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3 years ago

9. The three types of stress that act on Earth's rocks are compression, tension, and A. strain. B. shear. C. tephra. D. shale.

Physics

2 answers:

Evgen [1.6K]3 years ago

5 0

The three main types of stress in a rock are shearing, tension, and compression. hope. that helped

almond37 [142]3 years ago

4 0

The three tyres of stress that act on Earth's rocks are compression, tension, and shear.

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What is precision?what is precision?

RUDIKE [14]

Answer:

According to Oxford Dictionaries "Precision" means "the quality, condition, or fact of being exact and accurate."

Explanation:

Hope this helps! :)

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2 years ago

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A 40 cm wire with a radius of 3 cm is oriented along the y axis and carries a current of 2 A. What is the magnitude of the magne

ZanzabumX [31]

Answer:

a) B = 1.99 x 10⁻⁴ Tesla

b) B = 0.88 x 10⁻⁴ Tesla

Explanation:

According to Biot - Savart Law, the magnetic field due to a currnt carrying straight wire is given as:

B = μ₀ I L/4πr²

where,

μ₀ = permebility of free space = 1.25 x 10⁻⁶ H m⁻¹

I = current = 2 A

L = Length of wire = 40 cm = 0.4 m

r = radius of magnetic field = 2 cm = 0.02 m

Therefore,

B = (1.25 x 10⁻⁶ H m⁻¹)(2 A)(0.4 m)/4π(0.02 m)²

B = 1.99 x 10⁻⁴ Tesla

r = radius of magnetic field = 3 cm = 0.03 m

Therefore,

B = (1.25 x 10⁻⁶ H m⁻¹)(2 A)(0.4 m)/4π(0.03 m)²

B = 0.88 x 10⁻⁴ Tesla

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3 years ago

What is the weight of a dog with a mass of 12 kg? Explain how you obtained your answer.

butalik [34]

Answer: 117.60N

Explanation:

Weight is a force. Therefore, we can use the force formula to find weight.

$Formula: W=m*g$

W = weight

m = mass

g = acceleration due to gravity ( $9.8m/s^2$ )

$W=(12kg)(9.8m/s^2)\\W=117.60N$

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3 years ago

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6) Find the speed a spherical raindrop would attain by falling from 4.00 km. Do this:a) In the absence of air dragb) In the pres

sleet_krkn [62]

We are asked to determine the velocity of a rain drop if it falls from 4 km.

To do that we will use the following formula:

$2ah=v_f^2-v_0^2$

Where:

$\begin{gathered} a=\text{ acceleration} \\ h=\text{ height} \\ v_f,v_0=\text{ final and initial velocity} \end{gathered}$

If we assume the initial velocity to be 0 we get:

$2ah=v_f^2$

The acceleration is the acceleration due to gravity:

$2gh=v_f^2$

Now, we take the square root to both sides:

$\sqrt{2gh}=v_f$

Now, we substitute the values:

$\sqrt{2(9.8\frac{m}{s^2})(4000m)}=v_f$

solving the operations:

$280\frac{m}{s}=v$

Therefore, the velocity without air drag is 280 m/s.

Part B. we are asked to determine the velocity if there is air drag. To do that we will use the following formula:

$F_d=\frac{1}{2}C\rho_{air}Av^2$

Where:

$\begin{gathered} F_d=drag\text{ force} \\ C=\text{ constant} \\ \rho_{air}=\text{ density of air} \\ A=\text{ area} \\ v=\text{ velocity} \end{gathered}$

We need to determine the drag force. To do that we will use the following free-body diagram:

Since the velocity that the raindrop reaches is the terminal velocity and its a constant velocity this means that the acceleration is zero and therefore the forces are balanced: