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3 years ago

Yo 20 points no links plz

Physics

2 answers:

makvit [3.9K]3 years ago

8 0

Answer:

he movement of the plates creates three types of tectonic boundaries: ... Where plates serving landmasses collide, the crust crumples and buckles into mountain ranges. ... where two plates grind past each other along what are called strike-slip faults. These boundaries don't produce spectacular features like mountains or ...

Explanation:

asambeis [7]3 years ago

7 0

Answer: Describe how plates create mountain ranges like the Himalayas.

Answer- Most of the world's largest mountains form as plates collide at convergent plate boundaries. This creates mountains. Folding and faulting in these collision zones makes the crust thicker. The world's highest mountain range, the Himalayas, is growing as India collides with Eurasia

2. Diagram how pulling apart continental crust could create mountains and basins. What are the mountains and

basins called? Answer- When tensional stresses pull crust apart, it breaks into blocks that slide up and drop down along normal faults. The result is alternating mountains and valleys, known as a basin-and-range

3. Why don’t strike slip faults create mountains? Answer-When tensional stresses pull crust apart, it breaks into blocks that slide up and drop down along normal faults. The result is alternating mountains and valleys, known as a basin-and-range

Explanation: I'm really sorry that its a lot but I hope it helps :)

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The power of a purely resistive lead is always positive although the current and voltage are sometimes negative. explain

pickupchik [31]

Answer:

Current is in phase with voltage in a resistive circuit. Note that the wave form for power is always positive, never negative for this resistive circuit. This means that power is always being dissipated by the resistive load, and never returned to the source as it is with reactive loads.Explanation:

7 0

2 years ago

A small ball is attached to the lower end of a 0.800-m-long string, and the other end of the string is tied to a horizontal rod.

mestny [16]

Answer:

a = 17.68 m/s²

Explanation:

given,

length of the string, L = 0.8 m

angle made with vertical, θ = 61°

time to complete 1 rev, t = 1.25 s

radial acceleration = ?

first we have to calculate the radius of the circle

R = L sin θ

R = 0.8 x sin 61°

R = 0.7 m

now, calculating at the angular velocity

$\omega =\dfrac{2\pi}{T}$

$\omega =\dfrac{2\pi}{1.25}$

ω = 5.026 rad/s

now, radial acceleration

a = r ω²

a = 0.7 x 5.026²

a = 17.68 m/s²

hence, the radial acceleration of the ball is equal to 17.68 rad/s²

7 0

3 years ago

A satellite is revolving the earth 4km above the surface.find the orbital velocity of the satellite (R =6400km,g=9.8m/s^2)

Karo-lina-s [1.5K]

Answer:

v ≈ 7900 m/s

Explanation:

centripetal force will equal gravity force

mv²/R = mg

v²/R = g

v² = Rg

v = √(Rg)

v = √(6.4e6(9.8))

v = 7.91959...e+3

v ≈ 7900 m/s

of course, at those velocities and that deep into the atmosphere, the satellite would quickly burn up, slow down, and cause tremendous damage to buildings etc. with the sonic boom shock wave. It would also have to avoid a lot of mountains as 4000 m is not that high.

3 0

3 years ago

Air enters a turbine operating at steady state with a pressure of 75 Ibf/in.^2, a temperature of 800º R and velocity of 400 ft/s

Arturiano [62]

Answer:

(a) W/m = 49.334 Btu/lb

(b) $\frac{E_{d} }{m}$ = 22.12 Btu/lb

Explanation:

For the given problem, it can be assumed that the system is operating at steady state and the effects of potential energy can be neglected.

(a) Using the thermodynamic table for air.

At the temperature ( $T_{1}$ )of 800 ºR and pressure ( $P_{1}$ ) of 75 Ibf/in.^2, we can deduce that:

Specific enthalpy ( $h_{1}$ ) = 191.81 BTu/lb

Specific entropy ( $s_{1}$ ) = 0.6956 Btu/(lb.ºR)

At the temperature ( $T_{2}$ )of 600 ºR and pressure ( $P_{2}$ ) of 15 Ibf/in.^2, we can deduce that:

Specific enthalpy ( $h_{2}$ ) = 143.47 BTu/lb

Specific entropy ( $s_{2}$ ) = 0.6261 Btu/(lb.ºR)

The work done can be calculated using energy rate equation:

$\frac{W}{m} = \frac{Q}{m} + (h_{1} - h_{2}) + \frac{V_{1}^{2} - V_{2}^{2}}{2}$

Q/m = heat transfer = -2 Btu/lb

$V_{1}$ = 400 ft/s

$V_{2}$ = 100 ft/s

$\frac{W}{m} = -2 + (191.81 - 143.47) + \frac{400^{2} - 100^{2}}{2}*[tex]\frac{1}{2*32.2*778}$ [/tex] = -2 + 48.34 + 29.938 = 49.334 Btu/lb

(b) To calculate the exergy destruction, we will use the equation for exergy rate:

$\frac{E_{d} }{m} = [1-\frac{T_{o} }{T_{b} }](\frac{Q}{m}) - \frac{W}{m} + [(h_{1} - h_{2}) -T_{o}(s_{1} - s_{2}) + \frac{V^{2} _{1} - V_{2} ^{2}}{2}]$

The equation above is further simplified to:

$\frac{Ed}{m} = T_{o}[(s_{2} -s_{1}) - Rln\frac{P_{2} }{P_{1} } - \frac{Q/m}{T_{b} }]$

Using a reference temperature (To) = 500 °R

Average surface temperature (Tb = 620°R

$\frac{Ed}{m} = 500*[(0.6261 -0.6956) - (1.986/28.97)ln\frac{15 }{75 } - \frac{-2}{620}}]$

$\frac{E_{d} }{m}$ = 500*[-0.0695 +0.068688*1.609 +0.003225] = 22.12 Btu/lb

5 0

3 years ago

An AC generator has 80 rectangular loops on its armature. Each loop is 12 cm long and 8 cm wide. The armature rotates at 1200 rp

Sauron [17]

Answer: 28.96 V

Explanation:

Given

No of loops on the armature, N = 80

Length of the loop, l = 12 cm = 0.12 m

Width of the loop, b = 8 cm = 0.08 m

Speed of the armature, 1200 rpm

Magnetic field of the loop, B = 0.30 T

To solve this, we use the formula

V(max) = NBAω

Where,

A = area of loop

A = l*b = 0.12 * 0.08

A = 0.0096 m²

ω = 1200 rpm = 1200 * 2π/60 rad/s

ω = (1200 * 2 * 3.142) / 60

ω = 7540.8 / 60

ω = 125.68 rad/s

Substituting the values into the formula

V(max) = NBAω

V(max) = 80 * 0.30 * 0.0096 * 125.68

V(max) = 80 * 0.362

V(max) = 28.96 V

Therefore, the maximum output voltage of the generator would be 28.96 V

5 0

2 years ago