Resolve the vector shown below into its components.

The atmospheric pressure at the surface of titan is.

Harrizon [31]

Answer:

about 1.5 bars

Titan's atmosphere is similar to Earth's both in the predominance of nitrogen gas and in surface pressure, which is about 1.5 bars, or 50 percent higher than sea-level pressure on Earth.

Explanation:

8 0

1 year ago

Bowen’s reaction series illustrates relations between:

Finger [1]

C. Temperature, chemical composition and mineral structure

Explanation:

The Bowen's reaction series illustrates the relationship between temperature, chemical composition and mineral structure.

The series is made up of a continuous and discontinuous end through which magmatic composition can be understood as temperature changes.

The left part is the discontinuous end while the right side is the continuous series.
From the series, we understand that a magmatic body becomes felsic as it begins to cool to lower temperature.
A magma at high temperature is ultramafic and very rich in ferro-magnesian silicates which are the chief mineral composition of olivine and pyroxene. These minerals are predominantly found in mafic- ultramafic rocks. Also, we expect to find the calcic-plagioclase at high temperatures partitioned in the magma.
At a relatively low temperature, minerals with frame work structures begins to form . The magma is more enriched with felsic minerals and late stage crystallization occurs here.

Learn more:

Silicate minerals brainly.com/question/4772323

#learnwithBrainly

5 0

3 years ago

The earth has a vertical electric field at the surface, pointing down, that averages 100 N/C. This field is maintained by variou

Kipish [7]

Answer:

The excess charge on earth's surface was calculated to be 4.56 × 10⁵ C

Explanation:

Using the formula for an electric field;

E = kQ/r²

k = 1/(4πε₀) = 8.99 × 10⁹ Nm²/C²

E = 100N/C

r = radius of the earth = 6400 km = 6400000m

Q = Er²/k = 100 × (6400000)²/(8.99 × 10⁹)

Q = 455617.4 C = 4.56 × 10⁵ C

Hope this helps!!!

6 0

3 years ago

A light rope is attached to a block with mass 3.60 kg that rests on a frictionless, horizontal surface. The horizontal rope pass

Readme [11.4K]

Answer and Explanation:

(a) The fre-body diagrams for each block is shown below. In the block of mass 3.60 kg, there are 3 forces acting on it: horizontal force due to the rope ( $F_{t}$ ), vertical gravitational force ( $F_{g}$ ) and vertical normal force ( $F_{n}$ ), due to the surface. Since there is no vertical movement, $F_{g}$ and $F_{n}$ cancels it out. So, for this block, net force is horizontal due to the rope $F_{t}$ .

The block of mass m is hanging from the pulley, so there is the force of the rope ( $F_{t}$ ) and the gravitational force ( $F_{g}$ ). Both are vertical, because there is no surface "holding" block m.

(b) Since both blocks are attached to each other, the acceleration will be the same. To calculate it, we use the Second Law of Motion:

$F_{r}=m.a$

$a=\frac{F_{r}}{m}$

$a=\frac{18.8}{3.6}$

a = 5.22

The acceleration of either block is 5.22 m/s².

(c) Block m has 2 forces acting on it: tension and gravitational force. Gravitational force is the force of attraction the Earth does over an object. It is calculated as the product of mass and gravitational acceleration, which has magnitude g = 9.8 m/s².

Suppose positive referential is going up. To determine mass:

$F_{r}=m.a$

$F_{t}-F_{g}=m.a$

$F_{t}-m.g=m.a$

$18.8-9.8m=5.22m$

$15.02m=18.8$

m = 1.25

Block m has 1.25 kg.

(d) Gravitational force is also called weight. So, as described above: $F_{g}=m.g$ .

The weight for the hanging block is

$F_{g}=1.25*9.8$

$F_{g}=$ 12.25 N

Comparing tension and weight:

$\frac{12.25}{18.8}$ ≈ 0.65

We can see that, weight of the hanging block is almost 0.65 times smaller than the tension on the rope.

4 0

3 years ago

#3: a container has the dimensions of 30 cm x 50 mm x 0.2 m. the density of its contents is 2.5 g/cm3. what is the mass of the s

WARRIOR [948]

Good afternoon!

We calculate the volume of the container in cm³. To do that, we must put the units in cm:

30 cm → 30 cm
50 mm → 5 cm
0.2 m → 20 cm

The volume is:

V = 30 . 5 . 20

V = 3000 cm³

Now, we calculate the mas with the formula:

m = dV

m = 2.5 · 3000

m = 7500 g

Dividing by 1000, we have the mass in kg:

m = 7.5 kg

4 0

3 years ago

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