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2 years ago

Earth's lithospheric plates interact at____.

Physics

1 answer:

earnstyle [38]2 years ago

7 0

Explanation:

Tectonic plate interactions are classified into three basic types: Divergent boundaries are areas where plates move away from each other, forming either mid-oceanic ridges or rift valleys. These are also known as constructive boundaries. Convergent boundaries are areas where plates move toward each other and collide.

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What is the main purpose of trying to quickly cool heated food?

Ronch [10]

You can eat the food

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2 years ago

Read 2 more answers

An insulated rigid tank initially contains 1.4-kg saturated liquid water and water vapor at 200°C. At this state, 25 percent of

Elan Coil [88]

Solution:

Mass of liquid water and water vapor in the insulated tank initially = 1.4 kg

Temperature = 200 °C

And 25% of the volume by liquid water is steam.

State 1

$m=\frac{V}{v}$

$m=m_f+m_g$

$1.4=\frac{0.25V}{v_f}+\frac{0.75V}{v_g}$

$1.4=\frac{0.25V}{1.1565 \times 10^{-3}}+\frac{0.75V}{0.1274}$ (taking the value of $v_g$ and $v_g$ at 200°C )

$V=6.304 \times 10^{-3}$

Now quality of vapor

$x=\frac{m_g}{m}$

$=3.377 \times 10^{-3}$

Internal energy at state 1 can be found out by

$u_1=u_f+xu_{fg}$

$=850.65+3.377\times10^{-3}\times 1744.65$

= 856.54 kJ/kg

After heating with the resistor for 20 minutes, at state 2, the tank contains saturated water vapor $v_2=v_g \text { and }\ x=1$

Tank is rigid, so volume of tank is constant.

$v_g=v_2=\frac{V}{m}$

$v_g=\frac{6.304\times 10^{-3}}{1.4}$

$v_g=4.502 \times 10^{-3} \ m^3 /kg$

Now interpolate the value to get temperature at state 2 with specific volume value to get final temperature

$T_2=360+(374.14-360)\left(\frac{0.004502-0.006945}{0.003155-0.006945}\right)$

= 369.11° C

Internal energy at state 2

$u_2=2154.9 \ kJ/kg$

Now power rating of the resistor

$P=\frac{m(u_2-u_1)}{t}$

$P=\frac{1.4(2154.9-856.54)}{20 \times 60}$

= 1.51 kW

6 0

2 years ago

To initiate a nuclear reaction, an experimental nuclear physicist wants to shoot a proton into a 5.50-fm-diameter 12C nucleus. T

Fantom [35]

Answer:

$V_1= 3.4*10^7m/s$

Explanation:

From the question we are told that

Nucleus diameter $d=5.50-fm$

a 12C nucleus

Required kinetic energy $K=2.30 MeV$

Generally initial speed of proton must be determined,applying the law of conservation of energy we have

$K_2 +U_2=K_1+U_1$

where

$K_1$ =initial kinetic energy

$K_2$ =final kinetic energy

$U_1$ =initial electric potential

$U_2$ =final electric potential

mathematically

$U_2 = \frac{Kq_pq_c}{r_2}$

where

$r_f$ =distance b/w charges

$q_c$ =nucleus charge $=6(1.6*10^-^1^9C)$

$K$ =constant

$q_p$ =proton charge

Generally kinetic energy is know as

$K=\frac{1}{2} mv^2$

Therefore

$U_2 = \frac{Kq_pq_c}{r_2} + K_2=\frac{1}{2} mv_1^2 +U_1$

Generally equation for radius is $d/2$

Mathematically solving for radius of nucleus

$R=(\frac{5.50}{2}) (\frac{1*10^-^1^5m}{1fm})$

$R=2.75*10^-^1^5m$

Generally we can easily solving mathematically substitute into v_1

$q_p$ $=6(1.6*10^-^1^9C)$

$K_1=9.0*10^9 N-m^2/C^2$

$U_1= 0$

$R=2.75*10^-^1^5m$

$K=2.30 MeV$

$m= 1.67*10^-^2^7kg$

$V_1= (\frac{2}{1.67*10^-^2^7kg})^1^/^2 (\frac{(9.0*10^9 N-m^2/C^2)*(6(1.6*10^-^1^9C)(1.6*10^-^1^9C)}{2.75*10^-^1^5m+2.30 MeV(\frac{1.6*10^-^1^3 J}{1 MeV}) }$