This is a written question.

Give an example of nonpoint-source pollution

Ket [755]

Nonpoint source pollution is pollution that affects bodies of water, so remembering this can help in the future with this type of question. Some examples of nonpoint source pollution are; Excess fertilizers and herbicides, oil and toxic chemical runoff, as well as sediment from eroding stream banks. I hope this helps :)

8 0

3 years ago

How would you differentiate the theoretical and actual velocities of PE and KE in falling objects?

Solnce55 [7]

Answer: you've been trolled

Explanation:

8 0

3 years ago

A 0.500-kg potato is fired at an angle of 80.0° above the horizontal from a PVC pipe used as a "potato gun" and reaches a height

lions [1.4K]

Answer:

(a) $47.15ms^{-1}$

(b) $2470.13ms^{-2}$

(c) 1235.06N and 252.05 as a ratio

Explanation:

From Newton's second law of motion

F=ma where m is mass, a is acceleration and F is net force

Also from kinetic equation of motion, velocity and displacement are related using equation

$v^{2}=u^{2}+2sa$

Where v is final velocity, u is initial velocity, a is acceleration and s is displacement

From the free body diagram attached, final velocity at maximum height is 0 and initial velocity is $usin80^{o}$

Also, the vertical component can be written as

$v_{y}^{2} }=u_{y}^{2} } -2gs$ The negative sign before 2gs means displacement is opposite the gravitational force

Where g is acceleration due to gravity, $u_{y}$ is vertical component of initial velocity and $v_{y}$ is vertical component of final velocity

Since $v_{y}$ is 0

$u_{y}^{2} } =2gs$

s=110 and g is taken as 9.8

$u_{y}=\sqrt{2*9.8*110}=46.43275$

$u_{y}=46.43ms^{-1}$

Also, it's evident that the vertical component of initial velocity is $u_{y}=u_{i}sin \theta$ where $\theta$ is angle of projection and $u_{i}$ is resultant velocity

Making $u_{i}$ the subject we obtain $u_{i}=\frac {u_{y}}{sin \theta}$

Since $u_{y}$ and $\theta$ are known as $46.43ms^{-1}$ and $80^{o}$ respectively, then $u_{i}=\frac {46.43ms^{-1}}{sin 80^{o}}=47.15ms^{-1}$

Therefore, the velocity of potato is $47.15ms^{-1}$

(b)

Displacement depends on length of tube hence s=0.450m hence going back to kinetic equation $v^{2}=u^{2}+2sa$

The final velocity v is answer in part a which is $47.15ms^{-1}$ , initial velocity u is $0ms^{-1}$ hence the equation is re-written as

$v^{2}=2sa$ and making a the subject we obtain

$a=\frac {v^{2}}{2s}$

$a=\frac {47.15^{2}}{2*0.450}=2470.13ms^{-2}$

Therefore, average acceleration is $2470.13ms^{-2}$

(c)

From Newton's second law of motion, F=ma where m=0.500kg and a is $2470.13ms^{-2}$

Therefore, the average force of potato is

F=0.5*2470.13=1235.06N

F=1235.06N

The weight, W of potato is given by W=mg

Taking R as ratio of average force and weight of potato

$R=\frac {F}{W}=\frac {F}{mg}$ and since F=1235.06, m=0.500kg and g=9.8

$R=\frac {1235.06}{0.500*9.8}$ =252.05

Therefore, ratio of average force to weight is 252.05

8 0

3 years ago

Two particles, one with charge − 7.97 μC −7.97 μC and one with charge 7.75 μC, 7.75 μC, are 8.09 cm 8.09 cm apart. What is the m

mr Goodwill [35]

Answer:

Electric force, F = 84.93 N

Explanation:

Given that,

Charge on particle 1, $q_1=-7.97\ \mu C=-7.97\times 10^{-6}\ C$

Charge on particle 1, $q_2=7.75\ \mu C=7.75\times 10^{-6}\ C$

The distance between charges, d = 8.09 cm

To find,

The magnitude of the force that one particle exerts on the other.

Solution,

The electric force of attraction or repulsion is given by the formula as:

$F=\dfrac{kq_1q_2}{d^2}$

k is the electrostatic constant

$F=\dfrac{9\times 10^9\times 7.97\times 10^{-6}\times 7.75\times 10^{-6}}{(8.09\times 10^{-2})^2}$

F = 84.93 N

So, the magnitude of the force that one particle exerts on the other is 84.93 N.

4 0

3 years ago

Read 2 more answers

A bowling ball of mass m = 1.1 kg is resting on a spring compressed by a distance d = 0.35 m when the spring is released. At the

riadik2000 [5.3K]

The spring constant, k, in newtons per meter is 1,955.9 N/m.

<h3>Speed of the ball after the launch</h3>

h = v²sin²θ/2g

v = √[(2gh)/sin²θ]

v = √[(2 x 9.8 x 4.4)/ (sin 39)²]

v = 14.76 m/s

<h3>Energy of the ball at top </h3>

E = K.E + P.E

E = ¹/₂m(v cosθ)² + mgh

E = ¹/₂(1.1)(14.76 cos39)² + (1.1 x 9.8 x 4.4)

E = 119.8 J

<h3>Spring constant</h3>

E = ¹/₂kx²

k = 2E/x²

k = (2 x 119.8)/(0.35²)

k = 1,955.9 N/m

Thus, the spring constant, k, in newtons per meter is 1,955.9 N/m.

Learn more spring constant here: brainly.com/question/1968517

#SPJ1

7 0

2 years ago