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3 years ago

How do the offspring produced by asexual reproduction differ from those produced by sexual reproduction?

Physics

1 answer:

il63 [147K]3 years ago

7 0

Answer:

Asexual reproduction produces offspring that are genetically identical to the parent because the offspring are all clones of the original parent. During sexual reproduction the genetic material of two individuals is combined to produce genetically diverse offspring that differ from their parents.

Explanation:

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~I WILL MARK BRAINLYEST~ When using the metric system, what unit are the following measurements taken in? What tool is most comm

torisob [31]

Temperature: Kelvin or degree Celsius; thermometer

Length: meter and its fractions and multiples; ruler

Volume: liter or cubic meter, mm, cm, km etc.; ruler for regular solids or empty spaces, graduated cylinder or kitchen measuring cup for liquids and irregular solids

Mass: kilogram and its multiples and fractions; balance with calibrated samples, or scale with knowledge of local gravity, or methods of applying known force and measuring acceleration

8 0

4 years ago

The images below are models that represent the reactants and products of a chemical reaction.

Eva8 [605]

Explanation:

In a chemical reaction, when mass is conserved , the number of atoms or moles of the reactants must be equal to the number of moles or atoms in the products side.

From the diagram, we should carefully look to see if the number of atoms that makes up the reactants are equal to those on the product side.

For example:

A + B → AB

Here, mass is conserved because, on the reactant side, we have 1 atom of A and on the product side we have 1 atom of A

For B, on the reactant side, we have 1 atom of B and on the product side, we have 1 atom of B.

6 0

3 years ago

Read 2 more answers

A mass is attached to an ideal spring. At time t = 0 the spring is at its natural length and the mass is given an initial veloci

JulijaS [17]

Answer:

t = T/4

Explanation:

The power delivered to the mass by the spring is work done by the spring per second.

$P = \frac{dW}{dt}$

The work done by the spring is equal to the elastic potential energy stored in the spring.

$U = \frac{1}{2}kx^2$

The maximum energy stored in the spring is at the amplitude of the oscillation.

$U_{max} =\frac{1}{2}kA^2$

So the first time the mass reaches to its amplitude can be found by the following equation of motion:

$x = A\cos(\omega t + \phi)\\\phi = \pi/2 ~because ~at ~t= 0, ~ x = 0\\0 = A\cos(0 + \pi/2)\\x = A\cos(\omega t + \pi/2)$

When the mass reaches the amplitude:

$A = A\cos(\omega t + \pi/2)\\1 = \cos(\omega t + \pi/2)\\\omega t + \pi/2 = \pi$

because cos(π) = 1.

$\omega t = \pi/2$

Using ω = 2π/T,

$\omega t = \pi/2\\\frac{2\pi}{T}t = \pi/2\\t = \frac{T}{4}$

4 0

4 years ago

A 1423-kg car is moving along a level highway with a speed of 26.4 m/s. The driver takes the foot off the accelerator and the ca

Leona [35]

Answer:

Final speed, v = 28.81 m/s

Explanation:

Given that,

Mass of the car, m = 1423 kg

Initial speed of the car, u = 26.4 m/s

Force experience by the car, F = 901 N

Distance, d = 106 m

To find,

The speed of the car after traveling this distance.

Solution,

The force experienced by a car is equal to the product of mass and acceleration.

$F=ma$

$a=\dfrac{F}{m}$

$a=\dfrac{901}{1423}$

$a=0.63\ m/s^2$

Let v is the final speed of the car. Using third equation of motion to find it as :

$v^2-u^2=2ad$

$v^2=2ad+u^2$

$v^2=2\times 0.63\times 106+(26.4)^2$

v = 28.81 m/s

So, the final speed of the car is 28.81 m/s.

5 0

4 years ago

Two capacitors with capacitances of 1.0 m F and 0.50 m F, respectively, are connected in series. The system is connected to a 10

forsale [732]

Answer:

Therefore energy is stored in the 1.0 mF capacitor is 5.56×10⁻⁹ J

Explanation:

Series capacitor: The ending point of a capacitor is the starting point of other capacitor.

If C₁ and C₂ are connected in series then the equivalent capacitance is C.

where $\frac{1}{C} =\frac{1}{C_1}+\frac{1}{C_2}$

Given that,

C₁ = 1.0 mF=1.0×10⁻³F and C₂ = 0.50mF=0.50×10⁻³F

If C is equivalent capacitance.

Then $\frac{1}{C} =\frac{1}{1.0}+\frac{1}{0.5}$

$\Rightarrow \frac{1}{C} =\frac{3}{1}$

$\Rightarrow C=\frac{1}{3}$ mF

Again given that the system is connected to a 100-v battery.

We know that

q=Cv

q= charge

C= capacitor

v= potential difference

Therefore

$q=(\frac{1}{3} \times 10^{-3}\times10) C$

$=\frac{10^{-2}}{3} C$

The electrical potential energy stored in a capacitor can be expressed

$U=\frac{q^2C}{2}$

q= charge

c=capacitance of a capacitor

Therefore energy is stored in the 1.0 mF capacitor is

$U=\frac{q^2C_1}{2}$

$\Rightarrow U=\frac{(\frac{10^{-2}}{3})^2\times 10^{-3} }{2}$

$\Rightarrow U= 5.56\times 10^{-9}$ J

8 0

3 years ago