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Andrej [43]
3 years ago
12

Charles Darwin,

Physics
2 answers:
Roman55 [17]3 years ago
8 0
The answer is B. Artificial Selection :)
grandymaker [24]3 years ago
3 0
The answer is B artificial selection
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The three types of stress that act on Earth's rocks are compression, tension, and
Murrr4er [49]
The three types of stress that act on Earth's rocks are compression, tension and shear. Among all the options that are given in the question, the correct option is option "D". These kind of stress action creates the rocks to break in a natural way. The earth's rocks sometimes collide with one another, pull apart from each other and sometimes slide against each other. Whenever two rocks collide with each other, they create a compression force. When the two rocks of the earth tries to pull away from each other, it creates a tension force. The sliding of two earth's rocks creates a shearing force.
3 0
3 years ago
How hot can the desert get​
Art [367]

Answer:

134 f

Explanation:

The hottest temperature ever reliably measured in a desert was 134 degrees F, in Death Valley of the Mojave Desert in 1913.

8 0
3 years ago
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What is the value of ΔVBA=VB−VA if the charge on the plates is 1.00 x 10-9 C, the area of the plates is 2.00 m2 and the distanc
Murljashka [212]

Answer:

The Value is V_{AB} = 2.825V

Explanation:

The explanation is shown on the first  uploaded image

5 0
3 years ago
A block is at rest on a plank whose angle can be varied. The angle is gradually increased from 0 deg. At 31.8°, the block starts
galina1969 [7]

Answer:

\mu_s = 0.62

\mu_k = 0.415

The motion of the block is downwards with acceleration 1.7 m/s^2.

Explanation:

First, we will calculate the acceleration using the kinematics equations. We will denote the direction along the incline as x-direction.

x - x_0 = v_0t + \frac{1}{2}at^2\\3.4 = 0 + \frac{1}{2}a(2)^2\\a = 1.7~m/s^2

Newton’s Second Law can be used to find the net force applied on the block in the -x-direction.

F = ma\\F = 1.7m

Now, let’s investigate the free-body diagram of the block.

Along the x-direction, there are two forces: The x-component of the block’s weight and the kinetic friction force. Therefore,

F = mg\sin(\theta) - \mu_k mg\cos(\theta)\\1.7m = mg\sin(31.8) - \mu_k mg\cos(31.8)\\1.7 = (9.8)\sin(\theta) - mu_k(9.8)\cos(\theta)\\mu_k = 0.415

As for the static friction, we will consider the angle 31.8, but just before the block starts the move.

mg\sin(31.8) = \mu_s mg\cos(31.8)\\\mu_s = tan(31.8) = 0.62

5 0
3 years ago
When you take your 1900-kg car out for a spin, you go around a corner of radius 55 m with a speed of 15 m/s. The coefficient of
pentagon [3]

Answer:

7772.72N

Explanation:

When u draw your FBD, you realize you have 3 forces (ignore the force the car produces), gravity, normal force and static friction. You also realize that gravity and normal force are in our out of the page  (drawn with a frame of reference above the car). So that leaves you with static friction in the centripetal direction.

Now which direction is the static friction, assume that it is pointing inward so

Fc=Fs=mv²/r=1900*15²/55=427500/55=7772.72N

Since the car is not skidding we do not have kinetic friction so there can only be static friction. One reason we do not use μFn is because that is the formula for maximum static friction, and the problem does not state there is maximum static friction.

7 0
3 years ago
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