What are two conditions of equilibrium? write them in word

Physics

1 answer:

valentina_108 [34]3 years ago

6 0

In order for a system to be in equilibrium , two conditions must be met. Net force must be 0.

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"Giant Swing", the seat is connected to two cables as shown in the figure (Figure 1) , one of which is horizontal. The seat swin

Bingel [31]

The horizontal force is m*v²/Lh, where m is the total mass. The vertical force is the total weight (233 + 840)N.

Fx = [(233 + 840)/g]*v²/7.5 

v = 32.3*2*π*7.5/60 m/s = 25.37 m/s 

The horizontal component of force from the cables is Th + Ti*sin40º and the vertical component of force from the cable is Ta*cos40º 

Thh horizontal and vertical forces must balance each other. First the vertical components: 

233 + 840 = Ti*cos40º 

solve for Ti. (This is the answer to the part b) 

Horizontally 

[(233 + 840)/g]*v²/7.5 = Th + Ti*sin40º 

Solve for Th 

Th = [(233 + 840)/g]*v²/7.5 - Ti*sin40º 

using v and Ti computed above.

3 0

3 years ago

What is the equivalent resistance of the circuit?

raketka [301]

Answer:

I feel it would be D. 120.00 I'm really not sure

Explanation:

8 0

3 years ago

Suppose that Hubble's constant were H0 = 51 km/s/Mly (which is not its actual value). What would the approximate age of the univ

bija089 [108]

Given the Hubble's constant, the approximate age of the universe is 5.88 × 10⁹ Years.

Given the data in the question;

Hubble's constant; $H_0 = 51km/s/Mly$

Age of the universe; $t = \ ?$

We know that, the reciprocal of the Hubble's constant ( $H_0$ ) gives an estimate of the age of the universe ( $t$ ). It is expressed as:

$Age\ of\ Universe; t = \frac{1}{H_0}$

Now,

Hubble's constant; $H_0 = 51km/s/Mly$

We know that;

$1\ light\ years = 9.46*10^{15}m$

$1\ Million\ light\ years = [9.46 * 10^{15}m] * 10^6 = 9.46 * 10^{21}m$

Therefore;

$H_0 = 51\frac{km}{\frac{s}{Mly} } = 51000\frac{m}{s\ *\ Mly} \\\\H_0 = 51000\frac{m}{s\ *\ (9.46*10^{21}m)} \\\\H_0 = 5.39 *10^{-18}s^{-1}\\$

Now, we input this Hubble's constant value into our equation;

$Age\ of\ Universe; t = \frac{1}{H_0}\\\\t = \frac{1}{ 5.39 *10^{-18}s^{-1}} \\\\t = 1.855 * 10^{17}s\\\\We\ convert\ to\ years\\\\t = \frac{ 1.855 * 10^{17}}{60*60*24*365}yrs \\\\t = \frac{ 1.855 * 10^{17}}{31536000}yrs\\\\t = 5.88 *10^9 years$