An example of a point source water pollutant is a sewer pipe draining directly into a river.
According to the United States Environmental Protection Agency, a point source pollution is defined as; “any single identifiable source of pollution from which pollutants are discharged, such as a pipe, ditch, ship or factory smokestack.”
Hence, an example of a point source water pollutant is a sewer pipe draining directly into a river. The pollution is coming from an easily identifiable source which is a sewer pipe connected to a river directly.
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Which of the following methods of pest control used in IPM systems is effective because of the increase in biological diversity it provides? timed crop planting O weed suppression composted soil amendments deep tilling soil
Answer:
Decline in soil health is a serious worldwide problem that decreases complexity and stability of agricultural ecosystems, commonly making them more prone to outbreaks of herbivorous insect pests. Potato (Solanum tuberosum L., Solanaceae) and onion (Allium cepa L., Amaryllidaceae) production is currently characterized by high soil disturbance and heavy reliance on synthetic inputs, including insecticides. Evidence suggests that adopting soil conservation techniques often (but not always) increases mortality and decreases reproductive output for the major insect pests of these important vegetable crops. Known mechanisms responsible for such an effect include increases in density and activity of natural enemy populations, enhanced plant defenses, and modified physical characteristics of respective agricultural habitats. However, most research efforts focused on mulches and organic soil amendments, with additional research needed on elucidating effects and their mechanisms for conservation tillage, cover crops, and arbuscular mycorrhizae.
Introduction
Soil erosion is a serious problem worldwide (Amundson et al., 2015). Although it is often overshadowed in public discourse by other concerns, such as climate change and invasive species, soil deterioration receives considerable attention from the scientific community (Montgomery, 2007; Borrelli et al., 2017; Berhe et al., 2018). In addition to the loss of agricultural productivity, soil erosion has been linked to increased emissions of greenhouse gases and reduced water quality (Amundson et al., 2015; Berhe et al., 2018). Global soil erosion is forecasted to increase in the near future because of cropland expansion, especially in the least economically developed areas (Borrelli et al., 2017).
PLEASEEEEEEEE HELPPPPPPP
Answer: 62.25
Explanation: F = ma
F = 7.5 * 8.3
F = 62.25
Thus, the answer is 62.25 Newtons.
Would appreciate brainliest <3
7. A 1.0 kg metal head of a geology hammer strikes a solid rock with a velocity of 5.0 m/s. Assuming all the energy is retained by the hammer head, how much will its temperature increase
The increase in temperature of the metal hammer is 0.028 ⁰C.
The given parameters:
mass of the metal hammer, m = 1.0 kgspeed of the hammer, v = 5.0 m/sspecific heat capacity of iron, 450 J/kg⁰CThe increase in temperature of the metal hammer is calculated as follows;
[tex]Q = K.E\\\\mc \Delta T = \frac{1}{2} mv^2\\\\\Delta T = \frac{v^2}{2 c}[/tex]
where;
c is the specific heat capacity of the metal hammer
Assuming the metal hammer is iron, c = 450 J/kg⁰C
[tex]\Delta T = \frac{5^2}{2 \times 450} \\\\\Delta T = 0.028 \ ^0C[/tex]
Thus, the increase in temperature of the metal hammer is 0.028 ⁰C.
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Question below...............
Answer:
friction force
Explanation:
force of friction is opposite to the force applied it resist the motion
Danny is competing in the high jump. When he is in the air, his body has _______ energy due to its height, and it has _______ energy due to its motion.
Answer:
Gravitational potential energy
kinetic energy
Which of the following statements are true?
(a) An object can move even when no force acts on it.
(b) If an object isn't moving, no external forces act on it.
(c) If a single force acts on an object, the object accelerates.
(d) If an object accelerates, a force is acting on it.
(e) If an object isn't accelerating, no external force is acting on it.
(f) If the net force acting on an object is in the positive x-direction, the object moves only in the positive x-direction.
I can't understand how A is true.
Answer: a, c, d
Explanation: a is true because the object will continue to move even without any force because of inertia (so yh thats why a is true). c is true because an object can accelerate if a single force acts on it. to accelerate (not move), it needs a force to act on it
Would appreciate brainly <3
Explanation:
inertia ...
you push or throw something, and you apply some force to it at that moment, but then it moves and keeps moving even long after you have no more connection to it, and no more force is applied to it.
please consider : we are only talking about moving. not about acceleration.
so, yes, (a) is true.
