During most of a cell’s life cycle, DNA is found in a mass of loose strands called?

Answer:

twice as much

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).

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.

Answer:

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

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

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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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]

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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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.

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.