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Naddik [55]
3 years ago
9

What shoreline feature would develop if the global sea level were to rise considerably

Physics
1 answer:
valentinak56 [21]3 years ago
5 0
Wat does that mean I don't understand this
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What is the name of the plate to the west and directly adjacent to the plate on which Chile is located?
Ksenya-84 [330]
<h2>Answer: Nazca plate</h2>

Explanation:

The Nazca plate is an oceanic tectonic plate that is found in the Pacific Ocean off the west coast of South America, specifically in front of the north and central coast of Chile and the entire coastline of Peru, Ecuador and Colombia. This plate is in constant motion, which causes it to sink under the South American (phenomenon known as subduction).

It is mainly because of this subduction process that this region has a lot of seismic activity. Another important aspect is that thanks to these movements, the Andes mountain range and the Peruvian-Chilean fossa originated.

6 0
3 years ago
How does mass affect weight?​
devlian [24]
Answer: Mass affects the weight of an object with the effects of gravity.

Weight is the measure of the force of gravity on an object's mass, while mass is the measure of how much matter there is in an object.
5 0
3 years ago
The Moon and Earth rotate about their common center of mass, which is located about RcM 4700 km from the center of Earth. (This
erica [24]

To solve this problem it is necessary to apply the concepts related to gravity as an expression of a celestial body, as well as the use of concepts such as centripetal acceleration, angular velocity and period.

PART A) The expression to find the acceleration of the earth due to the gravity of another celestial body as the Moon is given by the equation

g = \frac{GM}{(d-R_{CM})^2}

Where,

G = Gravitational Universal Constant

d = Distance

M = Mass

R_{CM} = Radius earth center of mass

PART B) Using the same expression previously defined we can find the acceleration of the moon on the earth like this,

g = \frac{GM}{(d-R_{CM})^2}

g = \frac{(6.67*10^{-11})(7.35*10^{22})}{(3.84*10^8-4700*10^3)^2}

g = 3.4*10^{-5}m/s^2

PART C) Centripetal acceleration can be found throughout the period and angular velocity, that is

\omega = \frac{2\pi}{T}

At the same time we have that centripetal acceleration is given as

a_c = \omega^2 r

Replacing

a_c = (\frac{2\pi}{T})^2 r

a_c = (\frac{2\pi}{26.3d(\frac{86400s}{1days})})^2 (4700*10^3m)

a_c = 3.34*10^{-5}m/s^2

3 0
3 years ago
Which of the following tend to react by gaining 1 electron
Andrei [34K]
The correct answer should be C
8 0
3 years ago
Read 2 more answers
A 4-kg toy car with a speed of 5 m/s collides head-on with a stationary 1-kg car. After the collision, the cars are locked toget
mihalych1998 [28]

Kinetic energy lost in collision is 10 J.

<u>Explanation:</u>

Given,

Mass, m_{1} = 4 kg

Speed, v_{1} = 5 m/s

m_{2} = 1 kg

v_{2} = 0

Speed after collision = 4 m/s

Kinetic energy lost, K×E = ?

During collision, momentum is conserved.

Before collision, the kinetic energy is

\frac{1}{2} m1 (v1)^2 + \frac{1}{2} m2(v2)^2

By plugging in the values we get,

KE = \frac{1}{2} * 4 * (5)^2 + \frac{1}{2} * 1 * (0)^2\\\\KE = \frac{1}{2} * 4 * 25 + 0\\\\

K×E = 50 J

Therefore, kinetic energy before collision is 50 J

Kinetic energy after collision:

KE = \frac{1}{2} (4 + 1) * (4)^2 + KE(lost)

KE = 40J + KE(lost)

Since,

Initial Kinetic energy = Final kinetic energy

50 J = 40 J + K×E(lost)

K×E(lost) = 50 J - 40 J

K×E(lost) = 10 J

Therefore, kinetic energy lost in collision is 10 J.

4 0
3 years ago
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