What happens if two small positively charged particles of equal magnitude are placed close to each other? A) The particles will experience a repulsive force but remain stationary. B) The particles will experience an attractive force causing them to meet. C) The particles will experience a repulsive force that will drive them apart along the line joining their centers. D) The particles will experience a force that will cause them to move in a circular path of uniform motion with fixed distance of separation.

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

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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.

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The best metric is a combination of volume, frequency, soil moisture, evapotranspiration rate, and plant growth and health observation.

When measuring the amount of water given to a plant, it's crucial to choose a metric that accurately reflects the plant's needs and ensures optimal growth. The best metric to use for this purpose is:

1. **Volume of Water**: Measure water in milliliters or liters to quantify the actual amount provided to the plant.

2. **Frequency**: Track how often water is given, whether daily, weekly, or based on the plant's specific requirements.

3. **Moisture Content**: Use a soil moisture meter to gauge the water content in the soil around the plant's roots.

4. **Evapotranspiration Rate**: Consider the rate at which water evaporates from the soil and is transpired by the plant. This can help determine the water loss and replenishment needed.

5. **Plant Growth and Health**: Ultimately, the best measure is the plant's response. Monitor its growth, foliage appearance, and overall health to adjust watering practices accordingly.

By combining these metrics, you can ensure the plant receives the appropriate amount of water for optimal growth and health.

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:

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]

Makori can correct the error in his diagram by changing the label “Virtual image from objective lens” to “Real image from objective lens.”

Convex lens is used in light microscope.

Therefore, Makori can correct the error in his diagram by changing the label “Virtual image from objective lens” to “Real image from objective lens.”

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Answer: the wasp’s stinger - protection against predators

the peacock’s bright plumage - attraction to obtain a mate

the owl’s large eyes - help to search for food

the bear’s fur - maintain homeostasis

Explanation:

An adaptation can be define as any change that occurs in the morphology, physiology and genetic make up of the organism which makes it suitable for survival in it's natural environment even in adverse conditions and reproduce.

The following are the adaptations with their benefits.

the wasp’s stinger - protection against predators

The wasp stinger generally exhibit toxic substance or allergy causing agent which the wasp use when it finds a predator around it.

the peacock’s bright plumage - attraction to obtain a mate

The male peacock exhibit bright and showy feathers which is a phenotypic advantage for attraction of female mates for copulation. This phenomena is called as sexual selection.

the owl’s large eyes - help to search for food

The owl's large eyes allows the night vision. There are many rods and cones present in the eyes of the owl so as to allow better vision in night and allow it to observe and capture prey.

the bear’s fur - maintain homeostasis

The bear's fur are insulating in nature that means it does not allow the entry of external air. Hence, helps in retaining the heat in the body in otherwise cold environment. Therefore, helps in maintaining the homeostasis (the tendency of the body retain it's constant internal environment as compared to changing external environment).

Answer:

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

ANSWER:

C) Gravity & Inertia.

Choices

A) gravity only

B) inertia and air resistance

C) gravity and inertia

D) gravity and air resistance

Final answer:

The most suitable sensor for analyzing signals from outer space is radio wave sensors due to their economical production cost and efficiency in transmitting through interstellar space and atmospheres, thus providing valuable information about the sources of the waves.

Explanation:

In analyzing signals from outer space, the best sensor to use among sound waves, radio waves, and seismic waves would be radio wave sensors. Radio waves are the lowest-frequency and lowest-energy band of the electromagnetic spectrum, making them not only inexpensive to produce but also adept at traveling through interstellar dust, gas, and planetary atmospheres without significant absorption.

This capability is crucial for receiving unimpeded signals from space. The information collected through radio waves can reveal the chemistry and physical conditions of their sources. Moreover, radio waves are already extensively used in various communication systems on Earth. Therefore, in the context of SETI (Search for Extraterrestrial Intelligence) or any other form of cosmic study, radio waves are the most suitable and effective for gathering data from outer space.

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.