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10 months ago

Jan Baptista van Helmont's famous experiment incorrectly showed that the plants produce most of their mass from:

1 answer:

3 0

Jan Baptista van Helmont's famous experiment incorrectly showed that the plants produce most of their mass from water.

In physics, mass is used to express inertia, a property common to all matter. In essence, it is the resistance of a mass of matter to changing its course or speed in response to the application of a force. The more mass a body has, the less of a change an applied force makes. Using Planck's constant, the kilogram, the ISU's unit of mass, is equivalent to 6.62607015 1034 joule seconds (SI). One kilogram is multiplied by one square meter per second to produce one joule. Since the second and the meter have already been defined in terms of other physical constants, the kilogram is determined by precise measurements of Planck's constant.

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What is meant by the term illegal alien

anyanavicka [17]

Answer:

The term illegal alien refers to a foreign person who is living in a country to which they are not an official legal citizen.

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2 years ago

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a solid metal sphere of radius 3.00m carries a total charge of -5.50. what is the magnitude of the electric field at a distance

aivan3 [116]

Answer:

(a) Electric field at 0.250 m is zero.

(b) Electric field at 2.90 m is zero.

(c) Electric field at 3.10 m is - 5.15 x 10³ V/m.

(d) Electric field at 8.00 m is - 0.77 x 10³ V/m.

Explanation:

Let Q and R are the charge and radius of the solid metal sphere. The solid metal sphere behave as conductor, so total charge Q is on the surface of the sphere.

Electric field inside and outside the metal sphere is :

E = 0 for r ≤ R ( inside )

= $\frac{KQ}{r^{2} }$ for r > R ( outside )

Here K is electric constant and r is the distance from the center of the metal sphere.

(a) Electric field at 0.250 m is zero as r < R i.e. 0.250 m < 3 m from the above equation.

(b) Electric field at 2.90 m is zero as r < R i.e. 2.90 m < 3 m from the above equation.

E = $\frac{KQ}{r^{2} }$

Substitute 9 x 10⁹ N m²/C² for K, -5.50 μC for Q and 3.10 m for r in the above equation.

E = - $\frac{9\times10^{9}\times5.50\times10^{-6} }{3.10^{2} }$

E = - 5.15 x 10³ V/m

(d) Electric field at 8.00 m is given by the relation as r > R :

E = $\frac{KQ}{r^{2} }$

Substitute 9 x 10⁹ N m²/C² for K, -5.50 μC for Q and 8.00 m for r in the above equation.

E = - $\frac{9\times10^{9}\times5.50\times10^{-6} }{8^{2} }$

E = - 0.77 x 10³ V/m

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3 years ago

How do human population growth trends differ between developed nations and developing nations?

ololo11 [35]

Answer:

Replacement-Level Fertility

Another important population characteristic that differ btw develop nation and developing nations is relates to births is replacement-level fertility. Replacement-level fertility is the fertility rate that will result in the replacement of the parents in the population. Again, in an ideal world, the human replacement-level fertility rate would be exactly two. This would mean that each couple would produce two offspring that would replace them in the population. If this occurred, then the human population would stay at a stable rate

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2 years ago

It is important to maintain septic systems in order to

Andreyy89

It is important to maintain septic systems in order to <span>prevent untreated sewage from contaminating groundwater</span>

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3 years ago

At a given instant an object has an angular velocity. It also has an angular acceleration due to torques that are present. There

katen-ka-za [31]

a) Constant

b) Constant

Explanation:

We can answer this question by using the equivalent of Newton's second law of motion of rotational motion, which can be written as:

$\tau_{net} = I \alpha$ (1)

where

$\tau_{net}$ is the net torque acting on the object in rotation

I is the moment of inertia of the object

$\alpha$ is the angular acceleration

The angular acceleration is the rate of change of the angular velocity, so it can be written as

$\alpha = \frac{\Delta \omega}{\Delta t}$

where

$\Delta \omega$ is the change in angular velocity

$\Delta t$ is the time interval

So we can rewrite eq.(1) as

$\tau_{net}=I\frac{\Delta \omega}{\Delta t}$

In this problem, we are told that at a given instant, the object has an angular acceleration due to the presence of torques, so there is a non-zero change in angular velocity.

Then, additional torques are applied, so that the net torque suddenly equal to zero, so:

$\tau_{net}=0$

From the previous equation, this implies that

$\Delta \omega =0$

Which means that the angular velocity at that instant does not change anymore.

In this second case instead, all the torques are suddenly removed.

This also means that the net torque becomes zero as well:

$\tau_{net}=0$

Therefore, this means that

$\Delta \omega =0$

So also in this case, there is no change in angular velocity: this means that the angular velocity of the object will remain constant.

So cases (a) and (b) are basically the same situation, as the net torque is zero in both cases, so the object acts in the same way.

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3 years ago