1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
borishaifa [10]
3 years ago
13

Surfactants, or wetting agents, are chemicals that are used in detergents. These chemicals interfere with hydrogen bonding betwe

en water molecules to allow water to more easily penetrate the small spaces between the fibers in cloth.
If an accident released these substances into a pond, what danger would they most likely pose to living things in the pond?


A)The water in the pond would warm and cool more easily.

B)Plants on the bottom of the lake would die because ice would sink.

C)Insects would have trouble walking on the water.

D)The water would be able to dissolve fewer nutrients.
Physics
1 answer:
hichkok12 [17]3 years ago
8 0

Answer:

D)  Insects would have trouble walking on the water.

Explanation:

You might be interested in
Which illustration represent the substance at a higher temperature? Explain.
mamaluj [8]

The addition of heat energy to a system always causes the temperature of that system to increase. This is always true because you are adding heat of a substance to increase  its temperature. For example, you are going to drink a cup of coffee. And you wanted it hot to boost your attention. So you have to use hot water. In order for your water to become hot or warm, you need boil it in a kettle. Note that you are going to use an electric stove. The electric stove gets it energy from the source giving it a hotter temperature to the water in the kettle. You are applying heat energy to warm the water. So, the statement is true.

7 0
3 years ago
Read 2 more answers
Calculate the change in heat of the aluminum; show all calculations. Calculate the change in heat of the water; show all calcula
tatyana61 [14]
Calculate the change in heat of the aluminum; show all calculations. Calculate the change in heat of the water; show all calculations. Are the two values the same? Why or why not? See the attached picture for the numbers.

I got -3443.14 J for the aluminum and 3443.595 for the water
6 0
2 years ago
Difference between freefall and weightlessness.​
Margaret [11]

Answer:

Differences between freefall and weightlessness are as follows:

<h3><u>Freefall</u></h3>
  • When a body falls only under the influence of gravity, it is called free fall.
  • Freefall is not possible in absence of gravity.
  • A body falling in a vacuum is an example of free fall.

<h3><u>Weightlessness</u></h3>
  • Weightlessness is a condition at which the apparent weight of body becomes zero.
  • Weightlessness is possible in absence of gravity.
  • A man in a free falling lift is an example of weightlessness.

Hope this helps....

Good luck on your assignment....

6 0
3 years ago
Returning once again to our table top example of a horizontal mass on a low-friction surface with m = 0.254 kg and k = 10.0 N/m
Julli [10]

Explanation:

Given that,

Mass = 0.254 kg

Spring constant [tex[\omega_{0}= 10.0\ N/m[/tex]

Force = 0.5 N

y = 0.628

We need to calculate the A and d

Using formula of A and d

A=\dfrac{\dfrac{F_{0}}{m}}{\sqrt{(\omega_{0}^2-\omega^{2})^2+y^2\omega^2}}.....(I)

tan d=\dfrac{y\omega}{(\omega^2-\omega^2)}....(II)

Put the value of \omega=0.628\ rad/s in equation (I) and (II)

A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-0.628)^2+0.628^2\times0.628^2}}

A=0.0198

From equation (II)

tan d=\dfrac{0.628\times0.628}{((10.0^2-0.628)^2)}

d=0.0023

Put the value of \omega=3.14\ rad/s in equation (I) and (II)

A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-3.14)^2+0.628^2\times3.14^2}}

A=0.0203

From equation (II)

tan d=\dfrac{0.628\times3.14}{((10.0^2-3.14)^2)}

d=0.0120

Put the value of \omega=6.28\ rad/s in equation (I) and (II)

A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-6.28)^2+0.628^2\times6.28^2}}

A=0.0209

From equation (II)

tan d=\dfrac{0.628\times6.28}{((10.0^2-6.28)^2)}

d=0.0257

Put the value of \omega=9.42\ rad/s in equation (I) and (II)

A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-9.42)^2+0.628^2\times9.42^2}}

A=0.0217

From equation (II)

tan d=\dfrac{0.628\times9.42}{((10.0^2-9.42)^2)}

d=0.0413

Hence, This is the required solution.

5 0
3 years ago
Which pair of graphs represent the same motion?
GaryK [48]
Show me the graphs and i would be glad to help u
7 0
3 years ago
Other questions:
  • A firefighter with a weight of 707 N slides down a vertical pole with an acceleration of 2.79 m/s2, directed downward. (a) What
    7·1 answer
  • What processes formed the beaches of Florida?
    10·1 answer
  • Use the data table and information below to answer the following questions.
    15·1 answer
  • Remoras are small fish that attach themselves to the sides of sharks. They get protection and scraps of food from the sharks. Th
    11·1 answer
  • 1. Calculate the force of gravity in newtons if your weight is 110 lbs?
    5·2 answers
  • A steel tank of weight 600 lb is to be accelerated straight upward at a rate of 1.5 ft/sec2. Knowing the magnitude of the force
    11·1 answer
  • when you jump from an elevated position you usually bend your knees upon reaching the ground. by doing this, you make the time o
    7·1 answer
  • Which reverses the flow of current through<br> an electric motor?
    13·1 answer
  • Jonas and his family are moving to another part of the city. As Jonas, his brother, and his Dad were driving one of the trucks f
    12·2 answers
  • How do electromagnetic waves transfer energy?
    7·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!