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Dafna11 [192]
2 years ago
14

How does natural eutrophication affect a pond ecosystem?

Physics
2 answers:
mr Goodwill [35]2 years ago
6 0
Farm hinde puro kill
ch4aika [34]2 years ago
5 0
It increases nutrient to build up
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A 1000 kg weather rocket is launched straight up. The rocket motor provides a constant acceleration for 16 s, then the motor sto
Fed [463]

Answer:

a)   a = 34.375 m / s²,  b)    v_f = 550 m / s

Explanation:

This problem is the launch of projectiles, they tell us to ignore the effect of the friction force.

a) Let's start with the final part of the movement, which is carried out from t= 16 s with constant speed

             v_f = \frac{x-x_1}{t}

we substitute the values

             v_f = \frac{ 6600 -x_1}{4}  

The initial part of the movement is carried out with acceleration

             v_f = v₀ + a t

             x₁ = x₀ + v₀ t + ½ a t²

the rocket starts from rest v₀ = 0 with an initial height x₀ = 0

             x₁ = ½ a t²

             v_f = a t

we substitute the values

              x₁ = 1/2  a 16²

              x₁ = 128 a

              v_f = 16 a

let's write our system of equations  

               v_f = \frac{6600 - x_1}{4}

               x₁ = 128 a

               v_f = 16 a

we substitute in the first equation  

               16 a = \frac{6600 -128 a}{4}

               16 4 a = 6600 - 128 a

                a (64 + 128) = 6600

                a = 6600/192

                 a = 34.375 m / s²

b) let's find the time to reach this height

                x = ½ to t²

                t² = 2y / a

                t² = 2 5100 / 34.375

                t² = 296.72

                t = 17.2 s

We can see that for this time the acceleration is zero, so the rocket is in the constant velocity part

               v_f = 16 a

               v_f = 16 34.375

               v_f = 550 m / s

8 0
3 years ago
What process is used in this example? You have learned that all living things use energy. Your dog is a living thing. She must u
oksano4ka [1.4K]
Option B.) Deductive Reasoning
8 0
3 years ago
How much does a 20 kg rock weigh on Earth where g = 9.8 m/s2?
pentagon [3]

Answer:

\huge\boxed{196\:N}

Explanation:

Mass and weight ain't the same thing. So, for calculating weight we would use Newton's 2nd Law of Motion i.e. Force = Mass × Acceleration.

According to question,

The mass is 20kg

Acceleration is 9.8 m/s^2

Putting the given values into the formula,

F = 20 × 9.8

F = 196N

So, the weight of the rock is 196 N

Hope it helps!<3

3 0
2 years ago
1. Describe your egg drop apparatus and at least three (3) design features you implemented in order to try to help your egg surv
Olegator [25]

Answer:

Explanation:

There are three basic ways to increase the likelihood of safely dropping an egg:

Slow down the descent speed.

Parachutes are an obvious method for slowing the decent speed, as long as the design includes a way to keep the parachute open.

Cushion the egg so that something other than the egg itself absorbs the impact of landing.

The largest end of the egg has an area of air trapped between the egg's two membranes. This air space forms when the contents of the egg cool and contract after the egg is laid. It accounts for the crater you often see at the end of a hard-cooked egg. Upon impact the heavier spherical yolk continues moving towards the ground. The compression of the airspace acts like an air bag for the eggs' valuable contents. Building an artificial cushioning device will also help absorb the impact of landing.

The largest end of the egg has an area of air trapped between the egg's two membranes. This air space forms when the contents of the egg cool and contract after the egg is laid. It accounts for the crater you often see at the end of a hard-cooked egg. Upon impact the heavier spherical yolk continues moving towards the ground. The compression of the airspace acts like an air bag for the eggs' valuable contents. Building an artificial cushioning device will also help absorb the impact of landing.

Orient the egg so that it lands on the strongest part of the shell.

The arch structure at either end of the egg is stronger than its sides. Pressure is distributed down (or up) the arches so that less pressure acts on any one point. Orienting the arch downwards will increase the egg's survival.

Hope this helps you

5 0
2 years ago
In which sound wave application are sound waves absorbed?
ivann1987 [24]

Answer: it is ultrasonic cleaners

Explanation:

5 0
2 years ago
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