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3 years ago

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From this information, Jasmine concludes that "fertilizer b" is the best at increasing crop yields. Is Jasmine's conclusion scie

ntifically valid? A. Yes, because "fertilizer b" increased her crop yield. B. No, because the different sections of her land were not equal in area. C. No, because the different sections of her land have different characteristics. D. No, because the crop yields were all equal.

1 answer:

Lera25 [3.4K]3 years ago

7 0

Answer:

the answer is a

........................

hope this will maybe be helpful

You might be interested in

A thin spherical spherical shell of radius R which carried a uniform surface charge density σ. Write an expression for the volum

ozzi

Answer:

Explanation:

From the given information:

We know that the thin spherical shell is on a uniform surface which implies that both the inside and outside the charge of the sphere are equal, Then

The volume charge distribution relates to the radial direction at r = R

∴

$\rho (r) \ \alpha \ \delta (r -R)$

$\rho (r) = k \ \delta (r -R) \ \ at \ \ (r = R)$

$\rho (r) = 0\ \ since \ r< R \ \ or \ \ r>R---- (1)$

To find the constant k, we examine the total charge Q which is:

$Q = \int \rho (r) \ dV = \int \sigma \times dA$

$Q = \int \rho (r) \ dV = \sigma \times4 \pi R^2$

∴

$\int ^{2 \pi}_{0} \int ^{\pi}_{0} \int ^{R}_{0} \rho (r) r^2sin \theta \ dr \ d\theta \ d\phi = \sigma \times 4 \pi R^2$

$\int^{2 \pi}_{0} d \phi* \int ^{\pi}_{0} \ sin \theta d \theta * \int ^{R}_{0} k \delta (r -R) * r^2dr = \sigma \times 4 \pi R^2$

$(2 \pi)(2) * \int ^{R}_{0} k \delta (r -R) * r^2dr = \sigma \times 4 \pi R^2$

Thus;

$k * 4 \pi \int ^{R}_{0} \delta (r -R) * r^2dr = \sigma \times R^2$

$k * \int ^{R}_{0} \delta (r -R) r^2dr = \sigma \times R^2$

$k * R^2= \sigma \times R^2$

$k = R^2 --- (2)$

Hence, from equation (1), if k = $\sigma$

$\mathbf{\rho (r) = \delta* \delta (r -R) \ \ at \ \ (r=R)}$

$\mathbf{\rho (r) =0 \ \ at \ \ rR}$

To verify the units:

$\mathbf{\rho (r) =\sigma \ * \ \delta (r-R)}$

↓ ↓ ↓

c/m³ c/m³ × 1/m

Thus, the units are verified.

The integrated charge Q

$Q = \int \rho (r) \ dV \\ \\ Q = \int ^{2 \ \pi}_{0} \int ^{\pi}_{0} \int ^R_0 \rho (r) \ \ r^2 \ \ sin \theta \ dr \ d\theta \ d \phi \\ \\ Q = \int ^{2 \pi}_{0} \ d \phi \int ^{\pi}_{0} \ sin \theta \int ^R_{0} \rho (r) r^2 \ dr$

$Q = (2 \pi) (2) \int ^R_0 \sigma * \delta (r-R) r^2 \ dr$

$Q = 4 \pi \sigma \int ^R_0 * \delta (r-R) r^2 \ dr$

$Q = 4 \pi \sigma *R^2$ since $( \int ^{xo}_{0} (x -x_o) f(x) \ dx = f(x_o) )$

$\mathbf{Q = 4 \pi R^2 \sigma }$

6 0

3 years ago

What is uniforlmly acelerated motion and uniform motion

steposvetlana [31]

Uniformly Accelerated motion : When a body moves along a straight line and its velocity increases by equal amounts in equal interval of time then the motion is called uniformly accelerated motion. E.g.:- Motion of a freely falling body.

Uniform Motion : When a body covers equal distances in a straight line, equal intervals of time are called uniform motion. Example: a car moving at 20km/h in a straight line. When a body covers unequal distance in equal intervals of time in a straight line is called non-uniform Example : spinning wheel.

