It is possible for different covalent compounds to have the same empirical formula because empirical formulas represent

Yanka [14]

TLDR: R=30 Ohms, I=2 Amps, 12V, 28V, 20V, respectively.

Alright, let’s break this down. There are three resistors in this circuit, meaning that we have to find the equivalent resistance. Luckily, they are all in parallel with one another; this means we can add the resistances together without having to do inverses like in a series problem. This means that the equivalent resistance, Req, would equal:

Req=R1 + R2 + R3

Req=6 + 14+ 10

Req=30 Ohms

This means that we could theoretically replace all three resistors with a 30 Ohm resistor and accomplish the same goal. Now, the entire voltage of the system would normally be reduced to zero after passing through the resistors - in this case, the 60 Vs would be lost after passing through 30 Ohms. This means we’re losing 2V/Ohm; now we can figure out how much we’re losing at each resistor.

By losing 2V per Ohm, we’re losing 12 V at the first resistor, 28 V at the second resistor, and 20 V at the third resistor.

Finally, we can calculate the current through the circuit; for a series circuit, the current remains the same. Using V=IR, we can find that:

V=IR

60 V = I(30 Ohms)

I = 2 Amps

The current passing through the circuit is 2 Amps.

Hope this helps!

8 0

3 years ago

1. While driving to Palm Desert you notice that some of the windmills near the freeway are spinning 20 times per minute. The len

igomit [66]

Answer:

a) $\alpha = -0.233 rad/s^{2}$

b) the rotational acceleration will remain the same, $\alpha = -0.233 rad/s^{2}$

c) When r = 36 m, $a_{c} = 157.25 m/s^{2}$

When r = 18 m, $a_{c} = 78.63 m/s^{2}$

d) When r = 36 m, $a_{c} = 39.31 m/s^{2}$

When r = 36 m, $a_{c} = 19.66 m/s^{2}$

Explanation:

The windmills are spinning 20 times per minute, the number of spins in 1 second = 20/60 = 1/3 spins/sec

frequency, f = 1/3 spins/sec

there are 1/3 spins in 1 second,

there will be 50 spins in 50/ (1/3) seconds = 150 seconds

i.e time taken to make 50 spins = 150 seconds

Time interval between 20 and 50 spins, Δt = 150 - 60 = 90

Δt = 90

Angular frequency at 20 spins:

w₁ = 2 π f

w₁ = 2 π * 1/3 = 2π/3 = 2.09 rad/s

Since the blade stops at 50 spins, angular frequency at 50 spins, w₂ = 0

The rotational acceleration,

$\alpha = \frac{\triangle w}{\triangle t} \\\alpha = \frac{w_{2} - w_{1} }{\triangle t} \\\alpha = \frac{0 - 2.09 }{90} \\\alpha = -0.233 rad/s^{2}$

b) The magnitude of the rotational acceleration does not depend on the radius, r. It depends on the angular frequency, therefore the magnitude of the rotational acceleration doe not change.

c) The centripetal acceleration is given by the formula, $a_{c} = w^{2} r$

While the windmill spins at 20 times per minute, the angular speed gotten, w = 2.09 rad/s

When r = 36 m

$a_{c} = 2.09^{2} * 36\\a_{c} = 157.25 m/s^{2}$

At the halfway point, r = 18 m

$a_{c} = 2.09^{2} * 18\\a_{c} = 78.63 m/s^{2}$

d) If the angular velocity of the blades were cut in halves:

w = 2.09/2

w = 1.045 rad/s

When r = 36 m

$a_{c} = 1.045^{2} * 36\\a_{c} = 39.31 m/s^{2}$

When r = 18 m

$a_{c} = 1.045^{2} * 18\\a_{c} = 19.66 m/s^{2}$

8 0

3 years ago

The volume of a gas is 200.0 mL at 275 K and 92.1 kPa. Find its volume at STP.

ycow [4]

To solve this question we will use ideal gas equation:
$p*V=n*R*T$
Where:
p = pressure
V = volume
n = number of moles
R = gas constant
T = temperature

We can rearrange formula to get:
$\frac{p*V}{T} =n*R$
We are working woth same gas so we can write following formula. Index 1 stands for conditions before change and index 2 stands for conditions after change.
$\frac{ p_{1}*V_{1} }{T_{1}} = \frac{ p_{2}*V_{2} }{T_{2}}$

We are given:
p1=92.1kPa = 92100Pa
V1=200mL = 0.2L
T1=275K
p2= 101325Pa
T2=273K
V2=?

We start by rearranging formula for V2. After that we can insert numbers:
${ p_{1}*V_{1} *T_{2}} = { p_{2}*V_{2}*T_{1} } \\ \\ V_{2}= \frac{p_{1}*V_{1} *T_{2}}{ p_{2}*T_{1}} \\ \\ V_{2}= \frac{92100*0.2*273}{101325*275} \\ \\ V_{2}= 0.18L=180mL$

7 0

4 years ago

Which of the following statements about lunar phases is true? (A) Only one quarter of the first-quarter moon is illuminated by t

saul85 [17]

Answer:

A) False B) True C) False D) False E) False

Explanation:

A) At any point of time half of the moon is always illuminated by the sun. We are able to see only the part of the moon which is facing us.

B) Typically a Lunar cycle phases take 29.5 days to complete one cycle i.e. it will take 29.5 days for one event say full moon to reoccur. So except for February all the other month can have two full moons.

C) As explained in the above answer any month except February can have two full moons.

D) The tame taken between new moon is 29.5 days.

E) Full moon always rises at the sunset and sets at the sunrise

7 0

3 years ago

Describe how Bohr’s model used the work of Maxwell.

pochemuha

Answer:

The bohr's model is the primitive model for the hydrogen atom, comparatively to the atom of valence shell. And it is derived from the hydrogen atom of the first approximation by using the quantum mechanics.

Basically, the model state that the electron revolved around in circular orbit in atom around the central nucleus. And it can be fixed in the circular orbit at the set of discrete distance at the nucleus.

3 0

3 years ago