Select all that apply. Functional groups are a group of molecules attached to a carbon-based core of an organic molecule. Key functional groups are _____. aminos carbonyls proteins phosphates carbohydrates

Answer: A. Federal funding agencies typically rely on an institution(...)research misconduct.

Explanation: In its enforcement capacity, FINRA (Financial Industry Regulatory Authority) has the power to take disciplinary actions against registered individuals or firms that violate the industry´s rules. In 2018, for example, it reported that it initiated 921 disciplinary actions, levied fines totaling 61 million, and ordered restitution of 25.5 million to investors.

Answer:

D is the correct answer

Explanation:

Contractionist policies are a set of measures taken by the government, contractionary policies are measures to reduce disposable income to contain the economic growth of a certain sector, causing damage to the economy, but to know how they affect the economy, we must know this....

What is Contractionist Policy?Contractionist Policy is a set of measures taken by the Government, with the aim of reducing the high growth of the economic market, in addition to controlling inflation and currency circulation.

Contractionist Policy is a subgroup of some other policies that make up Economic Policy as a whole, such as Fiscal, Exchange and Monetary Policies, for example.

With this information, we can say that contractionary policies are measures to reduce disposable income to curb economic growth in a given sector, causing damage to the economy.

Answer:

(a) She should use specific and simple language.

Explanation:

The communication is the two-way process through which the sender and the receiver can communicate/ talk with each other. It is an exchange of ideas and information. It can be done in verbal or non verbally form.  

There are various modes of communication like - telephonic, emails, messaging, face to face, etc.  

In the given situation, the Serena should use the specific and the simple language while communicating this bad news so that everyone can listen properly, and the chances of argument and the aggressiveness of the people would be less occur.  

Answer:

A)  Rt = 1200  Ω B)  I = 0.100 A  the current in the two bulbs is the same  

Explanation:

Part A

As the bulbs are connected in series the equivalent resistance is

    Rt = R1 + R2

    Rt = 400 +800

    Rt = 1200  Ω

With this value we can find the current

    V = I Rt

     I = V / Rt

     I = 120/1200

     I = 0.100 A

In a circuit, the current is constant, so the current in the two bulbs is the same

Part B

The equation for power is

     P = I V = I² R

In the bulb of R = 400  Ω

    P₄₀₀ = 0.1² 400

    P₄₀₀ = 4.00 W

In the bulb of R = 800  Ω

    P₈₀₀ = 0.1²  800

    P₈₀₀ = 8.00 W

Part C

     Pt = I² Rt

     Pt = 0.1² 12

     Pt = 12.0 W

Part D.

The bulbs are in parallel .  In this case the voltage in each bulb is constant,

      V = I R

      i = V/R

     I₄₀₀ = V/R₄₀₀

     I₄₀₀= 120 /400

     I₄₀₀ = 0.300 A

     I₈₀₀ = V/R800

     I₈₀₀ = 120/800

     I₈₀₀= 0.150 A

The current in the bulb with the resistance of 400 Ω is 0.3 A, while the current in the bulb with the resistance of 800 Ω is 0.15 A.

What is power?

Power is the rate at which the energy is been transformed or transferred per unit time.

A.) As we know that the bulbs are connected in series, therefore, the equivalent resistance of both the bulbs can be written as,

R_e = R_1 + R_2\\\\R_e = 400 +800\\\\R_e = 1200\rm\ \Omega

As the voltage is known to be 120 V, while the equivalent resistance is 1200 Ω. therefore, the current can be written as,

V = I \times R_e\\\\I = \dfrac{V}{R_e}\\\\I = \dfrac{120}{1200}\\\\I = 0.100\rm\ A

Since the current in the circuit is constant, the current in the two bulbs is the same.

B.) We know that the equation of power for the circuit can be written as

\rm P = I \times V = I^2 \times R

Since the resistance in the first bulb is of R = 400 Ω, therefore, the power in the first bulb can be written as,

P_{400} = 0.1^2 \times 400\\\\P_{400}= 4.00\rm\ W

The power in the second bulb with resistance, R = 800Ω can be written as,

P_{800} = 0.1^2 \times 800\\\\P_{800}= 8.00\rm\ W

C.) The total power dissipated in both bulbs can be written as,

P_t = I^2 \times R_e\\\\P_t = 0.1^2 \times 12\\\\P_t = 12.0\rm\ W

D.) We know when the bulbs are connected in parallel. therefore, the voltage in each bulb is constant,

I = \dfrac{V}{R}

The current in the bulb with the resistance of 400 Ω,

I_{400} = \dfrac{V}{R_{400}}\\\\I_{400} = \dfrac{120}{400}\\\\I_{400} = 0.300\rm\ A

The current in the bulb with the resistance of 800 Ω,

I_{800} = \dfrac{V}{R_{800}}\\\\I_{400} = \dfrac{120}{800}\\\\I_{400} = 0.15\rm\ A

Hence, the current in the bulb with the resistance of 400 Ω is 0.3 A, while the current in the bulb with the resistance of 800 Ω is 0.15 A.