explain why electromagnetic radiation sometimes act as a wave and sometimes as a stream of particles?

To find the initial speed of a ball thrown horizontally from a 45m tall building, we first calculate the time of flight using the vertical motion equation and then determine the initial horizontal speed using the range and time of flight.

Finding the Initial Speed of a Horizontally Thrown Ball

To calculate the initial speed of a ball thrown horizontally from the roof of a building that is 45 meters tall and lands 24 meters from the base, we can use the kinematic equations of motion for projectile motion.

Step 1 - Time of Flight

To find the time it takes for the ball to hit the ground, we use the vertical motion equation:

 s = ut + 0.5*at^2

Since the ball is thrown horizontally, the initial vertical velocity (u) is 0 m/s. s is the height of the building (45 m) and a is the acceleration due to gravity (9.8 m/s2).

By substituting the known values, we can solve for t.

Step 2 - initial Horizontal Speed

Now, knowing the time of flight, we can find the initial horizontal speed(vx) using the horizontal motion equation:

 Range = vxt

The range is given as 24 m. Therefore, vx = Range / t.

Final Answer

After calculating, we get the initial horizontal speed of the ball, which is necessary to cover the horizontal distance within the derived time of flight.

Answer:

Explanation: A coil is an electric circuit that produces a magnetic field

Answer:

twice as much

Franklin must not drive through a flood for there may no road at all under the water, unless he is familiar with the road and the flowing water is below one foot. 

Moreover, if negotiating a flooded section of road, he must drive in the middle where the water will be at its shallowest and he must not drive through water against approaching vehicle to consider other drivers.

The electric potential of a 2.2 µC charge at a distance of 6.3 m is 3.14 × 10^6 V. At a distance of 99 m, the electric potential is 2.02 × 10^4 V.

To calculate the electric potential, we can use the equation V = kQ/r, where V is the electric potential, k is Coulomb's constant (8.99 × 10^9 N • V^-2 • m^2 • C^-2), Q is the charge, and r is the distance from the charge.

For a charge of 2.2 µC at a distance of 6.3 m, we can plug in the values to find:

V = (8.99 × 10^9 N • V^-2 • m^2 • C^-2)(2.2 × 10^-6 C)/(6.3 m) = 3.14 × 10^6 V

Similarly, for a distance of 99 m, the calculation would be:

V = (8.99 × 10^9 N • V^-2 • m^2 • C^-2)(2.2 × 10^-6 C)/(99 m) = 2.02 × 10^4 V

We have that Non examples of Asexual reproduction will be the reproduction that requires two cells to reproduce.

Examples

From the question we are told

What are non examples of Asexual reproduction

Generally

Therefore

non examples of Asexual reproduction will be the reproduction that requires two cells to reproduce.

Examples

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Answer:

500 Kg.

Explanation:

The problem is giving us the weight of the object the weight is different from the mass of an object

Weight is the mass of any object multiplied by gravity (the value of gravity is approximately  [tex]9.8\frac{m}{s^2}[/tex] ).

The unit of weight are newtons that equals

[tex]N= Kg.\frac{m}{s^2}[/tex]

[tex]W= m.g[/tex]

we cleared the mass of the formula

[tex]W=m.g\\ m=\frac{W}{g} \\ m= \frac{4900N}{9.8 \frac{m}{s^2} }= 500 Kg[/tex]

this is a true statement

True

Batteries carry acids and electrolytes which if comes in contact with human skin can cause severe effects and sometimes it can become fatal. Therefore a great care is needed while replacing batteries. To avoid any danger and harmful circumstances, batteries should always be replaced by a licenced battery haulers for recycling.

After 16 days, there will be approximately 0.24 grams of iodine-131 remaining in the sample.

The half-life of iodine-131 is 8 days, which means that every 8 days, the amount of iodine-131 in a sample will be reduced by half.

After 16 days, which is two half-lives, the amount of iodine-131 remaining can be calculated as follows:

First half-life: 0.96 g / 2 = 0.48 g remaining

Second half-life: 0.48 g / 2 = 0.24 g remaining

Therefore, after 16 days, there will be approximately 0.24 grams of iodine-131 remaining in the sample.

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The problems with solar energy include its intermittency and the high upfront cost of battery storage systems. The production of photovoltaic cells generates hazardous waste, and deployment of large-scale solar systems can harm undeveloped habitats. Also, solar energy cannot directly power all sectors like cars and airplanes due to its diffuse nature.

There are a number of challenges and problems associated with the use of solar energy. Some primary issues include its intermittency, as solar energy capture is not constant, and can be particularly low during the nighttime or in seasons with less sunshine. This inconsistency can be offset by battery storage, but such storage solutions add substantial cost, need regular maintenance, and require the environmentally unfriendly process of battery replacement.

Another concern is the environmental impact of photovoltaic (PV) cells, the main technology for generating solar energy. The process of manufacturing PV cells requires the use of chemicals and solvents, generating some hazardous waste. Besides, the deployment of large-scale solar arrays in undeveloped areas or fragile ecosystems like deserts can cause habitat destruction. Solar thermal systems, another form of solar energy, also pose problems as they use potentially hazardous fluids for heat transfer and need significant amounts of water for cleaning and cooling.

Finally, solar energy can't directly power some sectors and applications such as cars and airplanes due to its diffuse nature, making it not entirely substitutable with other energy sources. While the sun provides an abundant source of energy, transitioning to a solar-powered society poses significant challenges.

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