All categories

2 years ago

HELP PLZ ASAP... I'M CONFUSED :(

2 answers:

8 0

B. In forward-biased setup, large numbers of charge carriers will be pulled across the junction and result in a large current.

Snowcat [4.5K]2 years ago

6 0

D. In forward-biased setup, large numbers of charge carriers will be pulled across the junctions and result in a large current

You might be interested in

Why would the federal government and city government find it necessary regulate water standard?

Helga [31]

They do this so that when you drink it or use it the contaminates that used to be in it dont make you sick

4 0

3 years ago

An atom of boron has an atomic number of 5 and an atomic mass of 11. the atom contains

kirill [66]

The atom contain 6 neutrons, if that's the question

8 0

3 years ago

In a Dam, If we doubled the depth of the dam the hydrostatic force will be ?

AnnZ [28]

Answer: Option (A) is the correct answer.

Explanation:

Force acting on a dam is as follows.

F = $\frac{1}{2}\rho g\omega H^{2}$ .......... (1)

Now, when we double the depth then it means H is increasing 2 times and then the above relation will be as follows.

F' = $\frac{1}{2}\rho g\omega (2)^{2}$

F' = $\frac{1}{2}\rho g\omega 4$ ........... (2)

Now, dividing equation (1) by equation (2) as follows.

$\frac{F}{F'}$ = $\frac{\frac{1}{2}\rho g\omega H^{2}}{\frac{1}{2}\rho g\omega 4}$

Cancelling the common terms we get the following.

$\frac{F}{F'}$ = $\frac{1}{4}$

4F = F'

Thus, we can conclude that if doubled the depth of the dam the hydrostatic force will be 4F.

4 0

2 years ago

The reaction for the decomposition of ammonia (NH3) can be written as shown. If a student starts with 21.7 g of ammonia, how man

Eddi Din [679]

3.88 grams will be released I believe

3 0

3 years ago

Read 2 more answers

Given the value of the equilibrium constant (Kc) for the equation (a), calculate the equilibrium constant for equation (b)

matrenka [14]

Answer: The value of equilibrium constant for new reaction is $1.92\times 10^{-25}$

Explanation:

The given chemical equation follows:

$O_2(g)\rightarrow \frac{2}{3}O_3(g)+\frac{1}{2}O_2(g)$

The equilibrium constant for the above equation is $5.77\times 10^{-9}$

We need to calculate the equilibrium constant for the equation of 3 times of the above chemical equation, which is:

$3O_2(g)\rightarrow 2O_3(g)$

The equilibrium constant for this reaction will be the cube of the initial reaction.

If the equation is multiplied by a factor of '3', the equilibrium constant of the new reaction will be the cube of the equilibrium constant of initial reaction.

The value of equilibrium constant for reverse reaction is:

$K_{eq}'=(5.77\times 10^{-9})^3=1.92\times 10^{-25}$

Hence, the value of equilibrium constant for new reaction is $1.92\times 10^{-25}$

5 0

2 years ago