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borishaifa [10]
2 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]2 years ago
8 0

Answer:

D)  Insects would have trouble walking on the water.

Explanation:

You might be interested in
According to Charles's law, for a fixed quantity of gas at constant pressure, which of the given quantities is constant? Accordi
Naddik [55]
Charles's <span>law is also known as the law of volumes explained the relation between the volume of gas and its temperature.

Charles's law states that: " At constant pressure, the volume of a certain mass of gas is directly proportional to its temperature in kelvin".
This means that:
V = T*constant
V/T = constant

Based on this, the correct choice is:
</span><span>a) V/T</span>
7 0
3 years ago
Read 2 more answers
Consider the two-body situation at the right. A 300kg crate rests on an inclined plane and is connected by a cable to a 100 kg m
trasher [3.6K]

Answer:

a= 0.578 m/s

T = 1037.8 N

Explanation:

Data

m₁= 300 kg

m₂= 100 kg

inclined plane, θ =  30°

μk = 0.120

Newton's second law to m₁:

We define the x-axis in the direction parallel to the movement of the 300kg (m₁) crate on the ramp and the y-axis in the direction perpendicular to it.

∑F = m₁*a Formula (1)

Forces acting on m₁

W₁: m₁ weight : In vertical direction

N : Normal force : perpendicular to the inclined plane

f : Friction force: parallel to the inclined plane

T:  cable tension : parallel to inclined plane

Calculated of the W₁

W₁=m₁*g

W₁= 300kg* 9.8 m/s² = 2940 N

x-y weight components

W₁x= W₁sin θ =2940 N*sin(30)° =1470 N

W₁y= W₁cos θ =2940 N *cos(30)° =2156.4 N

Calculated of the N

We apply the formula (1)

∑Fy = m*ay    ay = 0

N - W₁y = 0

N = W₁y

N = 2156.4 N

Calculated of the f

f = μk* N= (0.120)*(2156.4 N)

f = 258.77 N

Newton's second law to m₁ in direction  x-axis :

∑Fx = m₁*ax   ,ax  =a

We assume that m₁ descends on the inclined plane and we positively take the direction of movement:

wx-f-T = m*a

wx - f - m*a =T

1470  -258.77 -300*a =T

T= 1211.23-300*a   Equation (1)

Newton's second law to m₂

∑Fy = m₂*ay   ,ay  =a

Forces acting on m₂

W₂: m₂ weight : In vertical direction

T:  cable tension:In vertical direction

Calculated of the W₂

W₂=m₂*g

W₂= 100kg* 9.8 m/s² = 980 N

∑Fy = m₂*a

Because we assume that m₁ descends on the inclined plane, then, m₂ ascends  vertically, we take positive the direction of movement:

T-W₂ = m₂*a

T-980 = 100*a

T = 980 + 100*a Equation (2)

Problem development

Equation (1) =  Equation (2) = T

1211.23-300*a= 980  + 100*a

1211.23- 980 = 100*a + 300*a

231.23 = 400*a

a= 231.23 / 400

a= 0.578 m/s

Because the acceleration tested positive then effectively m₁ descends on the inclined plane and m₂ ascends  vertically.

We replace a= 0.578 m/s in the equatión (2)

T = 980 + 100* (0.578 )

T = 1037.8 N

5 0
3 years ago
A proton is released in a uniform electric field, and it experiences an electric force of 2.04×10−14 N toward the south.
Lemur [1.5K]

Answer:

Part A:

E=127500N/C\\E=1.275*10^{5} N/C

Part B:

Option B (Towards the South)

Explanation:

Part A:

Magnitude if electric field E:

E=Force/charge

Force=2.04×10−14 N

Charge=1.6×10−19 C

E=\frac{2.04*10^{-14}}{1.6*10^{-19}} \\E=127500N/C\\E=1.275*10^{5} N/C

Part B:

Option B (Towards the South)

As electron is experiencing the force towards south,it means the direction of the electric field is towards the south because direction of field lines is from positive to negative, so proton is moving towards south it means negative charge is in south to which proton is attracted. So electric field is towards South.

7 0
3 years ago
A proton orbits a long charged wire, making 1.80 ×106 revolutions per second. The radius of the orbit is 1.20 cm What is the wir
Fantom [35]

Answer:

linear charge density = -9.495 × 10^{-34} C/m

Explanation:

given data

revolutions per second = 1.80 × 10^{6}

radius = 1.20 cm

solution

we know that when proton to revolve around charge wire then centripetal force is require to be in orbit of radius around provide by electric force

so

- q × E = m × w² × r     ..................1

- 9 × 10^{9}  × \frac{2*linear\ charge\ density}r} q =  m × w² × r   ............2

and w = \frac{2*\pi}{T}  

w = \frac{d\theta }{dt}

w = 1.80 × 10^{6} × \frac{2*\pi}{1}

w = 11304000 rad/s

so here from equation 2

- 9 × 10^{9}  × \frac{2*linear\ charge\ density}{0.012} 1.80 × 10^{6} =  1.672 × 10^{-27} × 11304000² × 0.0120  

linear charge density = -9.495 × 10^{-34} C/m

8 0
2 years ago
Write the ground-state electron configurations of the following ions. (a) Li+ (b) N3− (c) In3+ (Use the noble gas core electron
maria [59]

Li+  [He]

N³-  [Ne]

In³+  [Kr] 4d10

Tl+  [Xe] 4f14 5d10 6S2

6 0
3 years ago
Read 2 more answers
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