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lesantik [10]
3 years ago
10

A transverse wave is propagated in a string stretched along the x-axis. The equation of the wave, in SI units, is given by:

Physics
1 answer:
irina [24]3 years ago
6 0

Answer:

The wave speed in SI units, is closest to - 2.7 m/s

Explanation:

given information:

y = 0.005 cos [π(38t - 14x)]

we know that the wave function is

y =  A cos (kx ± ωt)

where

A = amplitude

w = angular velocity

k = wave number

thus,

y = 0.005 cos [(14πx - 38πt)]

ω = - 38π

k = 14π

to find the speed v, we use the equation below

v = ω/k

so,

v = - 38π/14π

  = - 2.7 m/s

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Answer:

T=6.75s

Explanation:

We must separate the motion into two parts, the first when the rocket's engines is on  and the second when the rocket's engines is off. So, we need to know the height (h_1) that the rocket reaches while its engine is on and we need to know the distance (h_2) that it travels while its engine is off.

For solving this we use the kinematic equations:

In the first part we have:

h_1=v_0T+\frac{1}{2}aT^2\\h_1=0*T+\frac{1}{2}(16\frac{m}{s^2})T^2\\h_1=8\frac{m}{s^2}T^2\\

and the final speed is:

v_f=v_0+aT\\v_f=0+16\frac{m}{s^2}T\\v_f=16\frac{m}{s^2}T

In the second part, the final speed of the first part it will be the initial speed, and the final speed is zero, since gravity slows it down the rocket.

So, we have:

v_f^2=v_0^2+2gh_2\\2gh_2=v_f^2-v_0^2\\h_2=\frac{v_f^2-v_0^2}{2g}\\h_2=\frac{0^2-(16\frac{m}{s^2}T)^2}{2(-9.8\frac{m}{s^2})}\\h_2=\frac{-256\frac{m^2}{s^4}T^2}{-19.6\frac{m}{s^2}}\\h_2=13.06\frac{m}{s^2}T^2

The sum of these heights will give us the total height, which is known:

h=h_1+h_2\\960m=8\frac{m}{s^2}T^2+13.06\frac{m}{s^2}T^2\\960m=21.06\frac{m}{s^2}T^2\\T^2=\frac{960m}{21.06\frac{m}{s^2}}\\T^2=45.58s^2\\T=\sqrt{45.58s^2}\\T=6.75s

This is the time that its needed in order for the rocket to reach the required altitude.

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Answer:

Explanation:

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In a laboratory experiment, a diffraction grating produces an interference pattern on a screen. If the number of slits in (inclu
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Answer:

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As we know that the fringe width is inversely proportional to the number of slits, so as the number of slits increases, the fringe width decreases, hence the fringes are narrower, bright and close together.

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A couple of astronauts agree to rendezvous in space after hours. Their plan is to let gravity bring them together. She has a mas
Oduvanchick [21]

Answer:

(a) Acceleration of female astronaut = 9.33*10^-12 m/s^2; Acceleration of male astronaut = 7.56*10^-12 m/s^2

(b) 27.013 days

(c) No, their accelerations would not be constant.

Explanation:

In the given question, we have:

Mass of female astronaut M_{1} = 60.0 kg

Mass of male astronaut M_{2} = 74.0 kg

Distance between them (S) = 23.0 m

(a) The free-body diagram is shown in the attached figure.

Using the equations below, the initial accelerations of the two astronauts can be calculated:

Force of gravity (F) = \frac{G*M_{1}*M_{2}}{S^{2} }

G = 6.67*10^-11 \frac{m^{3} }{kg*s^{2} }

F = (6.67*10^-11 *60*74)/23^2 = 5.598*10^-10 N

For the female astronaut, her initial acceleration = F/M_{1} = 5.598*10^-10/60 = 9.33*10^-12 m/s^2

For the male astronaut, his acceleration = F/M_{2} = 5.598*10^-10/74 = 7.56*10^-12 m/s^2

(b) Since the different between their mass is not much, we can deduce that:

a_{average} = \frac{a_{1}+a_{2}}{2} = (9.33*10^-12 + 7.56*10^-12)/2 = 8.445*10^-12 m/s^2

Using the equation below, we can calculate the the time:

S = ut + 1/2 (at^2)   where u = 0

23 = 1/2 (8.445*10^-12)*t^2

t^2 = 5.447*10^12

t = 2333883.044 s = 27.013 days

(c) No, their accelerations will not be constant. It will increase because their radii would be decreasing.

8 0
3 years ago
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