(c) is true.
(d) is true.
9
are things that you can achieve quickly.
O A.
Dreams
OB.
Long-term goals
O c.
Short-term goals
OD.
Plans
Answer:
d
Explanation:
A dreams
.......wkkwkwkwkwkwnwnsksk
Answer:
C. short term goal
Explanation:
took a midterm quiz on canvas, Hope this helps <3
In the popular game "Angry Birds" (refer Figure), for the red bird to hit the green bird with their range 55.0 m, what should be the launch angle (with respect to the horizontal) of the slingshot if the time of flight is 2.50 s?
Answer:
Explanation:
What are we to use as gravity in this imaginary world? I will ASSUME 9.81 m/s² and assume no air resistance.
horizontal velocity component
vx = 55.0/2.50 = 22 m/s
vertical initial velocity component
(12.0 - 10.0) = 0 + vy₀(2.50) + ½(-9.81)2.50²
vy₀ = 13.0625 m/s
θ = arctan(vy₀/vx) = arctan(13.0625/22) = 30.6997225... = 30.7°
When measuring the critical angle, in which medium do we need the refracted ray to be (air or glass)
Answer:
N1 sin theta1 = N2 sin theta2 Snell's Law
For the refracted ray to be reflected then sin theta2 = 1
N1 sin theta 1 = N2
Also N2 must be less than N1 for complete reflection
sin theta1 = N2/N1
If you are considering air and glass then N2 = 1 (for air)
sin theta1 = 1 / N2 where N2 must be for glass in this case
what do i do for #17
Answer:
whats the question...............
Explanation:
You throw an 8.75 g coin straight up into the air. The coin travels a distance of 337 cm upward. What was the initial speed of the coin when you released it
Answer:
hi
Explanation:
i like saying hi
A stone is allowed to fall from the top of the tower 100m high at the same moment another stone is projected vertically upward with a velocity of 25m/s. Where and when will the two cross each other
Answer:
both the stone will meet at a distance of 80 m from the top of tower.
Explanation:
let "t" = time after which both stones meet
"S" = distance travelled by the stone dropped from the top of tower
(100-S) = distance travelled by the projected stone.
◆ i) For stone dropped from the top of tower
-S = 0 + 1/2 (-10) t²
or, S = 5t²
◆ ii) For stone projected upward
(100 - S) = 25t + 1/2 (-10) t²
= 25t - 5t²
Adding i) and ii) , We get
100 = 25t
or t = 4 s
Therefore, Two stones will meet after 4 s.
◆ ¡¡¡) Put value of t = 4 s in Equation i) , we get
S = 5 × 16
= 80 m.
Thus , both the stone will meet at a distance of 80 m from the top of tower.
(Hope this helps can I pls have brainlist (crown)☺️)
Starting from rest, a racecar moves 112 m in the first 7 s of uniform acceleration. What is the car's acceleration?
Answer:
215mph
Explanation:
multiply 112milws by 2 and multiple that by seven, then divide it by two
2. An auditorium has 58 seats in the first row, 62 seats in the second row, 66 seats in the third row, and so
on.
a)Find the explicit formula of this arithmetic sequence.
B) find the number of seats in the twentieth row.
A. A small, closed chamber of gas is heated. When the gas in the
chamber expands, it does 5 J of work on a piston. The gas has
an initial energy of 8 J and a final energy of 30 J. Considering
the equation for the first law of thermodynamics (AU = Q + W.
does the work done in this scenario have a positive or negative
value? Explain.
B. How much heat is added to the gas in the chamber?
A.
The work has a negative value.
Since the gas in the chamber expands, it increases in volume and does positive work on the piston since the change in volume is positive.
Work, W = pΔV. Since the gas expands, ΔV > 0. So, W > 0. Thus Work done by the gas is positive.
Since in ΔU = Q + W, W here is work done by the surroundings, W is negative since it is the opposite of the work done on the surroundings by the gas..
So, the work has a negative value.
B.
The heat added to the gas chamber is 17 J
From the first law of thermodynamics, ΔU = Q + W where ΔU = internal energy change = U₂ - U₁ where U₁ = initial energy = 8 J and U₂ = final energy = 30 J, Q = heat added to the gas chamber and W = work on piston = - 5 J
ΔU = Q + W
U₂ - U₁ = Q + W
30 J - 8 J = Q + (-5 J)
22 J = Q - 5 J
Q = 22 J + 5 J
Q = 27 J
The heat added to the gas chamber is 27 J.
