According to the cladogram, which organisms have roundworms as a common ancestor? mollusks and flatworms sponges and chordates flatworms, sponges, and cnidarians mollusks, echinoderms, chordates, and annelids

Answer:

True

Explanation:

Recreational time is the time in which you do your recreation. Recreational activities are often done for enjoyment, amusement, or pleasure and are considered to be "fun". Which means it can come before or after school.

Hope I helped answer the question:)

Recreation encompasses all of the activities that people engage in to refresh their bodies and minds and make their leisure time more interesting and enjoyable. Walking, swimming, meditation, reading, playing games, and dancing are all examples of recreational activities.

True, recreational time can come before or after school.

What are the types of recreational activity?

•  Indoor activities are also included in the recreation program. Indoor recreation activities are carried out in the comfort of one's own home or, more specifically, indoors, and are intended to rejuvenate the mind and soul.
• Outdoor sports, such as the ones mentioned, also provide us with a way to exercise and are thus among the best forms of recreation. Nature walks, river rafting, cycling, camping, fishing, hiking, adventure parks, surfing, and sports are all examples of outdoor recreation sports.

Thus, we can conclude that the given statement is true.

Answer:

 λ = -47 nC / m  

Explanation:

The missing question is as follows:

" The potential difference between the surface of a 2.2 cm -diameter power line and a point 1.9 m distant is 3.8 kV. What is the magnitude of the line charge density on the power line? Express your answer using two significant figures.  "

Given:

- The Diameter of the power line D = 2.2 cm

- The distance between two ends of power line L = 1.9m

- The potential difference across two ends V = 3.8 KV

Find:

What is the magnitude of the line charge density on the power line?

Solution:

- The derivation of the line of charges for a length L oriented along any axis centered at origin and the potential difference between two ends is as follows:

                                      V = 2*k*λ*Ln( D / L )

Where,

k : Coulomb's Constant = 8.99*10^9

 λ : The line charge density

- Re-arrange and solve for λ:

                                     λ = V / 2*k*Ln( D / L )

Plug in the values:

                                    λ = 3800 / 2*8.99*10^9*Ln( 2.2 / 190 )

                                    λ = -4.74022*10^-8 C / m

                                   λ = -47 nC / m  

Answer: options 1 and 3

Explanation:

The function of a rheostat relates mainly:

- Limits the current in the circuit.

- it's used as a aid to adjusting the current in the circuit.

The purpose of the rheostat in this experiment is option 1 and option 3.

Information regarding the rheostat:

• It limits the current in the circuit.
• Also, it aids for adjusting the current in the circuit.
• So, the option 1 and option 3 should represent the purpose.
• Therefore, the option 2 and option 4 is wrong.

To develop this problem we will apply the concepts related to the conservation of momentum. For this purpose we will define that the initial moment must be preserved and be equal to the final moment. Since the system was at a stationary moment, the moment before the bullet is fired will be 0, while the moment after the bullet is fired will be equivalent to the sum of the products of the mass and the velocity of each object. , this is,

m_1u_1+m_2u_2 = m_1v_1+m_2v_2

Here,

m = mass of each object

u = Initial velocity of each object

v = Final velocity of each object

Our values are,

\text{Mass of hunter} ) m_1 = 75 kg = 75000 g

\text{Mass of bullet} = m_2 = 42 g

\text{Speed of bullet} = v_2 = 6200 m/s

\text{Recoil speed of hunter} =v_1 = ?

Applying our considerations we have that the above formula would become,

m_1v_1 + m_2v_2 = 0

Solving for speed 1,

v_1 = - \frac{m_2v_2}{m_1 }

v_1 = - \frac{42 * 620}{75000 }

v_1 = - 0.3472 m/s

Therefore  the recoil speed of the hunter is -0.3472m/s, said speed being opposite to the movement of the bullet.

The recoil speed of the hunter standing on the frictionless ice is 0.35m/s².

Given the data in the question;

• Mass of hunter; m_1 = 75kg
• Mass of bullet; m_2 = 42g = 0.042kg
• Velocity of bullet; v_2 = 620m/s

Recoil speed of the hunter; v_1 =\ ?

From the law of conservation of momentum:

MV = mv

Where m is mass and v is velocity

So

m_1v_1 = m_2v_2\\\\v_1 = \frac{m_2v_2}{m_1}

We substitute our given values into the equation

v_1 = \frac{ 0.042kg \ *\ 620m/s^2}{75kg} \\\\v_1 = \frac{26.04kg.m/s^2}{75kg} \\\\v_1 = 0.35m/s^2

Therefore, the recoil speed of the hunter standing on the frictionless ice is 0.35m/s².