again moves down following a parabolic path.

So it will reach the highest point at a time half the time it requires to follow teh parabolic path.

The time taken to reach the highest point is 4 sec

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Explanation:

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olganol [36]

Answer:

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θ = angle = ?

Using the equation

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y = 0.0667 m

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