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4 years ago

What do the subscript in the formula for ethane repersent

1 answer:

3 0

The formula for ethane is C2H6. (the 2 and 6 are supposed to be subscripts) they show the number of atoms per element. So there are two carbon atoms and six hydrogen atoms in ethane. The subscript in a chemical formula represents the amount of that atom in that compound's formula. In chemistry, no subscript indicates 1 of that particular atom. HCl contains 1 atom of hydrogen and 1 atom of chlorine, whereas H20 contains 2 atoms of hydrogen and 1 atom of oxygen.It should be noted, however that this does not necessarily hold true for empiracle formulas, since these have been reduced to the simplest whole number ratio and do not necessarily reflect the number of each element present in an actual molecule.

I hope this helps you! Good luck :)

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he deflection plates in an oscilloscope are 10 cm by 2 cm with a gap distance of 1 mm. A 100 volt potential difference is sudden

Lubov Fominskaja [6]

Answer:

Incomplete question

This is the complete question

The deflection plates in an oscilloscope are 10 cm by 2 cm with a gap distance of 1 mm. A 100 V potential difference is suddenly applied to the initially uncharged plates through a 1000 Ω resistor in series with the deflection plates. How long does it take for the potential difference between the deflection plates to reach 95 V?

Explanation:

Given that,

The dimension of 10cm by 2cm

0.1m by 0.02m

Then, the area is Lenght × breadth

Area=0.1×0.02=0.002m²

The distance between the plate is d=1mm=0.001m

Then,

The capacitance of a capacitor is given as

C=εoA/d

Where

εo is constant and has a value of

εo= 8.854 × 10−12 C²/Nm²

C= 8.854E-12×0.002/0.001

C=17.7×10^-12

C=17.7 pF

Value of resistor resistance is 1000ohms

Voltage applied is V = 100V

This Is a series resistor and capacitor (RC ) circuit

In an RC circuit, voltage is given as

Charging system

V=Vo[1 - exp(-t/RC)]

At, t=0, V=100V

Therefore, Vo=100V

We want to know the time, the voltage will deflect 95V.

Then applying our parameters

V=Vo[1 - exp(-t/RC)]

95=100[1-exp(-t/1000×17.7×10^-12)]

95/100=1-exp(-t/17.7×10^-9)

0.95=1-exp(-t/17.7×10^-9)

0.95 - 1 = -exp(-t/17.7×10^-9)

-0.05=-exp(-t/17.7×10^-9)

Divide both side by -1

0.05=exp(-t/17.7×10^-9)

Take In of both sides

In(0.05)=-t/17.7×10^-9

-2.996=-t/17.7×10^-9

-2.996×17.7×10^-9=-t

-t=-53.02×10^-9

Divide both side by -1

t= 53.02×10^-9s

t=53.02 ns

The time to deflect 95V is 53.02nanoseconds

5 0

3 years ago

Consult Multiple-Concept Example 11 for background material relating to this problem. A small rubber wheel on the shaft of a bic

Papessa [141]

Answer:

$Emf=24.4V$

Explanation:

From the question we are told that:

Angular speed $\omega=40*\omega'$

Radius $r=0.33m$

No. Turns $N=163$

Area $A= 3.23 * 10^{-3} m^2$

Magnetic field. $B=0.0948T$

Acceleration $a=0.60m/s^2$

Time $t=6.72s$

Generally the equation for momentum is mathematically given by

$\omega'=\omega_o+\frac[a}{r}t$

$\omega'=0+\frac{0.60}{0.33}*6.72$

$W=12.22rad/s^2$

Therefore

$\omega=Aw$

$\omega=12.22*40$

$\omega=488.7rads/s$

Generally the equation for Peak emf is mathematically given by

$Emf=NBA \omega$

$Emf=163*0.0948* 3.23 * 10^{-3} m^2*488.7rads/s$

$Emf=24.4V$

3 0

3 years ago

The position of a particle moving along the x axis depends on the time according to the equation x = ct² - bt³, where x is in me

Kamila [148]

Answer:

(a) $\frac{m}{s^2}$

(b) $\frac{m}{s^3}$

(d) 20 m

(e) 1 m

(f) $0\frac{m}{s}$

(g) $-12\frac{m}{s}$

(h) $-36\frac{m}{s}$

(i) $-72\frac{m}{s}$

(j) $-6\frac{m}{s^2}$

(k) $-18\frac{m}{s^2}$

(l) $-30\frac{m}{s^2}$

(m) $-42\frac{m}{s^2}$

Explanation:

Since x is measured in meters and t in seconds, constants a and b must have units that gives meters when multiplied by square and cubic seconds respectivly, so that would mean $\frac{m}{s^2}$ for a and $\frac{m}{s^3}$ for b.

We can get the velocity v equation by deriving the position with respect to t, which gives:

$v=6*t-6*t^2$

And the acceleration a equation by deriving again:

$a=6-12*t$

Now for getting the maximun position between 0 and 4, we must find to points where the positions first derivate is equal to cero and evaluate those points. That is v=0, which gives

$6*t-6*t^2=0\\6t*(1-t)=0\\t=0 or t=1$

For t = 0, x = 0 so the maximun position is archieved at 1 second, which gives x = 1 meter.

For obtaining it's displacement r, we can integrate the velocity from 0 seconds to 4 seconds, which gives the mean value of the position in that interval:

$r=\frac{1}{4}*(2*4^3-3*4^2)m\\r= 20m$

For the remaining questions, we just replace the values of t on the respective equations.

8 0

4 years ago

A 18-kg sled is being pulled along the horizontal snow-covered ground by a horizontal force of 30 N. Starting from rest, the sle

k0ka [10]

Answer:

Coefficient of kinetic friction = 0.146

Explanation:

Given:

Mass of sled (m) = 18 kg

Horizontal force (F) = 30 N

FInal speed (v) = 2 m/s

Distance (s) = 8.5 m

Find:

Coefficient of kinetic friction.

Computation:

Initial speed (u) = 0 m/s

v² - u² = 2as

2(8.5)a = 2² - 0²

a = 0.2352 m/s²

Nweton's law of :

F (net) = ma

30N - μf = 18 (0.2352)

30 - 4.2336 = μ(mg)

25.7664 = μ(18)(9.8)

μ = 0.146

Coefficient of kinetic friction = 0.146

6 0

4 years ago

Despite a very strong wind, a tennis player

Gnoma [55]

Answer:

Option 5. 1 and 3

Solution:

The only forces acting on the tennis ball after it has left contact with the racquet and the instant before it touches the ground are the force of gravity in the downward direction and the force by the air exerted on the ball.

The ball after it left follows the path of trajectory and as it moves forward in the horizontal direction the force of the air acts on it.

In the whole projectile motion of the ball, the acceleration due to gravity acts on the ball thus the force of gravity acts on the ball in the downward direction before it hits the ground.

6 0

4 years ago