All categories

3 years ago

7.

Physics

2 answers:

Usimov [2.4K]3 years ago

7 0

Answer:

Explanation:

Tpy6a [65]3 years ago

4 0

Answer:

Explanation:

Magma rising up from the mantle at a divergent boundary, I got this question and I got it right.

You might be interested in

A 50 g copper calorimeter contains 250 g of water at 20 C. How much steam be condensed into the water to make the final temperat

Nostrana [21]

Answer:

Approximately $13\; \rm g$ of steam at $100\; \rm ^\circ C$ (assuming that the boiling point of water in this experiment is $100\; \rm ^\circ C\!$ .)

Explanation:

Latent heat of condensation/evaporation of water: $2260\; \rm J \cdot g^{-1}$ .

Both mass values in this question are given in grams. Hence, convert the specific heat values from this question to $\rm J \cdot g^{-1}$ .

Specific heat of water: $4.2\; \rm J \cdot g^{-1}\cdot \rm K^{-1}$ .

Specific heat of copper: $0.39\; \rm J \cdot g^{-1}\cdot K^{-1}$ .

The temperature of this calorimeter and the $250\; \rm g$ of water that it initially contains increased from $20\; \rm ^\circ C$ to $50\; \rm ^\circ C$ . Calculate the amount of energy that would be absorbed:

$\begin{aligned}& Q(\text{copper}) \\ =\;& c \cdot m \cdot \Delta t \\ =\;& 0.39\; \rm J \cdot g^{-1}\cdot K^{-1} \times 50\; \rm g \times (50\;{\rm ^\circ C} - 20\;{\rm ^\circ C}) \\ =\; & 585\; \rm J \end{aligned}$ .

$\begin{aligned}& Q(\text{cool water}) \\ =\;& c \cdot m \cdot \Delta t \\ =\;& 4.2\; \rm J \cdot g^{-1}\cdot K^{-1} \times 250\; \rm g \times (50\;{\rm ^\circ C} - 20\;{\rm ^\circ C}) \\ =\; & 31500\; \rm J \end{aligned}$ .

Hence, it would take an extra $585\; \rm J + 31500\; \rm J = 32085\; \rm J$ of energy to increase the temperature of the calorimeter and the $250\; \rm g$ of water that it initially contains from $20\; \rm ^\circ C$ to $50\; \rm ^\circ C$ .

Assume that it would take $x$ grams of steam at $100\; \rm ^\circ C$ ensure that the equilibrium temperature of the system is $50\; \rm ^\circ C$ .

In other words, $x\; \rm g$ of steam at $100\; \rm ^\circ C$ would need to release $32085\; \rm J$ as it condenses (releases latent heat) and cools down to $50\; \rm ^\circ C$ .

Latent heat of condensation from $x\; \rm g$ of steam: $2260\; {\rm J \cdot g^{-1}} \times (x\; {\rm g}) = (2260\, x)\; \rm J$ .

Energy released when that $x\; {\rm g}$ of water from the steam cools down from $100\; \rm ^\circ C$ to $50\; \rm ^\circ C$ :

$\begin{aligned}Q = \;& c \cdot m \cdot \Delta t \\ =\;& 4.2\; {\rm J \cdot g^{-1}\cdot K^{-1}} \times (x\; \rm g) \times (100\;{\rm ^\circ C} - 50\;{\rm ^\circ C}) \\ =\; & (210\, x)\; \rm J \end{aligned}$ .

These two parts of energy should add up to $32085\; \rm J$ . That would be exactly what it would take to raise the temperature of the calorimeter and the water that it initially contains from $20\; \rm ^\circ C$ to $50\; \rm ^\circ C$ .

$(2260\, x)\; {\rm J} + (210\, x)\; {\rm J} = 32085\; \rm J$ .

Solve for $x$ :

$x \approx 13$ .

Hence, it would take approximately $13\; \rm g$ of steam at $100\; \rm ^\circ C$ for the equilibrium temperature of the system to be $50\; \rm ^\circ C$ .

4 0

3 years ago

What is a circumpolar star? A)a star that is close to the north celestial pole a star that is close to the south celestial pole

Vedmedyk [2.9K]

Answer:

A star that always remains above your horizon and appears to rotate around the celestial pole.

Explanation:

A) a star that is close to the north celestial pole: a circumpolar star could be close to the north celestial pole, but this answer is omitting the south celestial pole.

B) a star that is close to the south celestial pole: a circumpolar star could be close to the south celestial pole, but this answer is omitting the north celestial pole.

C) a star that always remains above your horizon and appears to rotate around the celestial pole: this is the definition of a circumpolar star.

D) a star that makes a daily circle around the celestial sphere: every star does this.

E) a star that is visible from the Arctic or Antarctic circles : there are many starts visible from there that are not circumpolar.

6 0

4 years ago

A projectile is launched at an angle of 29 degrees above the horizontal with an initial velocity of 36.6 at an unknown height.

alex41 [277]

The magnitude of the unknown height of the projectile is determined as 16.1 m.

<h3>Magnitude of the height</h3>

The magnitude of the height of the projectile is calculated as follows;

H = u²sin²θ/2g

H = (36.6² x (sin 29)²)/(2 x 9.8)

H = 16.1 m

Thus, the magnitude of the unknown height of the projectile is determined as 16.1 m.

Learn more about height here: brainly.com/question/1739912

#SPJ1

3 0

2 years ago

Over millions of years weathered rock soil dead plant and animal remains are pressed and cemented together under the ground form

Eduardwww [97]

Answer:

true

Explanation:

The statement being made is completely true. This layer of rock is called a Sedimentary Rock level and is slowly formed over millions of years with minerals and organic remains from the bottom of the Oceans that may no longer be covered in water anymore. Since it is made up of all these minerals and remains, it is studied widely by Geologists and Archeologists to better understand the Earth's past.

5 0

3 years ago

WILL MARK BRAINLIST

ryzh [129]

Answer:

Definition of envelop

transitive verb

1: to enclose or enfold completely with or as if with a covering

2: to mount an attack on (an enemy's flank)

3 0

2 years ago