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Grace [21]
3 years ago
8

What drives plate tectonics in the earths interior

Physics
2 answers:
xxTIMURxx [149]3 years ago
4 0
The convection going on with the magma in the asthenosphere
Gala2k [10]3 years ago
4 0
The athenosphere and/or mantel are constantly moving. the plates lie on them. The movement s are very small so we do not feel them. thats all i can say for now.
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Calculate the change in length of a 90.5 mm aluminum bar that has increased in temperature by from -14.4 oC to 154.6 oC
nignag [31]

Answer:

 ΔL = 3.82 10⁻⁴ m

Explanation:

This is a thermal expansion exercise

          ΔL = α L₀ ΔT

          ΔT = T_f - T₀

where ΔL is the change in length and ΔT is the change in temperature

Let's reduce the length to SI units

          L₀ = 90.5 mm (1m / 1000 mm) = 0.0905 m

let's calculate

          ΔL = 25.10⁻⁶ 0.0905 (154.6 - (14.4))

          ΔL = 3.8236 10⁻⁴ m

     

using the criterion of three significant figures

          ΔL = 3.82 10⁻⁴ m

5 0
3 years ago
The volume of liquid flowing per second is called the volume flow rate Q and has the dimensions of [L]3/[T]. The flow rate of li
melamori03 [73]

Answer: n=4

Explanation:

We have the following expression for the volume flow rate Q of a hypodermic needle:

Q=\frac{\pi R^{n}(P_{2}-P_{1})}{8\eta L}  (1)

Where the dimensions of each one is:

Volume flow rate Q=\frac{L^{3}}{T}

Radius of the needle R=L

Length of the needle L=L

Pressures at opposite ends of the needle P_{2} and P_{1}=\frac{M}{LT^{2}}

Viscosity of the liquid \eta=\frac{M}{LT}

We need to find the value of n whicha has no dimensions, and in order to do this, we have to rewritte (1) with its dimensions:

\frac{L^{3}}{T}=\frac{\pi L^{n}(\frac{M}{LT^{2}})}{8(\frac{M}{LT}) L}  (2)

We need the right side of the equation to be equal to the left side of the equation (in dimensions):

\frac{L^{3}}{T}=\frac{\pi}{8} \frac{ L^{n}}{LT}  (3)

\frac{L^{3}}{T}=\frac{\pi}{8} \frac{ L^{n-1}}{T}  (4)

As we can see n must be 4 if we want the exponent to be 3:

\frac{L^{3}}{T}=\frac{\pi}{8} \frac{ L^{4-1}}{T}  (5)

Finally:

\frac{L^{3}}{T}=\frac{\pi}{8} \frac{ L^{3}}{T}  (6)

8 0
3 years ago
Gravity on earth is 9.8 m/s squared, and gravity on the moon is 1.6 m/s squared. So if the mass of an object on earth is 40 kilo
garik1379 [7]

The mass of an object on Earth is the same as its mass on the Moon. The weight is different.

Weight = m * g

Weight ( Moon ) = 40 kg * 1.6 m/s² = 64 N

If the mass of an object on Earth is 40 kg, its mass on the Moon is 40 kg and its weight on the Moon is 64 N. 

7 0
3 years ago
A current runs from the left to the right in a long, straight wire. How does the magnetic field at point X compare with the magn
vitfil [10]
I think the answer is <span>D. The magnetic field at point X points into the page, and the magnetic field at point Y points out of the page.</span>
7 0
3 years ago
a vector has an x-component of 19.5m and a y-component of 28.4m. Find the magnitude and direction of the vector
Delicious77 [7]

Answer:

magnitude=34.45 m

direction=55.52\°

Explanation:

Assuming the initial point P1 of this vector is at the origin:

P1=(X1,Y1)=(0,0)

And knowing the other point is P2=(X2,Y2)=(19.5,28.4)

We can find the magnitude and direction of this vector, taking into account a vector has a initial and a final point, with an x-component and a y-component.

For the magnitude we will use the formula to calculate the distance d between two points:

d=\sqrt{{(Y2-Y1)}^{2} +{(X2-X1)}^{2}} (1)

d=\sqrt{{(28.4 m - 0 m)}^{2} +{(19.5 m - 0m)}^{2}} (2)

d=\sqrt{1186.81 m^{2}} (3)

d=34.45 m (4) This is the magnitude of the vector

For the direction, which is the measure of the angle the vector makes with a horizontal line, we will use the following formula:

tan \theta=\frac{Y2-Y1}{X2-X1}  (5)

tan \theta=\frac{24.8 m - 0m}{19.5 m - 0m}  (6)

tan \theta=\frac{24.8}{19.5}  (7)

Finding \theta:

\theta= tan^{-1}(\frac{24.8}{19.5})  (8)

\theta= 55.52\°  (9) This is the direction of the vector

3 0
3 years ago
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