If your bedroom is cold, you might feel warmer with several thin blankets than with one thick one

A ball having mass 2 kg is connected by a string of length 2 m to a pivot point and held in place in a vertical position. A cons

Virty [35]

Complete Question

A ball having mass 2 kg is connected by a string of length 2 m to a pivot point and held in place in a vertical position. A constant wind force of magnitude 13.2 N blows from left to right. Pivot Pivot F F (a) (b) H m m L L If the mass is released from the vertical position, what maximum height above its initial position will it attain? Assume that the string does not break in the process. The acceleration of gravity is 9.8 m/s 2 . Answer in units of m.What will be the equilibrium height of the mass?

Answer:

$H_m=1.65m$

$H_E=1.16307m$

Explanation:

From the question we are told that

Mass of ball $M=2kg$

Length of string $L= 2m$

Wind force $F=13.2N$

Generally the equation for $\angle \theta$ is mathematically given as

$tan\theta=\frac{F}{mg}$

$\theta=tan^-^1\frac{F}{mg}$

$\theta=tan^-^1\frac{13.2}{2*2}$

$\theta=73.14\textdegree$

Max angle = $2*\theta= 2*73.14=>146.28\textdegree$

Generally the equation for max Height $H_m$ is mathematically given as

$H_m=L(1-cos146.28)$

$H_m=0.9(1+0.8318)$

$H_m=1.65m$

Generally the equation for Equilibrium Height $H_E$ is mathematically given as

$H_E=L(1-cos73.14)$

$H_E=0.9(1+0.2923)$

$H_E=1.16307m$

8 0

3 years ago

What particles bond together to

yanalaym [24]

Answer:

atoms bond together and forms molecules

5 0

3 years ago

Read 2 more answers

Please help!!30 points and will give the brainliest answer for a clear explanation.I need it really fast. Thank you!

kompoz [17]

I would rather be hit by the deflated ball because it wouldn't hurt as bad because it wouldn't have a lot of weight to hurt me in anyway

4 0

3 years ago

Michael Jordan, el célebre basquetbolista, ganó el torneo de clavadas de la NBA en 1988. Para lograr la hazaña saltó 1.35 metros

kozerog [31]

(a) 0.40 s

First of all, let's find the initial speed at which Jordan jumps from the ground.

The maximum height is h = 1.35 m. We can use the following equation:

$v^2-u^2=2gh$

where

v = 0 is the velocity at the maximum height

u is the initial velocity

$g=-9.8 m/s^2$ is the acceleration of gravity

Solving for u,

$u=\sqrt{-2gh}=\sqrt{-2(-9.8)(1.35)}=5.14 m/s$

The time needed to reach the maximum height can now be found by using the equation

$v=u+gt$

Solving for t,

$t=\frac{v-u}{g}=\frac{0-5.14}{-9.8}=0.52s$

Now we can find the velocity at which Jordan reaches a point 20 cm below the maximum height, so at a height of

h' = 1.35 - 0.20 = 1.15 m

Using again the equation

$v'^2-u^2=2gh'$

we find

$v'=\sqrt{u^2+2gh}=\sqrt{5.14^2+2(-9.8)(1.15)}=1.97 m/s$

And the corresponding time is

$t'=\frac{v'-u}{g}=\frac{1.97-5.14}{-9.8}=0.32s$

So the time to go from h' to h is

$\Delta t = t-t'=0.52-0.32=0.20 s$

And since we have also to take into account the fall down (after Jordan reached the maximum height), which is symmetrical, we have to multiply this time by 2 to get the total time of permanence in the highest 20 cm of motion:

$\Delta t=2\cdot 0.20 = 0.40 s$