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Lostsunrise [7]
3 years ago
7

What are some types of landforms on Earth’s surface? PLS ANSWER QUICK 11 POINTS

Physics
2 answers:
Brrunno [24]3 years ago
5 0

Answer:

plateau, mountains, hills, plains

DochEvi [55]3 years ago
4 0
Plateau, butte, canyon, valley, mountain, hill
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The sun is 1.50x10^11 m from earth. How long does it take the suns light to reach earth? How long
galina1969 [7]

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3 years ago
1. What are three examples of how invasive species spread?
Darya [45]

Answer:

Their primary way of spreading is from human activities, they can quickly travel around the world for example these new "murder" hornets that can kill a large bee hive with one sting, they traveled all the way from Asia. Invasive species can also be through people's luggage, small boats, planes and large shipment like cargo carriers. I hope this helps. :  )

Explanation:

8 0
3 years ago
(5 pt) You tie a cord to a pail of water, and your swing the pail in a vertical circular 0.700 m. What is the minimum speed must
Bas_tet [7]

Answer:

The minimum speed required  is 2.62m/s

Explanation:

The value of  gravitational acceleration = g = 9.81 m/s^2

Radius of the vertical circle = R = 0.7 m

Given the mass of the pail of water = m

The speed at the highest point of the circle = V

The centripetal force will be needed must be more than the weight of the pail of water in order to not spill water.

Below is the calculation:

\frac{mV^{2}}{R} = mg

V = \sqrt{gR}

V = \sqrt{9.81 \times 0.7}

V = 2.62 m/s

6 0
3 years ago
A parallel-plate capacitor has plates with an area of 451 cm2 and an air-filled gap between the plates that is 2.51 mm thick. Th
Nostrana [21]

To solve this problem we will apply the concepts related to Energy defined in the capacitors, as well as the capacitance and load. From these three definitions we will build the solution to the problem by defending the energy with the initial conditions, the energy under new conditions and finally the change in the work done to move from one point to the other.

Energy in a capacitor can be defined as

E = \frac{1}{2}CV^2 = \frac{1}{2}\frac{Q^2}{C}

Here,

V = Potential difference across the capacitor plates

Q = Charge stored on the capacitor plates

At the same time capacitance can be defined as,

C = \epsilon_0 (\frac{A}{d})

Here,

\epsilon_0 =  Vacuum permittivity constant

A = Area

d = Distance

Replacing with our values we have that,

C = (8.85*10^{-12})(\frac{0.0451}{2.51*10^{-3}})

C = 1.5901*10^{-10}F

PART A) Energy stored in the capacitor is

E = \frac{1}{2} CV^2

E = \frac{1}{2} (1.5901*10^{-10})(575)^2

E = 2.628*10^{-5}J

PART B) We know first that everything that the load can be defined as the product between voltage and capacitance, therefore

Q = CV

Q = (1.59*10^{-10})(575)

Q = 9.1425*10^{-8}C

Now if d = 10.04*10^{-3}m we have that the capacitance is

C = \epsilon_0 (\frac{A}{d})

C = (8.85*10^{-12})(\frac{0.0451}{10.04*10^{-3}})

C = 3.9754*10^{-11}F

Then the energy stored is

E = \frac{1}{2} \frac{Q^2}{C}

E = \frac{1}{2} (\frac{(9.1425*10^{-8})^2}{3.9754*10^{-11}})

E = 1.051*10^{-4} J

PART C) The amount of work or energy required to carry out this process is the difference between the energies obtained, therefore

W = 1.051*10^{-4} J -2.628*10^{-5}J

W = 7.882*10^{-5} J

6 0
3 years ago
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