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Georgia [21]
3 years ago
10

A wave on a string is described by Y(x, t) : 15.0 sin( nclS - 4rrt),, where x and y are in centimeters and / is in seconds. (a)

What is the transverse speed for a point on the string at x : 6.00 cm when t - 0.250 s? (b) What is the maximum transverse speed of any point on the string? (c) What is the magnitude of the transverse acceleration for a poi
Physics
1 answer:
KIM [24]3 years ago
4 0

Answer:

dsf

Explanation:

fsdfs

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If the time for F18 to take off doubles, the acceleration will ____.
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as time increases acceleration decreases and vice versa

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Which event in the “The Medicine Bag” is most symbolic of Martin beginning to connect with his Sioux heritage?
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A- Martin brings his friends home to meet grandpa.

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A car of m = 1200. kg collides with a tree while traveling 60.0 mph. The collision occurs over a time period of 0.0500 seconds.
MrRissso [65]
You may know linear momentum is given by P= mass.velocity. Initially car is moving with some velocity so you know initial momentum of the car. Finally it comes to rest i.e final momentum of the car is 0. According to Newton's second law : Force = change in momentum /time. Applying this you'll get answer as 642840N. Hope it helped you. Revert back to me if you have any questions. Please check out the calculation it might be wrong!
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Assume that a satellite orbits mars 150km above its surface. Given that the mass of mars is 6.485 X 10^23kg, and the radius of m
Kisachek [45]
<span>3598 seconds The orbital period of a satellite is u=GM p = sqrt((4*pi/u)*a^3) Where p = period u = standard gravitational parameter which is GM (gravitational constant multiplied by planet mass). This is a much better figure to use than GM because we know u to a higher level of precision than we know either G or M. After all, we can calculate it from observations of satellites. To illustrate the difference, we know GM for Mars to within 7 significant figures. However, we only know G to within 4 digits. a = semi-major axis of orbit. Since we haven't been given u, but instead have been given the much more inferior value of M, let's calculate u from the gravitational constant and M. So u = 6.674x10^-11 m^3/(kg s^2) * 6.485x10^23 kg = 4.3281x10^13 m^3/s^2 The semi-major axis of the orbit is the altitude of the satellite plus the radius of the planet. So 150000 m + 3.396x10^6 m = 3.546x10^6 m Substitute the known values into the equation for the period. So p = sqrt((4 * pi / u) * a^3) p = sqrt((4 * 3.14159 / 4.3281x10^13 m^3/s^2) * (3.546x10^6 m)^3) p = sqrt((12.56636 / 4.3281x10^13 m^3/s^2) * 4.458782x10^19 m^3) p = sqrt(2.9034357x10^-13 s^2/m^3 * 4.458782x10^19 m^3) p = sqrt(1.2945785x10^7 s^2) p = 3598.025212 s Rounding to 4 significant figures, gives us 3598 seconds.</span>
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What is the difference between the initial position and the final position of an object?
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The initial is where you are starting and the final postion is where the object ends up
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