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Tems11 [23]
3 years ago
5

Set up differential equation of angular S.H.M​

Physics
1 answer:
3241004551 [841]3 years ago
3 0

Answer:

{ \sf{ \omega \: is \: the \: angular \: velocity}} \\ { \sf{ \theta \: is \: the \: angular \: displacement}}

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A little boy is standing at the edge of a cliff 1000 m high. He throws a ball straight downward at an initial speed of 20 m/s, a
marin [14]

Answer:

The ball will be at 700 m above the ground.

Explanation:

We can use the following kinematic equation

y(t) = \ y_0 \ + \ v_0 \ t \ + \frac{1}{2} \ a \ t^2.

where y(t) represent the height from the ground. For our problem, the initial height will be:

y_0 \ = \ 1000 m.

The initial velocity:

v_0 = - 20 \frac{m}{s},

take into consideration the minus sign, that appears cause the ball its thrown down.  The same minus appears for the acceleration:

a=-10\frac{m}{s}

So, the equation for our problem its:

y(t) = \ 1000 m \ - \ 20 \ \frac{m}{s} \ t \ - \frac{1}{2} \ 10 \frac{m}{s^2} \ t^2.

Taking t=6 s:

y(6 \ s) = \ 1000 m \ - \ 20 \ \frac{m}{s} \ * \ 6 \ s \ - \frac{1}{2} \ 10 \frac{m}{s^2} \ * \ (6 s)^2.

y(6 \ s) = \ 1000 m \ - 120 m - \frac{1}{2} \ 10 \frac{m}{s^2} \ * \ 36 s^2.

y(6 \ s) = \ 1000 m \ - 120 m - 180 m.

y(6 \ s) = \ 1000 m \ - 300 m.

y(6 \ s) = \ 700 m.

So this its the height of the ball 6 seconds after being thrown.

6 0
3 years ago
A 60 kilogram student jumps down from a laboratory counter. At the instant he lands on the floor hus speed is 3 meters per secon
erastovalidia [21]

As per Newton's law rate of change in momentum is net force

so we can write it as

F = \frac{dP}{dt}

F = \frac{m(v_f - v_i)}{\Delta t}

now we know that

m = 60 kg

v_f = 3 m/s

v_i = 0

\Delta t= 0.2 s

from above equation

F = \frac{60(3 - 0)}{0.2} = 900 N

so he will experience 900 N force in above case

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2 years ago
Help me please........................
stepan [7]

Answer:

Left

Explanation:

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2 years ago
Explain the origin of the magnitude designation for determining the brightness of stars. Why does it seem to go backward, with s
Mashcka [7]

Answer:

Hipparchus was an ancient Greek who classified stars based on the brightness in 129 B.C. He grouped the brightest stars and ranked them 1 (first magnitude) and dimmest stars as 6 (sixth magnitude). Thus, the smaller numbers indicated brighter stars. Now, the scale extends in negative axis as well. More the negative number, brighter is the star. For example, Sun has magnitude -26.74.

This the apparent magnitude which means the classification is based on the brightness of the star as it appears from the Earth.

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