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3 years ago

Between what depths does Earth's temperature increase the slowest?

1 answer:

8 0

Temperature increases at a rate of approximately 20o C/km in the upper crust (first 10 km) but then the rate decreases to only approximately 0.3o C/km below 200km due to the homogenizing effect of mantle convection. A good approximation of the rate of pressure increase with depth is: 30 MPa/km in the crust and 35 MPa/km in the mantle.

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Assume that you can drive at a constant speed of 100 kilometers per hour. suppose you started driving from the sun. how long wou

belka [17]

The Sun is 149.6 million kilometers from the earth.
There are 8760 hours in a year.
876000 km are traveled in a year
It would take 170.776 years to reach the sun, or 171 years rather

4 0

3 years ago

A student is trying to calculate the density of a can. He already knows the mass, but he needs to determine the volume as well.

elena-14-01-66 [18.8K]

Answer:

V=Bh

Explanation:

B h is used for rectangular solids and cylinders

6 0

3 years ago

Continuous and aligned fiber-reinforced composite with cross-sectional area of 340 mm2 (0.53 in.2) is subjected to a longitudina

Alecsey [184]

(a) 23.4

The fiber-to-matrix load ratio is given by

$\frac{F_f}{F_m}=\frac{E_f V_f}{E_m V_m}$

where

$E_f = 131 GPa$ is the fiber elasticity module

$E_m = 2.4 GPa$ is the matrix elasticity module

$V_f=0.3$ is the fraction of volume of the fiber

$V_m=0.7$ is the fraction of volume of the matrix

Substituting,

$\frac{F_f}{F_m}=\frac{(131 GPa)(0.3)}{(2.4 GPa)(0.7)}=23.4$ (1)

(b) 44,594 N

The longitudinal load is

F = 46500 N

And it is sum of the loads carried by the fiber phase and the matrix phase:

$F=F_f + F_m$ (2)

We can rewrite (1) as

$F_m = \frac{F_f}{23.4}$

And inserting this into (2):

$F=F_f + \frac{F_f}{23.4}$

Solving the equation, we find the actual load carried by the fiber phase:

$F=F_f (1+\frac{1}{23.4})\\F_f = \frac{F}{1+\frac{1}{23.4}}=\frac{46500 N}{1+\frac{1}{23.4}}=44,594 N$

(c) 1,906 N

Since we know that the longitudinal load is the sum of the loads carried by the fiber phase and the matrix phase:

$F=F_f + F_m$ (2)

Using

F = 46500 N

$F_f = 44594 N$

We can immediately find the actual load carried by the matrix phase:

$F_m = F-F_f = 46,500 N - 44,594 N=1,906 N$

(d) 437 MPa

The cross-sectional area of the fiber phase is

$A_f = A V_f$

where

$A=340 mm^2=340\cdot 10^{-6}m^2$ is the total cross-sectional area

Substituting $V_f=0.3$ , we have

$A_f = (340\cdot 10^{-6} m^2)(0.3)=102\cdot 10^{-6} m^2$

And the magnitude of the stress on the fiber phase is

$\sigma_f = \frac{F_f}{A_f}=\frac{44594 N}{102\cdot 10^{-6} m^2}=4.37\cdot 10^8 Pa = 437 MPa$

(e) 8.0 MPa

The cross-sectional area of the matrix phase is

$A_m = A V_m$

where

$A=340 mm^2=340\cdot 10^{-6}m^2$ is the total cross-sectional area

Substituting $V_m=0.7$ , we have

$A_m = (340\cdot 10^{-6} m^2)(0.7)=238\cdot 10^{-6} m^2$

And the magnitude of the stress on the matrix phase is

$\sigma_m = \frac{F_m}{A_m}=\frac{1906 N}{238\cdot 10^{-6} m^2}=8.0\cdot 10^6 Pa = 8.0 MPa$

(f) $3.34\cdot 10^{-3}$

The longitudinal modulus of elasticity is

$E = E_f V_f + E_m V_m = (131 GPa)(0.3)+(2.4 GPa)(0.7)=41.0 Gpa$

While the total stress experienced by the composite is

$\sigma = \frac{F}{A}=\frac{46500 N}{340\cdot 10^{-6}m^2}=1.37\cdot 10^8 Pa = 0.137 GPa$

So, the strain experienced by the composite is

$\epsilon=\frac{\sigma}{E}=\frac{0.137 GPa}{41.0 GPa}=3.34\cdot 10^{-3}$

3 0

3 years ago

A long solenoid that has 810 turns uniformly distributed over a length of 0.380 m produces a magnetic field of magnitude 1.00 10

IRINA_888 [86]

Answer:

0.037 A

Explanation:

Magnetic field = B = 1.00 e-4 T

Length = L = 0.380 m

Number of turns = 810

B = μ₀ N I / L

⇒ Current = I = B L / μ₀ N = ( 1 e-4) ( 0.380) / (4π × 10⁻⁷)(810)

= 0.037 A = 37.3 mA

5 0

3 years ago

Five identical quintuplets leave earth when they reach the age of 21, in the year 2121. Each quintuplet goes on a spaceship jour

Elena-2011 [213]

Answer:

Explanation:

This is a problem based on time dilation , a theory given by Albert Einstein .

The formula of time dilation is as follows .

t₁ = $\frac{t}{\sqrt{1-\frac{v^2}{c^2} } }$

t is time measured on the earth and t₁ is time measured by man on ship .

A ) Given t = 20 years , t₁ = ? v = .4c

$\frac{20}{\sqrt{1-\frac{.16c^2}{c^2} } }$

=1.09 x 20

t₁= 21.82 years

B ) Given t = 5 years , t₁ = ? v = .2c

$\frac{5}{\sqrt{1-\frac{.04c^2}{c^2} } }$

=1.02 x 5

t₁= 5.1 years

C ) Given t = 10 years , t₁ = ? v = .8c

$\frac{10}{\sqrt{1-\frac{.64c^2}{c^2} } }$

=1.67 x 10

t₁= 16.7 years

D ) Given t = 10 years , t₁ = ? v = .4c

$\frac{10}{\sqrt{1-\frac{.16c^2}{c^2} } }$

=1.09 x 10

t₁= 10.9 years

E ) Given t = 20 years , t₁ = ? v = .8c

$\frac{20}{\sqrt{1-\frac{.64c^2}{c^2} } }$

=1.67 x 20

t₁= 33.4 years

7 0

3 years ago