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Which would most likely form a homogenous mixture?
Answer:
B) a pinch of sugar mixed with cup of water
Answer:
A pinch of sugar mixed with a cup of water
Explanation:
A homogeneous mixtures have a uniform appearance where the parts are pretty evenly spaced out throughout the mixture. Picking out the individual pieces should be hard, ,which is why sugar in water is the best choice.
what is a capacitor and capacitance.
Answer:
Capacitor is a device which used to store energy .
Capacitance is the ratio of the change in an electric charge in a system to the corresponding change in its electric potential.
Explanation:
builder places a 3kg hammer on the top of a ladder, which is 4m above the ground. Calculate the gravitational potential energy of the hammer while on the ladder.
Answer:
[tex]E=mgh[/tex]
[tex]m=3kg[/tex]
[tex]h=4m[/tex]
[tex]g=9.8m/s^{2}[/tex]
[tex]E= 3*4*9.8=117.6J[/tex]
Explanation:
Only substitute amounts to formula.
Hope this helps ;)
Cheers :D
Can you please do this for me I’ll do 75 points
Answer:
Answer below, hope this helps!
Explanation:
1. Chemical - e
2.Light - b
3. Sound - g
4. Mechanical - f
5. Kinetic - a
6. Potential - d
7. Energy - j
8. Thermal - c
9. Nuclear - i
10. Electrical - h
Answer:
Chemical - e
Light - b
Sound - g
Mechanical - f
Kinetic - a
Potential - d
Energy - j
Thermal - c
Nuclear - i
Electrical - h
Explanation:
i’m confused on this worksheet can someone help
Answer:
well for C the horse is at rest at 10 to 15
D is 0 to 10
E is 15 to 25
Explanation:
If you search up examples it will help with this worksheet.
If I am work then I appologize
 A wooden box with a mass of 10.0 kg rest on a ramp that is incline at an angle of 25° to the horizontal. A rope attached to the box runs parallel to the ramp and then passes over a frictionless bully. A bucket with a mass of M hangs at the end of the rope. The coefficient of static friction between the ramp in the box is 0.50. The coefficient of Connecticut friction between the ramp in the box is 0.35.
Suppose the box remains at rest relative to the ramp. What is the maximum magnitude of the friction force exerted on the box by the ramp?
The maximum magnitude of the friction force exerted on the box by the ramp is 44.41 N.
The given parameters;
Mass of the box, m = 10 kgInclination of the ramp, θ = 25⁰Coefficient of static friction, μ = 0.5 Coefficient of kinetic friction, μk = 0.35The normal force on the wooden box is calculated as follows;
[tex]F_n = mg \times cos(\theta)\\\\F_n = 10 \times 9.8 \times cos(25)\\\\F_n = 88.8 2 \ N[/tex]
The maximum magnitude of the friction force exerted on the box by the ramp is calculated as follows;
[tex]F_f = \mu \times F_n\\\\F_f = 0.5 \times 88.82 \\\\F_f = 44.41 \ N[/tex]
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A large water tank is 3.70 m high and filled to the brim, the top of the tank open to the air. A small pipe with a faucet is attached to the side of the tank, 0.580 m above the ground. If the valve is opened, at what speed (in m/s) will water come out of the pipe
h =(3.7 - .58)m = 3.12m
Now put PE into KE and we have to use the formula:
√2gh (g = gravity and h = height) therefor:
√2 x 9.8 x 3.12
= 7.82m/s
I hope this helps!
What’s Newton’s second law? Explain and mention some examples in daily life
Answer:
Newton's second law states that .
The rate of change of linear momentum is directly proportional to the force applied.Formulically
F=maF=Force
m=mass
a=acceleration
The best example is hitting a tennis ball.
What is First Aid.
I mark u brainliest answer
Answer:
First aid refers to the emergency or immediate care you should provide when a person is injured or ill until full medical treatment is available.
Explanation:
What is the volume of 150g of a substance that has a density of 150g of a substance that has a density of 11.3g/cm3
Answer:
25.0 cm3
Explanation:
The volume is 25.0 cm3 .
