Built an atom of Krypton which has an atomic number of 36?

Nepal has high potential for producing hydroelectricity however it is difficult too.

Stels [109]

Answer:

I'm not a scholar of hydroelectric power in Nepal, so consider my answers carefully, below.

Explanation:

High Potential: Hydroelectric power comes from the potential energy stored in a mass that is above Earth's surface. As the word "hydro" implies, the mass in this case is water. Water from snow and glacier melt, and from normal precipitation (rain) in mountainous regions eventually cascades down the mountains in fast-flowing rivers or waterfalls. Often, there are lakes or man-made reservoirs to collect and store the water before it flows down. Mt. Everest is 8848 meters tall (about 29,000 feet). If a lake forms at just 2,000 meters, one can calculate the amount of energy in each kilogram of water stored in the lake that represents the potential energy available at that altitude. 1 kg of water at 2,000 meters has potential energy, PE, according to the equation: PE = mgh, where m is the mass in kg, g is Earth's acceleration due to gravity (9.8 m/sec^2), and h is height, in meters.

PE = mGH

PE = (1 kg)*(9.8 m/sec^2)*(2,000 meters) = 19,000 kg*m/sec^2

1 kg*m/sec^2 is the SI unit for 1 Joule, a measure of energy.

This potential energy can be converted into electrical energy by releasing the water so that it can flow down to a water-powered turbine that spins magnets and coils of wire that produce electricity. The 19,000 Joules of water potential energy can be converted to electrical power, less any inefficiency in the system, such as friction.

Nepal has the natural advantage in that it has many high mountain ranges with water flows that can be used for generating electrical power. The result is low operating costs (the fuel is the flowing water) and no greenhouse gas emissions

The difficulty in developing hydroelectric power in Nepal is due to the same factor that gives it an advantage: it is difficult constructing large hydroelectric plants in such rough terrain, and the power lines that are needed to transport the power to its destination are expensive and difficult to maintain and repair.

8 0

2 years ago

an 269 kg object is moved a distance of 1.9 m by a force if 580 j of work is done on the object what is the object acceleration

IrinaK [193]

Answer:

Explanation:

w=f*d=580*1.5=870J

7 0

2 years ago

Define 1 pascal pressure

Rasek [7]

Answer:For example, standard atmospheric pressure (or 1 atm) is defined as 101.325 kPa. The millibar, a unit of air pressure often used in meteorology, is equal to 100 Pa. (For comparison, one pound per square inch equals 6.895 kPa.)

Explanation:A pascal is a pressure of one newton per square metre, or, in SI base units, one kilogram per metre per second squared.

I hope this helps.... I'm sorry if it doesn't

5 0

2 years ago

Strontium 3890Sr has a half-life of 28.5 yr. It is chemically similar to calcium, enters the body through the food chain, and co

patriot [66]

Answer:

Thus the time taken is calculated as 387.69 years

Solution:

As per the question:

Half life of $^{3890}Sr\, t_{\frac{1}{2}}$ = 28.5 yrs

Now,

To calculate the time, t in which the 99.99% of the release in the reactor:

By using the formula:

$\frac{N}{N_{o}} = (\frac{1}{2})^{\frac{t}{t_{\frac{1}{2}}}}$

where

N = No. of nuclei left after time t

$N_{o}$ = No. of nuclei initially started with

$\frac{N}{N_{o}} = 1\times 10^{- 4}$

(Since, 100% - 99.99% = 0.01%)

Thus

$1\times 10^{- 4} = (\frac{1}{2})^{\frac{t}{28.5}}}$

Taking log on both the sides:

$- 4 = \frac{t}{28.5}log\frac{1}{2}$

$t = \frac{-4\times 28.5}{log\frac{1}{2}}$

t = 387.69 yrs

5 0

3 years ago

A(n) 55.5 g ball is dropped from a height of 53.6 cm above a spring of negligible mass. The ball compresses the spring to a maxi

Serggg [28]

Answer:

The spring force constant is $k=243\ \frac{N}{m}$ .

Explanation:

We are told the mass of the ball is $m=0.0555\ kg$ , the height above the spring where the ball is dropped is $h=0.536\ m$ , the length the ball compresses the spring is $d=0.04897\ m$ and the acceleration of gravity is $9.8\ \frac{m}{s^{2}}$ .

We will consider the initial moment to be when the ball is dropped and the final moment to be when the ball stops, compressing the spring. We supose that there is no friction so the initial mechanical energy $E_{mi}$ is equal to the final mechanical energy $E_{mf}$ :

$E_{mf}=E_{mi}$

Initially there is only gravitational potential energy because the force of the spring isn't present and the speed is zero. In the final moment there is only elastic potential energy because the height is zero and the ball has stopped. So we have that:

$\frac{1}{2}kd^{2}=mgh$

If we manipulate the equation we have that:

$k=\frac{2mgh}{d^{2} }$

$k=\frac{2\ 0.0555\ kg\ 9.8\frac{m}{s^{2}}\ 0.536\ m}{(0.04897)^{2}m^{2}}$

$k=\frac{0.58306\ \frac{kgm^{2}}{s^{2}}}{2.398x10^{-3}m^{2}}$

$k=243\ \frac{N}{m}$

5 0

3 years ago