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valina [46]
3 years ago
10

Enzo investigated the effect of different types of fertilizers on the growth of bean plants. He hypothesized that fertilizer B w

ould result in the greatest amount of growth. However, the results showed that the control plants without fertilizer had the greatest growth. Why was Enzo's data still valuable even though it did not support his hypothesis?
Physics
1 answer:
Mama L [17]3 years ago
4 0

Answer:

The answer is "In this information, Enzo will be confident that if any fertilizer is required only for plants".

Explanation:

The Beans plants were members of the group of legumes. Legume flowering plant under which the roots form nodules. These nodules require a specific bacterium called bacteria, which fix nitrogen. Its bacterium transforms nitrogen from the atmosphere into the soil's nitrate. Consequently, for some of its growth, beans do not require fertilizer.

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A groups beliefs, values, and patterns of behavior as known as its
Nady [450]

I believe the answer is Nonmaterial Culture.

8 0
3 years ago
4. The flow of electric charge is know as
vredina [299]

Answer:

electrons

Explanation:

An electric current is said to exist when there is a net flow of electric charge through a region. In electric circuits this charge is often carried by electrons moving through a wire. It can also be carried by ions in an electrolyte, or by both ions and electrons such as in an ionized gas (plasma).

5 0
3 years ago
a person with a mass of 75kg jumps on the trampoline the trampoline creates a fprce of 375n on them what is the acceleration of
Anarel [89]

Newton's second law allows calculating the response for the person's acceleration while leaving the trampoline is:

            -4.8 m / s²

Newton's second law says that the net force is proportional to the product of the mass and the acceleration of the body

            F = m a

Where the bold letters indicate vectors, F is the force, m the masses and the acceleration

The free body diagram is a diagram of the forces without the details of the body, in the attached we can see the free body diagram for this system

 

               F_t -W = m a

Whera F_t is the trampoline force

Body weight is

                W = mg

We substitute

              F_t - mg = ma

              a =\frac{F_t - m g}{m}

Let's calculate

              a = \frac{375 - 75 \ 9.8 }{75}

              a = -4.8 m / s²

The negative sign indicates that the acceleration is directed downward.

In conclusion using Newton's second law we can calculate the acceleration of the person while leaving the trampoline is

            -4.8 m / s²

Learn more here:  brainly.com/question/19860811

7 0
3 years ago
The half-life of Iodine-131 is 8.0252 days. If 14.2 grams of I-131 is released in Japan and takes 31.8 days to travel across the
MakcuM [25]

Answer:

Explanation:

Half-life problems are modeled as exponential equations.  The half-life formula is P=P_o\left (\dfrac{1}{2} \right)^{\frac{t}{k}} where P_o is the initial amount, k is the length of the half-life, t is the amount of time that has elapsed since the initial measurement was taken, and P is the amount that remains at time t.

P=14.2\left (\dfrac{1}{2} \right)^{\frac{t}{8.0252}}

<u>Deriving the half-life formula</u>

If one forgets the half-life formula, one can derive an equivalent equation by recalling the basic an exponential equation, y=a b^{t}, where t is still the amount of time, and y is the amount remaining at time t.  The constants a and b can be solved for as follows:

Knowing that amount initially is 14.2g, we let this be time zero:

y=a b^{t}

(14.2)=ab^{(0)}

14.2=a *1

14.2=a

So, a=14.2, which represents out initial amount of the substance, and our equation becomes: y=14.2 b^{t}

Knowing that the "half-life" is 8.0252 days (note that the unit here is "days", so times for all future uses of this equation must be in "days"), we know that the amount remaining after that time will be one-half of what we started with:

\left(\frac{1}{2} *14.2 \right)=14.2 b^{(8.0252)}

\dfrac{7.1}{14.2}=\dfrac{14.2 b^{8.0252}}{14.2}

0.5=b^{8.0252}

\sqrt[8.0252]{\frac{1}{2}}=\sqrt[8.0252]{b^{8.0252}}

\sqrt[8.0252]{\frac{1}{2}}=b

Recalling exponent properties, one could find that  \left ( \frac{1}{2} \right )^{\frac{1}{8.0252}}=b, which will give the equation identical to the half-life formula.  However, recalling this trivia about exponent properties is not necessary to solve this problem.  One can just evaluate the radical in a calculator:

b=0.9172535661...

Using this decimal approximation has advantages (don't have to remember the half-life formula & don't have to remember as many exponent properties), but one minor disadvantage (need to keep more decimal places to reduce rounding error).

So, our general equation derived from the basic exponential function is:

y=14.2* (0.9172535661)^t  or y=14.2*(0.5)^{\frac{t}{8.0252}} where y represents the amount remaining at time t.

<u>Solving for the amount remaining</u>

With the equation set up, substitute the amount of time it takes to cross the Pacific to solve for the amount remaining:

y=14.2* (0.9172535661)^{(31.8)}          y=14.2*(0.5)^{\frac{(31.8)}{8.0252}}

y=14.2* 0.0641450581                    y=14.2*(0.5)^{3.962518068}

y=0.9108598257                              y=14.2* 0.0641450581

                                                        y=0.9108598257

Since both the initial amount of Iodine, and the amount of time were given to 3 significant figures, the amount remaining after 31.8days is 0.911g.

8 0
1 year ago
You need to focus a 10 mW, 632.8 nm Gaussian laser beam that is 5.0 mm in diameter into a sample. You have access to a lens with
Anna [14]

Answer:

ee that the lens with the shortest focal length has a smaller object

           

Explanation:

For this exercise we use the constructor equation or Gaussian equation

        \frac{1}{f}  = \frac{1}{p} + \frac{1}{q}

where f is the focal length, p and q are the distance to the object and the image respectively.

Magnification a lens system is

          m = \frac{h'}{h} = - \frac{q}{p}

             h ’= -\frac{h q}{p}

In the exercise give the value of the height of the object h = 0.50cm and the position of the object p =∞

Let's calculate the distance to the image for each lens

f = 6.0 cm

           \frac{1}{q} = \frac{1}{f }  - \frac{1}{p}

as they indicate that the light fills the entire lens, this indicates that the object is at infinity, remember that the light of the laser rays is almost parallel, therefore p = inf

          q = f = 6.0 cm

for the lens of f = 12.0 cm q = 12.0 cn

to find the size of the image we use

           h ’= h q / p

where p has a high value and is the same for all systems

           h ’= h / p q

Thus

f = 6 cm h ’= fo 6 cm

 

f = 12 cm h ’= fo 12  cm

therefore we see that the lens with the shortest focal length has a smaller object

8 0
3 years ago
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