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evablogger [386]
3 years ago
15

According to the principle of segregation, _____.

Physics
1 answer:
Anastaziya [24]3 years ago
3 0

Answer:

The principles that govern heredity were discovered by a monk named Gregor Mendel in the 1860s. One of these principles, now called Mendel's Law of Segregation, states that allele pairs separate or segregate during gamete formation, and randomly unite at fertilization. ... Organisms inherit two alleles for each trait.

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A concrete column has a diameter of 380 mm and a length of 2.6 m . if the density (mass/volume) of concrete is 2.45 mg/m3, deter
grigory [225]

The volume of the column is

(π) · (r²) · (length) =

(π) · (0.19 meter)² · (2.6 meters) =

(π) · (0.036 m²) · (2.6 m) =

0.294 m³ .

The density is 2,450 kg/m³ (VERY very dense, heavy concrete)

so the weight of the column is (mass)·(gravity) or

(density) · (volume) · (gravity) =

(2,450 kg/m³) · (0.294 m³) · (9.81 m/s²) =

(2,450 · 0.294 · 9.81) (kg · m³· m) / (m³ · s²) =

7,066 kg-m/s² = 7,066 Newtons .

But 9.81 Newtons = 2.20462 pounds on Earth (the weight of 1 kilogram of mass), so we have

(7,066 N) · (2.205 pound/9.81 N) =

(7,066 · 2.205 / 9.81) pounds =

1,588 pounds .

5 0
3 years ago
If the change in enthalpy is -5074.1 kj , how much work is done during the combustion?
baherus [9]
Hence, 
<span>∆H = ∆U + W </span>
<span>=> </span>
<span>W = ∆H - ∆U </span>
<span>= -5074 kJ - (-5084.) kJ </span>
<span>= 10 kJ</span>
8 0
3 years ago
A block (6 kg) initially compresses spring #1 (k = 2000 N/m) by 60 cm from its equilibrium point. When the block is released, it
mart [117]

Answer:

block K = 29.39 J and spring #1   Ke = 360 J

Explanation:

In this problem we have that the elastic energy of the spring becomes part kinetic energy and the part in work against the force of friction, so, to use the law of conservation of energy, the decrease in energy is the rubbing force work

        W_{fr}= Ef - E₀

Let's look for the energies

Initial

        E₀ = Ke = ½ k₁  x₁²

Final, this is just before starting to compress the spring

        Ef = Ke = ½ m v²

The work of the rubbing force is

       W_{fr}= -fr x

Let's write Newton's second law the y axis

       N-W = 0

      N = W

      fr = μ N

      fr = μ mg

Let's replace

      -μ mg x = ½ m v² - ½ k₁ x₁²

       v² = 2/m (½ k₁ x1₁² -μ mg x)

       v² = 2/6  (½ 2000 0.6²2 - 0.5  6  9.8 1) = 1/3 (360 - 29.4)

       v = 3.13 m / s

With this value we calculate the energy of the block

       K = ½ m v²

       K = ½  6  3.13²

       K = 29.39 J

Calculate eenrgy of the spring ke 1

      Ke = ½ k₁ x₁²

      Ke = ½ 2000 0.60²

      Ke = 360 J

4 0
3 years ago
WILL GIVE 40 PTS AND BRAINLIEST IF ANSWERED ASAP AND CORRECTLY!!! When a glass falls and breaks on the floor, potential energy i
bekas [8.4K]

Answer:

Explanation:

1.C

2.B

3.D

idk the rest

6 0
3 years ago
Read 2 more answers
What is the free body for the platform
AURORKA [14]
Universal gravitional field free body for the platform
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