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Natali5045456 [20]
3 years ago
13

According to the dichotomous key, stentor would be a member of the kingdom

Physics
1 answer:
dalvyx [7]3 years ago
5 0

D) Protista

<u>Explanation:</u>

  • The Dichotomous key is a organized set of couplets of mutual characteristics of biological organisms. This key begins with general characteristics and leads couplets indicating specific characteristics.
  • Dichotomous key is a key that allow the user to identify type of items in natural world. Stentor is one of the largest protozoa found in water.
  • Stentor is a single cell. An organism can be of 2 mm in length making them visible to human eye and it is even larger than many multi celled organisms. Stentor is eukaryote means it has nuclear membrane. It is commonly found in fresh water ponds.  
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What is a distraction that a pedestrian may engage in while crossing the street?
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Answer:

a dog walking or their phone rings or heard a neighbor talking to them

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Frequency<br> In excercising
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Frequency. This refers to how often you exercise. The point is to meet your goals without overtraining the body. When it comes to cardio: As a general rule of thumb, aim for a minimum of three cardio sessions per week.

3 0
3 years ago
You position two plane mirriors at right angles to each other a light ray strikes one mirrior at an angle of 60 to the normal an
Dafna1 [17]

Answer:

30 degrees

Explanation:

Reflects off of mirror 1  at   60 degrees....this makes it incident to second mirror at 30 degrees ....then angle of reflection equals this angle of incidence = 30 degrees

See atached diagram

4 0
2 years ago
The volume electric charge density of a solid sphere is given by the following equation: The variable r denotes the distance fro
qwelly [4]

Answer:

62.8 μC

Explanation:

Here is the complete question

The volume electric charge density of a solid sphere is given by the following equation: ρ = (0.2 mC/m⁵)r²The variable r denotes the distance from the center of the sphere, in spherical coordinates. What is the net electric charge (in μC) of the sphere if the radius of the sphere is 0.5 m?

Solution

The total charge on the sphere Q = ∫∫∫ρdV where ρ = volume charge density = 0.2r² and dV = volume element in spherical coordinates = r²sinθdθdrdΦ

So,  Q =  ∫∫∫ρdV

Q =  ∫∫∫ρr²sinθdθdrdΦ

Q =  ∫∫∫(0.2r²)r²sinθdθdrdΦ

Q =  ∫∫∫0.2r⁴sinθdθdrdΦ

We integrate from r = 0 to r = 0.5 m, θ = 0 to π and Φ = 0 to 2π

So, Q =  ∫∫∫0.2r⁴sinθdθdrdΦ

Q =  ∫∫∫0.2r⁴[∫sinθdθ]drdΦ

Q =  ∫∫0.2r⁴[-cosθ]drdΦ

Q =  ∫∫0.2r⁴-[cosπ - cos0]drdΦ

Q =  ∫∫∫0.2r⁴-[-1 - 1]drdΦ

Q =  ∫∫0.2r⁴-[- 2]drdΦ

Q =  ∫∫0.2r⁴(2)drdΦ

Q =  ∫∫0.4r⁴drdΦ

Q =  ∫0.4r⁴dr∫dΦ

Q =  ∫0.4r⁴dr[Φ]

Q =  ∫0.4r⁴dr[2π - 0]

Q =  ∫0.4r⁴dr[2π]

Q =  ∫0.8πr⁴dr

Q =  0.8π∫r⁴dr

Q =  0.8π[r⁵/5]

Q = 0.8π[(0.5 m)⁵/5 - (0 m)⁵/5]

Q = 0.8π[0.125 m⁵/5 - 0 m⁵/5]

Q = 0.8π[0.025 m⁵ - 0 m⁵]

Q = 0.8π[0.025 m⁵]

Q = (0.02π mC/m⁵) m⁵

Q = 0.0628 mC

Q = 0.0628 × 10⁻³ C

Q = 62.8 × 10⁻³ × 10⁻³ C

Q = 62.8 × 10⁻⁶ C

Q = 62.8 μC

3 0
3 years ago
A NASA explorer spacecraft with a mass of 1,000 kg takes off in a positive direction from a stationary asteroid. If the velocity
Jet001 [13]

Before the launch, the momentum of the (spacecraft + asteroid) was zero.  So after the launch, the momentum of the (spacecraft + asteroid) has to be zero.

Momentum = (mass) x (velocity)

Momentum after the launch:

Spacecraft:  (1,000 kg) x (250 m/s) = 250,000 kg-m/s

Asteroid: (mass) x (-25 m/s)

Their sum:  250,000 - 25(mass) .

Their sum must be zero, so  250,000 kg-m/s = (25 m/s) x (mass)

Divide each side by  25 :  10,000 kg-m/s = (1 m/s) x (mass)

Divide each side by (1 m/s) :  10,000 kg = mass


3 0
3 years ago
Read 2 more answers
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