Which of these choices is the smallest matter

Determine weather or not the equation below is balanced.

Liula [17]

Answer:

Yep.. It's balanced and its a combination reaction

Explanation:

Reactants : S₈ + 24F₂

Product 8SF₆

8 0

3 years ago

A "seconds pendulum" is one that moves through its equilibrium position once each second. (The period of the pendulum is precise

Alex73 [517]

<h2>Ratio of free fall acceleration of Tokyo to Cambridge = 0.998</h2>

Explanation:

We know the equation

$T=2\pi \sqrt{\frac{l}{g}}$

where l is length of pendulum, g is acceleration due to gravity and T is period.

Rearranging

$g= \frac{4\pi^2l}{T^2}$

Length of pendulum in Tokyo = 0.9923 m

Length of pendulum in Cambridge = 0.9941 m

Period of pendulum in Tokyo = Period of pendulum in Cambridge = 2s

We have

$\frac{ g_{\texttt{Tokyo}}}{ g_{\texttt{Cambridge}}}= \frac{\frac{4\pi^2 l_{\texttt{Tokyo}}}{ T_{\texttt{Tokyo}}^2}}{\frac{4\pi^2 l_{\texttt{Cambridge}}}{ T_{\texttt{Cambridge}}^2}}\\\\\frac{ g_{\texttt{Tokyo}}}{ g_{\texttt{Cambridge}}}=\frac{\frac{0.9923}{2^2}}{\frac{0.9941}{2^2}}=0.998$

Ratio of free fall acceleration of Tokyo to Cambridge = 0.998

6 0

3 years ago

You are conducting an experiment inside a train car that may move along level rail tracks. A load is hung from the ceiling on a

allochka39001 [22]

Answer:

b.

c.

e.

h.

Explanation:

If we consider the ball,the tension is not balance by weight then the force acting will be non zero.It means that train car in not inertial frame of reference.And the same time the train car must accelerated with respected to the earth.It is not moving straight path.When train is in rest position but the string will be at the constant angle.

So the following option are correct:

b.

c.

e.

h.

6 0

3 years ago

which machine changes the direction of force needed to do a task without changing the size of the force needed?

Evgen [1.6K]

<span>A pulley is able to do that.</span>

7 0

4 years ago

A particle of mass 3m is located 1.00 m from a particle of mass m. (a) Where should you put a third mass M so that the net gravi

Airida [17]

Explanation:

It is given that net gravitational force on M is exactly equal to zero. Hence, distance to M from the bigger mass is 3m. Therefore, expression for net force will be as follows.

$F_{net} = F_{1} + F_{2} = 0$

So,

$\frac{-G(3m)(M)}{x^{2}} + \frac{G(m)(M)}{(1 - x)^{2}} = 0$

The first term is negative as the third mass is located between the other two masses. This means that 3 m will be pulling it leftwards (negative x direction) and m will be pulling it rightwards (positive x direction).

$\frac{G(m)(M)}{(1 - x)^{2}} = \frac{G(3m)(M)}{(x)^{2}}$

On dividing both sides of the equation by G.m.M, we get the following.

$\frac{1}{(1 - x)^{2}} = \frac{3}{x^{2}}$

$x^{2} = 3 - 6x + 3x^{2}$

0 = $3 - 6x + 2x^{2}$

Using the formula, $\frac{-b \pm \sqrt{(b)^{2} - 4ac}}{2a}$ the value of x comes out to be equal to +2.37 (not usabale) and -0.634 (usable).

Hence, we can conclude that the third mass will be located 0.634 meters away from the 3 m mass.

7 0

3 years ago