All categories

3 years ago

How can the process of climate change best be described?

Physics

2 answers:

uysha [10]3 years ago

6 0

Answer:

The correct answer is Option C, natural and continuous

Explanation:

The variation in climate is predominantly a natural activity which is governed by various natural processes/cycles/activities/components such as blowing of wind – its direction and speed, the solar radiation, the rotation and revolution of earth, the behavior of atmosphere , various biogeochemical cycles, varying heat content in the earth’s atmosphere etc.

All these processes have been occurring continuously since the origin of planet earth and occur till today in the same manner and on the same time scale, thus, climatic variation can be considered as a natural and continuous process.

seropon [69]3 years ago

3 0

It is either A or C. i hope i could help. if not im really sorry

You might be interested in

Can you answer this math homework? Please!

steposvetlana [31]

Using the count data and observational data you acquired, calculate the number of CFUs in the original sample

5 0

3 years ago

You launch a cannonball at an angle of 35° and an initial velocity of 36 m/s (assume y = y₁=

velikii [3]

Answer:

Approximately $4.2\; {\rm s}$ (assuming that the projectile was launched at angle of $35^{\circ}$ above the horizon.)

Explanation:

Initial vertical component of velocity:

$\begin{aligned}v_{y} &= v\, \sin(35^{\circ}) \\ &= (36\; {\rm m\cdot s^{-1}})\, (\sin(35^{\circ})) \\ &\approx 20.6\; {\rm m\cdot s^{-1}}\end{aligned}$ .

The question assumed that there is no drag on this projectile. Additionally, the altitude of this projectile just before landing $y_{1}$ is the same as the altitude $y_{0}$ at which this projectile was launched: $y_{0} = y_{1}$ .

Hence, the initial vertical velocity of this projectile would be the exact opposite of the vertical velocity of this projectile right before landing. Since the initial vertical velocity is $20.6\; {\rm m\cdot s^{-1}}$ (upwards,) the vertical velocity right before landing would be $(-20.6\; {\rm m\cdot s^{-1}})$ (downwards.) The change in vertical velocity is:

$\begin{aligned}\Delta v_{y} &= (-20.6\; {\rm m\cdot s^{-1}}) - (20.6\; {\rm m\cdot s^{-1}}) \\ &= -41.2\; {\rm m\cdot s^{-1}}\end{aligned}$ .

Since there is no drag on this projectile, the vertical acceleration of this projectile would be $g$ . In other words, $a = g = -9.81\; {\rm m\cdot s^{-2}}$ .

Hence, the time it takes to achieve a (vertical) velocity change of $\Delta v_{y}$ would be:

$\begin{aligned} t &= \frac{\Delta v_{y}}{a_{y}} \\ &= \frac{-41.2\; {\rm m\cdot s^{-1}}}{-9.81\; {\rm m\cdot s^{-2}}} \\ &\approx 4.2\; {\rm s} \end{aligned}$ .

Hence, this projectile would be in the air for approximately $4.2\; {\rm s}$ .

8 0

2 years ago

Read 2 more answers

Why does wave height increase in shallow water??

Vitek1552 [10]

As the water russhes toward the shore, it rises because it is pushing against it.<span />

3 0

3 years ago

A 24-V battery is powering a light bulb with a resistance of 3.0 ohms. What is the current flowing through the bulb? A) 7.20 A B

Usimov [2.4K]

According to Ohm's law for a portion of the circuit we have:

U=RI=>I=U/R=24/3=8 A

The correct answer is B

3 0

3 years ago

When a golf ball is hit, it travels at 41 meters per second. The mass of a golf ball is 0.045kg. Calculate the kinetic energy of

Fittoniya [83]

Answer:

75.645 J

Explanation:

The kinetic energy is related to the mass and velocity by the formula ...

KE = 1/2mv²

For the given mass of 0.045 kg, and velocity of 41 m/s, the kinetic energy is ...

KE = 1/2(0.045 kg)(41 m/s)² = 75.645 J

The unit of energy, joule, is a derived unit equal to 1 kg·m²/s².

4 0

3 years ago