1 question, brainliest for whoever gets it right.

Input Voltage = 220V

muminat

Answer:

Correct option is

A

450 V and 15 A

P

in

=I

p

V

p

⇒T

p

=

V

p

P

in

=

200N

3000W

=15A

Efficciencyofthetransformer

η=

P

in

P

out

=

V

p

I

p

V

s

I

s

⇒

100

90

=

3000

6V

s

V

s

=

100×6

90×5000

=450V

Hence,

option (A) is correct answer.

4 0

3 years ago

Why do we add alcohol/ethanol to the leaf once it is boiled?

hichkok12 [17]

Answer:

Complete answer: We boil the leaf in alcohol when we are checking it for starch to eradicate chlorophyll, which is the green pigment present in leaves. ... Hence to dissolve the chlorophyll or the green pigment present in the leaf we boil the leaf in alcohol when we are testing it for starch.

Hope this helps! Have a great day!

4 0

3 years ago

25 POINTS FOR CORRECT ANSWER

castortr0y [4]

No. I do not agree with Stefan. Quite the contrary. I disagree
with his description of "<span>angle of incidence" as the angle between
the surface of the mirror and the incoming ray.

The correct description of "angle of incidence" is </span><span>the angle between
the NORMAL TO the surface of the mirror and the incoming ray.

Thus, the true angle of incidence is the complement of the angle that
Stefan calculates or measures.</span>

5 0

3 years ago

Read 2 more answers

An uncharged series RC circuit is to be connected across a battery. For each of the following changes, determine whether the tim

slavikrds [6]

a) Increase

b) Unchanged

c) Increase

Explanation:

a)

The charge on a capacitor charging in a RC circuit connected to a battery follows the exponential equation:

$Q(t)=Q_0 (1-e^{-\frac{t}{RC}})$

where

$Q_0 = CV$ is the final charge stored in the capacitor, where C is the capacitance and V is the voltage of the battery

t is the time

R is the resistance of the circuit

The capacitor reaches 90% of its final charge when

$Q(t)=0.90Q_0$

Substituting and re-arranging the equation, we find:

$0.90Q_0 = Q_0(1-e^{-\frac{t}{RC}})\\0.90=1-e^{-\frac{t}{RC}}\\e^{-\frac{t}{RC}}=0.10\\-\frac{t}{RC}=ln(0.10)\\t=-RCln(0.10)=2.30RC$

We see that if we double the RC constant, then $(RC)'=2(RC)$

So the time taken will double as well:

$t'=2.30(RC)'=2.30(2RC)=2(2.30RC)=2t$

So, the answer is "increase"

b)

In this second part, the battery voltage is doubled.

According to the equation written in part a),

$Q_0 =CV$

this means also that the final charge stored on the capacitor will also double.

However, the equation that gives us the time needed for the capacitor to reach 90% of its full charge is

$t=2.30 RC$

We see that this equation does not depend at all on the voltage of the battery.

Therefore, if the battery voltage is doubled, the final charge on the capacitor will double as well, but the time needed for the capacitor to reach 90% of its charge will not change.

So the correct answer is

"unchanged"

c)

In this case, a second resistor is added in series with the original resistor of the circuit.

We know that for two resistors in series, the total resistance of the circuit is given by the sum of the individual resistances:

$R=R_1+R_2$

Since each resistance is a positive value, this means that as we add new resistors, the total resistance of the circuit increases.

Therefore in this problem, if we add a resistor in series to the original circuit, this means that the total resistance of the circuit will increase.

The time taken for the capacitor to reach 90% of its final charge is still

$t=2.30 RC$

As we can see, this time is directly proportional to the resistance of the circuit, R: therefore, if we add a resistor in series, the resistance of the circuit will increase, and therefore this time will increase as well.

So the correct answer is

"increase"

8 0

3 years ago

PLS HELP ME!!!

natka813 [3]

Hmmmmmmmmm i don’t know

3 0

3 years ago