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3 years ago

Charles Darwin's groundbreaking book, The Origin of Species, described the process of

1 answer:

3 0

B)natural selection

Darwin discovered this idea of natural selection while on a voyage to places such as the Galapagos Islands. Essentially, the idea is that certain animals developed traits that allowed them to survive better than others.

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Anyone want to help a brother out

s344n2d4d5 [400]

I would say B but I’m not 100%

5 0

3 years ago

Hooke's law. The distance d when a spring is stretched by a hanging objective varies directly as the weight w of the object. If

Fofino [41]

Hooke's law is F=-kx, which means the elastic force contained by the spring is a product of the distance it stretches and its spring constant, but the direction of the force is opposite that of the displacement. We calculate as follows:

(3 kg)(9.8 m/s^2) = -k(-0.38 m)
k = 77.4

Then use k to find the new displacement, again using Hooke's law:
(7 kg)(9.8 m/s^2) = -(77.4)x
x = -0.89 m

7 0

3 years ago

he membrane that surrounds a certain type of living cell has a surface area of 6.0 x 10-9 m2 and a thickness of 1.6 x 10-8 m. As

k0ka [10]

Answer:

$1.54481175\times 10^{-12}\ C$

Explanation:

$\epsilon_0$ = Permittivity of free space = $8.85\times 10^{-12}\ F/m$

A = Area = $6\times 10^{-9}\ m^2$

d = Thickness = $1.6\times 10^{-8}\ m$

k = Dielectric constant = 5.4

V = Voltage = 86.2 mV

Charge is given by

$Q=CV\\\Rightarrow Q=k\epsilon\dfrac{A}{d}V\\\Rightarrow Q=5.4\times 8.85\times 10^{-12}\times \dfrac{6\times 10^{-9}}{1.6\times 10^{-8}}\times 86.2\times 10^{-3}\\\Rightarrow Q=1.54481175\times 10^{-12}\ C$

The charge on the outer surface is $1.54481175\times 10^{-12}\ C$

4 0

2 years ago

During a solar eclipse, which of the following is true?

kaheart [24]

Answer:

The Moon blocks the suns light from hitting the surface of the earth

6 0

3 years ago

Read 2 more answers

Suppose that a nascar race car is moving to the right with a constant velocity of +86 m/s. What is the average acceleration of t

Stels [109]

(a) As the car is moving with constant velocity, it means the rate change of velocity does not change, therefore the average acceleration of the car is zero.

Thus, there is no acceleration, when velocity is constant.

(b) Average acceleration,

$a =\frac{ v-u}{\Delta t}$

Here, v is final velocity and u is the initial velocity and $\Delta t$ is the time interval.

As twelve seconds later, the car is halfway around the track and traveling in the opposite direction with the same speed, therefore

$a =\frac{ -86 \ m/s-86 \ m/s }{12 \ s} = - 14 .3 \ m/s^2$

Thus, the average acceleration of the car is $14 .3 \ m/s^2$ in the direction to the left.

6 0

3 years ago