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3 years ago

Which rock is made mostly of dark, fine-grained silicate minerals, chiefly plagioclase feldspar and pyroxene, and magnetite?

1 answer:

8 0

A dark-colored (igneous rock,) commonly extrusive from volcanic eruptions and composed primarily of the minerals of calcic plagioclase and pyroxene, and sometimes olivine. Basalt is the fine-grained equivalent of gabbro.

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Kai takes a hamburger off of the grill and puts a piece of cheese on it. Explain how and why the cheese melts when Kai puts it o

Yanka [14]

Because the hamburger is still hot from the grill, the cheese melts because of that heat.

3 0

3 years ago

Read 2 more answers

Find a parametric representation for the surface. The plane through the origin that contains the vectors i - j and j - k

boyakko [2]

Answer:

parametric representation: x = u, y = v - u , z = - v

Explanation:

Given vectors :

i - j , j - k

represent the vector equation of the plane as:

r ( u, v ) = r₀ + <em>u</em>a + vb

where: r₀ = position vector

u and v = real numbers

a and b = nonparallel vectors

expressing the nonparallel vectors as :

a = i -j , b = j - k , r = ( x,y,z ) and r₀ = ( x₀, y₀, z₀ )

hence we can express vector equation of the plane as

r(u,v) = ( x₀ + u, y₀ - u + v, z₀ - v )

Finally the parametric representation of the surface through (0,0,0) i.e. origin = 0

( x, y , z ) = ( x₀ + u, y₀ - u + v, z₀ - v )

x = 0 + u ,

y = 0 - u + v

z = 0 - v

∴ parametric representation: x = u, y = v - u , z = - v

3 0

2 years ago

I need help with answer 51.

Troyanec [42]

Answer:

Explanation:

5 0

2 years ago

A traffic light is weighing 200N hangs from a vertical cable tied to two other cables that are fastened to a support. The upper

Levart [38]

Answer:

T₁ = 93.6 N , T₂ = 155.6 N , T₃ = 200 N

Explanation:

This is a balance exercise where we must apply the expressions for translational balance in the two axes

∑ F = 0

Suppose that cable t1 goes to the left and the angles are 41º with respect to the horizontal and cable t2 goes to the right with angles of 63º

decompose the tension of the two upper cables

cos 41 = T₁ₓ / T1

sin 41 = T₁y / T1

T₁ₓ = T₁ cos 41

T₁y= T₁ sin 41

for cable gold

cos 63 = T₂ / T₂

sin 63 = $T_{2y}$ / T₂

We apply the two-point equilibrium equation: The junction point of the three cables and the point where the traffic light joins the vertical cable.

Let's start by analyzing the point where the traffic light meets the vertical cable

T₃ - W = 0

T₃ = W

T₃ = 200 N

now let's write the equations for the single point of the three wires

X axis

- T₁ₓ + T₂ₓ = 0

T₁ₓ = T₂ₓ

T1 cos 41 = T2 cos 63

T1 = T2 cos 63 / cos 41 (1)

y Axis

$T_{1y}$ + T_{2y} - T3 = 0

T₁ sin 41 + T₂ sin 63 = T₃ (2)

to solve the system we substitute equation 1 in 2

T₂ cos 63 / cos 41 sin 41 + T₂ sin 63 = W

T₂ (cos 63 tan 41 + sin 63) = W

T₂ = W / (cos 63 tan 41 + sin 63)

We calculate

T₂ = 200 / (cos 63 tan 41 + sin 63)

T₂ = 200 / 1,2856

T₂ = 155.6 N

we substitute in 1

T₁ = T₂ cos 63 / cos 41

T₁ = 155.6 cos63 / cos 41

T₁ = 93.6 N

therefore the tension in each cable is

T₁ = 93.6 N

T₂ = 155.6 N

T₃ = 200 N

6 0

2 years ago

If I wanted to generate a maximum emf of 20 V, what angular velocity (radians/sec, aka Hz) would be required given a circular co

sergij07 [2.7K]

Answer:

Angular velocity, $\omega=35.36\ rad/s$

Explanation:

It is given that,

Maximum emf generated in the coil, $\epsilon=20\ V$

Diameter of the coil, d = 40 cm

Radius of the coil, r = 20 cm = 0.2 m

Number of turns in the coil, N = 500

Magnetic field in the coil, $B=9\times 10^{-3}\ T$

The angle between the area vector and the magnet field vector varies from 0 to 2 π radians. The formula for the maximum emf generated in the coil is given by :

$\epsilon=NBA\omega$

$\omega=\dfrac{\epsilon}{NBA}$

$\omega=\dfrac{20\ V}{500\times 9\times 10^{-3}\ T\times \pi (0.2\ m)^2}$

$\omega=35.36\ rad/s$

So, the angular velocity of the circular coil is 35.36 rad/s. Hence, this is the required solution.

6 0

3 years ago