g A mass of 2 kg is attached to a spring whose constant is 7 N/m. The mass is initially released from a point 4 m above the equilibrium position with a downward velocity of 10 m/s, and the subsequent motion takes place in a medium that imparts a damping force numerically equal to 10 times the instantaneous velocity. What is the differential equation for the mass-spring system.

Answers

Answer 1

Answer:

mass 20 times of an amazing and all its motion

Related Questions

The smallest shift you can reliably measure on the screen is about 0.2 grid units. This shift corresponds to the precision of positions measured with the best Earth-based optical telescopes. If you cannot measure an angle smaller than this, what is the maximum distance at which a star can be located and still have a measurable parallax

Answers

Answer:

The distance is [tex]d = 1.5 *10^{15} \ km[/tex]

Explanation:

From the question we are told that

The smallest shift is [tex]d = 0.2 \ grid \ units[/tex]

Generally a grid unit is [tex]\frac{1}{10}[/tex] of an arcsec

This implies that 0.2 grid unit is [tex]k = \frac{0.2}{10} = 0.02 \ arc sec[/tex]

The maximum distance at which a star can be located and still have a measurable parallax is mathematically represented as

[tex]d = \frac{1}{k}[/tex]

substituting values

[tex]d = \frac{1}{0.02}[/tex]

[tex]d = 50 \ parsec[/tex]

Note [tex]1 \ parsec \ \to 3.26 \ light \ year \ \to 3.086*10^{13} \ km[/tex]

So [tex]d = 50 * 3.08 *10^{13}[/tex]

[tex]d = 1.5 *10^{15} \ km[/tex]

A soccer ball is released from rest at the top of a grassy incline. After 2.2 seconds, the ball travels 22 meters. One second later, the ball reaches the bottom of the incline. (Assume that the acceleration was constant.) How long was the incline

Answers

Answer:

x = 46.54m

Explanation:

In order to find the length of the incline you use the following formula:

[tex]x=v_ot+\frac{1}{2}at^2[/tex] (1)

vo: initial speed of the soccer ball = 0 m/s

t: time

a: acceleration

You first use the the fact that the ball traveled 22 m in 2.2 s. Whit this information you can calculate the acceleration a from the equation (1):

[tex]22m=\frac{1}{2}a(2.2s)^2\\\\a=9.09\frac{m}{s^2}[/tex] (2)

Next, you calculate the distance traveled by the ball for t = 3.2 s (one second later respect to t = 2.2s). The values of the distance calculated is the lenght of the incline:

[tex]x=\frac{1}{2}(9.09m/s^2)(3.2s)^2=46.54m[/tex] (3)

The length of the incline is 46.54 m

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