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3 years ago

How does a branching tree diagram divide organism into groups

Physics

2 answers:

Sliva [168]3 years ago

8 0

Answer:

kingdom

phylum

class

order

family

genus

species

its kind of like how a library is organized

Explanation:

Galina-37 [17]3 years ago

7 0

Answer:

Because it shows how closely related two organisms are and when they had a common ancestor or evolved into a different species

Explanation:

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ROY G BIV is an acronym for what? What part of the EMR spectrum does it refer to?

mel-nik [20]

Answer:

ROYGBIV or Roy G. Biv is an acronym for the sequence of hues commonly described as making up a rainbow: red, orange, yellow, green, blue, indigo and violet. The initialism is sometimes referred to in reverse order, as VIBGYOR.

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3 years ago

2.10-kg box is moving to the right with speed 8.50 m>s on a horizontal, frictionless surface. At t = 0 a horizontal force is

larisa86 [58]

Explanation:

Below is an attachment containing the solution.

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3 years ago

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What is the potential difference across one wire of a 70 m extension cord made of 16 gauge copper wire carrying a current of 4 A

Butoxors [25]

Answer:

V = 3.6 volts

Explanation:

From Ohm's Law, we know that:

V = IR

but,

R = ρL/A

Therefore,

V = IρL/A

where,

V = Potential Difference = ?

I = Current = 4 A

ρ = resistivity of copper = 1.68 x 10⁻⁸ Ω.m

L = Length = 70 m

A = Cross-sectional Area = πd²/4 = π(1.29 x 10⁻³ m)²/4 [16 gauge wire has a diameter of 1.29 mm]

A = 1.31 x 10⁻⁶ m²

V = (4 A)(1.68 x 10⁻⁸ Ω.m)(70 m)/(1.31 x 10⁻⁶ m²)

<u>V = 3.6 volts</u>

3 0

3 years ago

Freeeeeeee points!!!

Andrej [43]

Answer:

thanks so much lol can I get brainliest

Explanation:

7 0

3 years ago

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Find the useful power output (in W) of an elevator motor that lifts a 2600 kg load a height of 30.0 m in 12.0 s, if it also incr

Annette [7]

Answer:

P = 251,916.667 W

Cost = 2,267.25 cents

Explanation:

To solve this question we will use the Work Energy Theorem, which is

$W = dP + dK\\$

Where

$dP$ = Change in Potential Energy

$dK$ = Change in Kinetic Energy

Change in Potential Energy

$P_{i} = mgh_{i}\\ P_{f} = mgh_{f}$

Where

$P_{i}$ = Initial Potential Energy

$P_{f}$ = Final Potential Energy

$m$ = Mass of System = 10,000 kg

$g$ = Acceleration due to gravity = 9.81 m/s

$h_{i}$ = Initial Height = 0

$h_{f}$ = Final Height = 30 m

Inputting the values we get the answer for $dP$

$dP = P_{f} - P_{i}\\dP= mgh_{f} - mgh_{i}\\ dP= 10000(9.81)(30) - 0\\ dP= 2943000$

Change in Kinetic Energy

$K_{i} = \frac{1}{2} mv_{i} ^2\\ K_{f} = \frac{1}{2} mv_{f} ^2$

Where

$K_{i}$ = Initial Kinetic Energy

$K_{f}$ = Final Kinetic Energy

$m$ = Mass of System = 10,000 kg

$g$ = Acceleration due to gravity = 9.81 m/s

$v_{i}$ = Initial Velocity = 0 m/2

$v_{f}$ = Final Velocity = 4 m/s

Inputting the values we get the answer for $dK$

$dK = K_{f} - K_{i}\\ dK = \frac{1}{2} mv_{f} ^2 - \frac{1}{2} mv_{i} ^2\\ dK = \frac{1}{2} (10000)(4)^2 - 0 \\ dK = 80000$

Total Work

$W = dP + dK\\$

Inputting the values

$W = 2943000 + 80000$

$W = 3,023,000$

a) Finding the useful Power Output

$P = \frac{W}{t}$

Where

$P$ = Power Output

$W$ = Work Done = 3,023,000J

$t$ = Time = 12s

Inputting the values

$P = \frac{3,023,000}{12}\\ P = 251,916.667$

P = 251,916.667 W

b) Finding the Total Cost

Cost = $0.0900 x P/1000

Cost = $0.0900 x (251,916.667/1000)

Cost = $22.67 or 2,267.25 cents

4 0

3 years ago