All categories

3 years ago

I DONT CARE WHAT SUBJECT THIS IS ON BUT PLS HELP THIS IS SCIENCE plz help

Physics

2 answers:

abruzzese [7]3 years ago

6 0

Answer:

The northern end of the Earth's axis is tilted towards the sun

Dima020 [189]3 years ago

4 0

Answer:

5)Plants use carbon dioxide to create food

7)Curved path

You might be interested in

A solid uniform cylinder is rolling without slipping. What fraction of its kinetic energy is rotational?

tester [92]

Answer:

Explanation:

Let m be the mass of cylinder and r be the radius. It is moving with velocity v and angular velocity is ω. Let I be the moment of inertia of the cylinder.

I = 0.5 mr²

Total kinetic energy, T = 0.5 mv² + 0.5 Iω²

T = 0.5 (mv² + 0.5 mr²ω²)

v = rω

So, T = 0.5 (mv² + 0.5 mv²) = 0.75 mv²

Rotational kinetic energy is

R = 0.5 Iω² = 0.5 x 0.5 mr²ω²

R = 0.25 mv²

So, R / T = 0.25 / 0.75 = 1/3

5 0

3 years ago

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

How do gases respond to changes in pressure and temperature?

kakasveta [241]

They compress or expand depending on amount of pressure or depending on the temperature

6 0

3 years ago

Read 2 more answers

Which of the following rocks would most likely be found on the ocean floor

luda_lava [24]

Is there a graph or a picture

7 0

4 years ago

Read 2 more answers

Imagine that Earth stops orbiting the Sun but continues to rotate in place about its own axis at its current rate. In this case,

otez555 [7]

Answer:

The length of the solar day will get shorter.

Explanation:

The blue planet Earth not only rotates around it's own axis but also rotates around the Sun and everyday it moves a little bit around the axis.
Since the speed of the Earth's rotation on it's own axis and around the Sun is constant we don't feel the effects of the rotation.We can only feel the motion if the earth changes it's rotation speed.
If by any means or chance the Earth stopped spinning (stopped rotation) then the atmosphere surrounding the Earth would be in motion and all the Earth's land would be scoured clean.

7 0

3 years ago