<h3>Difference About Uniformly Accelerated motion And Uniform Motion</h3>

In Uniform Velocity Motion a body will be moving with a constant /unchanging velocity moving in a particular direction and thus acceleration will be zero whereas in Uniform accelerated motion a body will move at constant acceleration and its velocity will keep changing with time at a constant/ steady rate.

<h2>IF YOU THINK MY ANSWER IS WRONG, CHANGE IT THANK YOU. </h2>

8 0

3 years ago

What forces are being used when walking a dog and how ?

natta225 [31]

There are tons of forces that balance out on your body while you walk. Subsequent physics classes will tell you about each and how they are represented. Here are a few in order of how people usually learn them.

Gravity: The earth exerts a gravitational force on each particle in your body that has mass. Overall, this can be represented as a single force that pulls directly toward the center of the earth from the point called your center of mass.

Normal Force: The contact between your feet/shoes and the ground exerts a force normal (straight out from) the ground. If you are on flat ground, this force is directly opposite the force of gravity, and in most cases will be equal to it such that you have no vertical net force.

Friction: Friction between your shoes/feet and the ground, pointing parallel to the ground and in the direction of your walking motion creates the force necessary for you to move. The microscopic peaks and valleys of the ground and your feet/shoes create small normal forces that can sum into a direction of motion.

Air Buoyancy: Since you are in a fluid, the mass of the fluid you displace creates an upward force away from the center of the earth. Since the density of air is miniscule, this force is generally neglected except in the most precise of circumstances.

Drag and Air resistance: While you walk, as you move through a fluid, that fluid exerts friction on your body in the form of drag. It is usually small unless you’re moving very fast relative to the fluid.

Air pressure, blood pressure, body tensions: Your body has a balance of blood pressure, muscle tensions, which oppose outside air pressures which equalize out to form the shape your body is in.

Internal forces: Many forces act within you such as air pressure, other muscle tensions, and internal stresses which balance out. Usually in physics these are lumped under internal forces.

6 0

3 years ago

The tonga trench in the pacific ocean is 36,000 feet deep. assuming that sea water has an average density of 1.04 g/cm3, calcula

MakcuM [25]

1110 atm

Let's start by calculating how many cm deep is 36,000 feet.

36000 ft * 12 in/ft * 2.54 cm/in = 1097280 cm

Now calculate how much a column of water 1 cm square and that tall would mass.

1097280 cm * 1.04 g/cm^3 = 1141171.2 g/cm^2

We now have a number using g/cm^2 as it's unit and we desire a unit of Pascals ( kg/(m*s^2) ).

It's pretty obvious how to convert from g to kg. But going from cm^2 to m is problematical. Additionally, the s^2 value is also a problem since nothing in the value has seconds as an unit. This indicates that a value has been omitted. We need something with a s^2 term and an additional length term. And what pops into mind is gravitational acceleration which is m/s^2. So let's multiply that in after getting that cm^2 term into m^2 and the g term into kg.

1141171.2 g/cm^2 / 1000 g/kg * 100 cm/m * 100 cm/m = 11411712 kg/m^2

11411712 kg/m^2 * 9.8 m/s^2 = 111834777.6 kg/(m*s^2) = 111834777.6 Pascals

Now to convert to atm

111834777.6 Pa / 1.01x10^5 Pa/atm = 1107.2750 atm

Now we gotta add in the 1 atm that the atmosphere actually provides (but if you look closely, you'll realize that it won't affect the final result).

1107.274 atm + 1 atm = 1108.274 atm

And finally, round to 3 significant figures since that's the accuracy of our data, giving 1110 atm.

8 0

3 years ago

¿Por qué es útil comprender tu condición física?

finlep [7]

Answer:

Es útil porque al saberlo evitas sobrexponerte a problemas como <em><u>enfermedades, desgarros o daños a largo plazo de tu cuerpo.</u></em>

Explanation:

When a person decides to start exercising or some type of physical activity, it is necessary for him to know how much his body can resist that activity during a defined period of time.

If in some way or another your body does not resist physical activity, you can hurt yourself, tear a ligament or muscle or get sick.

One of the most common diseases that almost nobody notices is the increase in circulatory and heart problems.

It is best to know your physical condition, how much weight you can carry with weights, or how long you can last doing physical activities.

4 0

3 years ago