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ad-work [718]
3 years ago
10

A student reads that nanosilver particles are used in cosmetics to allow nutrients to penetrate deeper into the skin. When the c

osmetics are washed off, the nanosilver particles can enter freshwater ecosystems. These particles may decrease survival in many aquatic organisms. The student determines that Daphnia magna, a tiny freshwater crustacean widely found in North America, might make a good test subject for an experiment.
Which statement below is the most likely testable hypothesis based on the information provided?
A. When the concentration of nanosilver particles in water and the temperature of water are both increased, the population of Daphnia magna will decrease.
B. When the concentration of nanosilver particles in water and the temperature of water are both increased, the population of Daphnia magna will decrease.
C. When the concentration of nanosilver particles in water is increased, the population of Daphnia magna will adapt to new conditions and become a new species.
D. When the concentration of nanosilver particles in water is decreased, the population of aquatic organisms that feed on Daphnia magna will increase.
Physics
1 answer:
Vsevolod [243]3 years ago
7 0

Answer: When the concentration of nanosilver particles in water is increased, the population of Daphnia magna will decrease

Explanation:

The statement below that is the most likely testable hypothesis based on the information provided is that when there is an increase in the concentration of nanosilver particles in water, the population of Daphnia magna will decrease.

Since we are informed that the particles may decrease survival in many aquatic organisms, it simply means that an increase in the particles in water will bring about the decrease in the Daphnia magna, which is the tiny freshwater crustacean.

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Answer:

A, B Y E

Explanation:

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3 years ago
A circular disc of mass 20kg and radius 15cm is mounted in an horizontal cylindrical axle of radius
disa [49]

Using the concepts of energy, rotational Newton's second law and rotational kinematics we can find the kinematic energy of the system formed by the disk and the cylindrical axis

          KE = 0.23 J

given parameters

  • Disk radius R = 15 cm = 0.15 m
  • Cylinder radius r = 1.5 cm = 0.0015 m
  • Disk mass M = 20 kg
  • Time t = 1.2 s
  • Force F = 12 N

to find

  • Kinetic energy (KE)

This exercise must be solved in parts:

1st part. Endowment kinetic energy is the energy due to the circular motion of an object and is described by the equation

         KE = ½ I w²

Where KE is the kinetic energy, I the moment of inertia and w the angular velocity

The moment of inertia is a magnitude that measures the inertia for rotational movement, it is a scalar quantity, therefore it is additive. In this system it is composed of two bodies, the disk and the cylindrical axis, for which the total moment of inertia it is

         I_{ total} = I_{ disk} + I_{ cylinder}

the moments of inertia with respect to an axis passing through the center of mass are tabulated

disk          I_{disk} = ½ M R²

cylinder   I_{cylinder} = ½ m r²

where M and m are the masses of the disk and cylinder respectively, R and r their radii

         I_{total} = ½ (M R² + m r²) = ½ M R² ( 1 + \frac{m}{M} \ (\frac{r}{R})^2 )

         I_{total} = ½ M R² ( 1+ \frac{m}{20}  (\frac{0.015}{0.15} )^2 ) = \frac{1}{2} M R² (1 + 0.005 m)

As the shaft mass  is much lighter than the disk mass , the last term is very small, which is why we despise it.

         I_{total} = ½ M R²

2nd part. Let's use Newton's second law for endowment motion

        τ = I α

        α = \frac{\tau }{I_{total}}l

        τ = F R

        α = \frac{F \ R}{I_{total}}

With the rotational kinematics expressions, we assume that the system starts from rest (w₀ = 0)

        w = w₀ + α  t

where w is the angular velocity, alpha is the angular acceleration and t is the time

        w = 0 + \frac{\tau }{I_{total}} \ t

we substitute in the kinetic energy equation

        KE = ½ I_{total}  ( \frac{ \tau }{I_{total}} \ t )²

        KE = ½ \frac{ \tau^2 }{I_{total}} \ t^2

let's substitute

        KE = \frac{F^2 \ R^4}{M \ R^2 } \ t^2

        KE = F² R² t² / M

let's calculate

        KE = 12² 0.15² 1.2² / 20

        KE = 0.23 J

With the concepts of energy and rotational kinematics we can find the kinetic energy of the system is

       KE = 0.23 j

learn more about rotational kinetic energy here:

brainly.com/question/20261989

4 0
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Help with these three
Elenna [48]
The first: alright, first: you draw the person in the elevator, then draw a red arrow, pointing downwards, beginning from his center of mass. This arrow is representing the gravitational force, Fg.
You can always calculate this right away, if you know his mass, by multiplying his weight in kg by the gravitational constant
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let's do it for this case:
f_{g}  = m \times g \\ f _{g}  = 65kg \times 9.81 \frac{m}{s {}^{2} }  = 637.65
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so, first step solved, Fg is 637.65N
Fg is a field force by the way, and at the same time, the elevator is pushing up on him with 637.65N, so you draw another arrow pointing upwards, ending at the tip of the downwards arrow.
now let's calculate the force of the elevator
f = m \times a \\ f = 65 \times 5 \frac{m}{s {}^{2} }  \\ f = 325n
so you draw another arrow which is pointing downwards on him, because the elevator is accelating him upwards, making him heavier
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A= 40 *0.1 = 4mm² -----convert to m²

A= 4e⁻⁶ m²

P= 4000000 pa

F= P/A = 4000000/4e⁻⁶

F= 403428793.493 pa

F= 403429 kpa

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mafiozo [28]
I believe it is D. 2
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2 years ago
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