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Akimi4 [234]
3 years ago
5

How does competition affect population size? Use the terms carrying capacity and limiting factor with your example.

Physics
2 answers:
svetoff [14.1K]3 years ago
6 0

Limiting factors are resources or other factors in the environment that can lower the population growth rate. Competition for resources like food and space cause the growth rate to stop increasing, so the population levels off. This flat upper line on a growth curve is the carrying capacity.

cestrela7 [59]3 years ago
3 0

Answer:

they can eat all of the food and kill off the population

Explanation:

if the competition eats all of the food there's no food for the population and they will die off.

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A weight lifter lifts a 1.0 x 102 kg mass 1.5 m in 2.0 s. The power is?
Serga [27]

Answer:

\boxed{p =  {7.4 \times 10}^{2} W} \to \: option \: c.

Explanation:

his \: power \: is \to \\ p =  \frac{mgh}{t}  \\ p =  \frac{1.0 { \times 10}^{2}  \times 1.5 \times 9.8}{2.0}   \\  p= 735 \\  \boxed{p =  {7.4 \times 10}^{2} }

6 0
3 years ago
A 0.10 kg baseball travelling at 40 m s−1 hits straight back to the pitcher at 55 m s−1. The contact time is 0.01 seconds. What
Ilia_Sergeevich [38]

Answer:

1.5 kgms⁻¹

Explanation:

Momentum can be defined as "<em>mass in motion</em>."  

The amount of momentum that an object has is dependent upon two factors

  • mass of the moving object  
  • speed of motiom 

when there is a change in the velocity , it creates a change in momentum also

when we consider that we can mathematically show this,In terms of an equation,

Change in momentum    (ΔΡ) = m(Δv)

where (Δv) - change in velocity

<em>(Δv) = final velocity - initial velocity</em>

Change in momentum (ΔΡ) = m(Δv)

                                             = 0.1×([55-40])

                                             = 1.5 kgms⁻¹

     

7 0
3 years ago
One watt is A. equal to one kilogram per second B. the power that can be delivered by an average horse C. the same as one-foot p
Naddik [55]

Answer:

D. the proper replacement unit for one joule per second

Explanation:

When energy is divided by the time the energy was used we get power

P=\dfrac{E}{t}\\\Rightarrow P=\dfrac{mv^2}{t}\\\Rightarrow P=\dfrac{kg(\dfrac{m^2}{s^2})}{s}\\\Rightarrow P=kg(\dfrac{m^2}{s^2})}\times \dfrac{1}{s}\\\Rightarrow P=\dfrac{kgm^2}{s^3}

kg\dfrac{m^2}{s^2}=Joule

P=\dfrac{kgm^2}{s^3}

So, the answer is D. the proper replacement unit for one joule per second

8 0
3 years ago
What is the definition of electrical power?
Nonamiya [84]
An electric power measure the rate of electrical energy transfer by an electric circuit per unit of time.
6 0
2 years ago
A girl rolls a ball up an incline and allows it to re- turn to her. For the angle and ball involved, the acceleration of the bal
zalisa [80]

Answer:

3.28 m

3.28 s

Explanation:

We can adopt a system of reference with an axis along the incline, the origin being at the position of the girl and the positive X axis going up slope.

Then we know that the ball is subject to a constant acceleration of 0.25*g (2.45 m/s^2) pointing down slope. Since the acceleration is constant we can use the equation for constant acceleration:

X(t) = X0 + V0 * t + 1/2 * a * t^2

X0 = 0

V0 = 4 m/s

a = -2.45 m/s^2 (because the acceleration is down slope)

Then:

X(t) = 4*t - 1.22*t^2

And the equation for speed is:

V(t) = V0 + a * t

V(t) = 4 - 2.45 * t

If we equate this to zero we can find the moment where it stops and begins rolling down, that will be the highest point:

0 = 4 - 2.45 * t

4 = 2.45 * t

t = 1.63 s

Replacing that time on the position equation:

X(1.63) = 4 * 1.63 - 1.22 * 1.63^2 = 3.28 m

To find the time it will take to return we equate the position equation to zero:

0 = 4 * t - 1.22 * t^2

Since this is a quadratic equation it will have to answers, one will be the moment the ball was released (t = 0), the other will eb the moment when it returns:

0 = t * (4 - 1.22*t)

t1 = 0

0 = 4 - 1.22*t2

1.22 * t2 = 4

t2 = 3.28 s

7 0
3 years ago
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