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STatiana [176]
3 years ago
12

No force is necessary to

Physics
1 answer:
igor_vitrenko [27]3 years ago
4 0
No force is necessary to keep a moving object moving (in a straight line at a constant speed).
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10.6. Go back to the original functions but change one of the minus signs to a plus sign (so now f(x,t) = 2.0*sin(x+t) and g(x,t
Pavel [41]

Answer:

The period of the wave does not change looting the value that accompanies the time, the wavelength does not change since it is the constant that accompanies x.

We see that the amplitude is twice the amplitude of the incident waves. Since the wave is stationary the velocity is zero

Explanation:

In this exercise we are given the equation of two traveling waves, it is asked to find the resulting wave

    u = f + g

    u = 2 sin (x + t) + 2 sin (x-t)

we will develop double angled breasts

    u = 2 [(sin x cos t + sin t cos x) + (sin x cos t - sin t cos x)]

    u = 2 [2 sin x cos t]

    u = 4 sin x cos t

The period of the wave does not change looting the value that accompanies the time, the wavelength does not change since it is the constant that accompanies x.

We see that the amplitude is twice the amplitude of the incident waves. Since the wave is stationary the velocity is zero

8 0
3 years ago
A stone is dropped at t = 0. A second stone, with 6 times the mass of the first, is dropped from the same point at t = 59 ms. (a
xxTIMURxx [149]

Answer:y_{com}=0.707 m

v_{com}=3.713 m/s

Explanation:

Given

first stone mass is m

second stone mass is 6 m

distance traveled by  first stone in 430 ms

y_1=ut+\frac{at^2}{2}

y_1=0+\frac{g(0.43)^2}{2}

y_1=0.9069 m

Distance traveled by stone 2 in t=430-59=371 ms

y_2=ut+\frac{at^2}{2}

y_2=0+\frac{g(0.0.371)^2}{2}

y_2=0.674 m

velocity of first stone after t=0.43 s

v_1=u+at

v_1=0+9.8\times 0.43=4.214 m/s

velocity of second stone after t=0.371 s

v_2=u+at

v_2=0+9.8\times 0.371=3.63 m/s

Position of Center of mass of system

y=\frac{y_1m_1+y_2m_2}{m_1+m_2}

y=\frac{0.9069\times m+0.674\times 6m}{m+6m}

y=\frac{4.95m}{7m}=0.707 m

Velocity of COM

v_{com}=\frac{v_1m_1+v_2m_2}{m_1+m_2}

v_{com}=\frac{4.214\times m+3.63\times 6m}{m+6m}

v_{com}=3.713 m/s

6 0
4 years ago
3. A car going 22m/s accelerates to pass a truck. Five seconds late the car is going 35m/s. Calculate the acceleration of the ca
DedPeter [7]

Answer:

57

Explanation:

6 0
3 years ago
The frequency is slowly increased. Once it passes the critical value fb, the student hears the ball bounce. There is now enough
vitfil [10]

Answer: g = acceleration = A*w^2 = A*(2*pi*fb)^2.

Explanation:

The ball bounces when the acceleration of the ball exceeds that of gravity. If A and fb are measured at that point, g = acceleration = A*w^2 = A*(2*pi*fb)^2.

3 0
3 years ago
PLEASE HELP I NEED TO TURN IT IN IN AN HOUR ITLL GIVE YOU POINTS PLS PLEASE
miv72 [106K]

Answer:

  1. 17.95025
  2. 172.3995

Explanation:

1.

\mathrm{The\:solution\:for\:Long\:Division\:of}\:\frac{2.995}{0.16685}\:\\\mathrm{is}\:17.95025

2.

\mathrm{Multiply\:without\:the\:decimal\:points,\:then\:put\:the\:decimal\:point\:in\:the\:answer}\\\\910\times\:18945=17239950\\\\910\mathrm{\:has\:}0\mathrm{\:decimal\:places}\\0.18945\mathrm{\:has\:}5\mathrm{\:decimal\:places}\\\\\mathrm{Therefore,\:the\:answer\:has\:}5\mathrm{\:decimal\:places}\\\\=172.39950\\\\Refine\\=172.3995

5 0
3 years ago
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