If it takes 50.0 seconds to lift 10.0 Newtons of books to a height of 7.0
meters, calculate the power required to lift it?
During the experiment it is determined that, as the cart rolls between two points on the track, the work done on the cart by the hanging masses and other forces present is 0.91 J. At the initial time of observation the cart moves with speed 0.61 m/s. Determine the speed at the second point of observation. The mass of the cart is 0.80 Kg.
Answer:
Explanation:
If the work done on the cart is NET work
Then the work will result in an increase in kinetic energy
KE₀ + W = KE₁
½mv₀² + W = ½mv₁²
½(0.80)(0.61²) + 0.91 = ½(0.80)v₁²
v₁ = 1.626991...
v₁ = 1.6 m/s
The speed of the cart at the second point of observation is equal to 1.62 m/s.
What is the work done?Work can be demonstrated as the energy utilized when a force is exerted to make an object move through a particular displacement. Work done by this force is calculated from the product of the magnitude of the exerted force (F) and the distance (d) covered by the body
W= F.d
Where 'F' is the exerted force and 'd' is the displacement and W is work done.
Work and energy have a direct relationship with each other. Work done by an object can be expressed as a change in kinetic energy:
[tex]W =\frac{1}{2}mv_2^2- \frac{1}{2}mv_1^2[/tex]
Where m is the mass of the object, v₂ is the final velocity (m/s), and v₁ is the initial velocity (m/s).
Given, the work done by the cart, W = 0.91 J
The mass of the cart, m = 0.80 Kg
The speed of the cart at the first point, v₁ = 0.61 m/s
[tex]0.91 =\frac{1}{2}\times 0.80 \times v_2^2- \frac{1}{2}0.80 \times (0.61)^2[/tex]
0.40 v₂² = 1.06
v₂² = 2.65
v₂ = 1.62 m/s
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what are people words in english
Answer: Plenty
Explanation:
some words are Hi, Banana, Dude and many more
Figure 1 shows that the two vectors in different direction respectively. Determine the resultant for the two vectors
Answer:
15-45/2 = -7.5 downwards
Explanation:
cos(60) x 45 = z
z = 45/2
sin (30) x 70 = y
y = 15
15-45/ 2 = -7.5 downwards
Entra vapor a una tobera adiabática con un flujo másico de 250 kg/h. Al entrar el vapor
tiene una energía interna específica de 2510 kJ/kg, una presión de 1378 kPa, un
volumen específico de 0.147 m3
/kg y una velocidad de 5 m/s. Las condiciones de salida
son P= 138.7 kPa, volumen específico de 1.099 m3
/kg y energía interna específica de
2263 kJ/kg. Determine la velocidad de salida.
Este problema describe el funcionamiento de una tobera adiabática (sin flujo de calor), la cual tiene una corriente de entrada que difiere de la de salida espacial y energéticamente, pero que conservan el mismo flujo másico de 250 kg/h. De este modo, usamos un balance de energía con el fin the determinar la velocidad a la que sale el fluido, según es requerido en el problema:
[tex]mu_1+mP_1v_1+mgh_1+\frac{1}{2} mv_1^2=mu_2+mP_2v_2+mgh_2+\frac{1}{2} mv_2^2[/tex]
En la que se tiene que la incógnita es la velocidad de salida, [tex]v_2[/tex], y es posible simplificar el flujo másico, [tex]m[/tex], al estar como factor común en ambos lados y despreciar la energía potencial (mgh), ya que no hay diferencia de altura significativa entre la entrada y salida de la tobera.
De este modo, es posible reemplazar los valores dados para obtener la siguiente expresión:
[tex]2510\frac{kJ}{kg} +1378kPa*0.147\frac{m^3}{kg} +\frac{1}{2} (5\frac{m}{s} )^2=2263\frac{kJ}{kg}+138.7kPa*1.099\frac{m^3}{kg} +\frac{1}{2} v_2^2[/tex]
Y así, hallar la velocidad de salida como sigue:
[tex]2725.066\frac{kJ}{kg}=2415.431\frac{kJ}{kg} +\frac{1}{2} v_2^2\\\\v_2=\sqrt{2(2725.066\frac{kJ}{kg}-2415.431\frac{kJ}{kg} )} \\\\v_2=24.9\frac{m}{s}[/tex]
Para revisar:
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