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3 years ago

If the solid, purple line represents a prey population

Physics

1 answer:

worty [1.4K]3 years ago

3 0

Answer:

There is no correlation that can be made between the predator and prey populations.

Explanation:

Ermrmrm.

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Use the values from PRACTICE IT to help you work this exercise. Suppose the same two vehicles are both traveling eastward, the c

Mariulka [41]

Answer:

A. $v_{3}=12.17m/s$

B. $v_{car}=6.3m/s\\v_{truck}=-6.3m/s$

C. ΔK $=-4.13x10^3J$

Explanation:

From the exercise we know that the car and the truck are traveling eastward. I'm going to name the car 1 and the truck 2

$v_{1}=5.79m/s\\m_{1}=102kg\\v_{2}=18.5m/s\\m_{2}=103kg$

A. Since the two vehicles become entangled the final mass is:

$m_{3}=102kg+103kg=205kg$

From linear momentum we got that:

$p_{1}=p_{2}$

$m_{1}v_{1}+m_{2}v_{2}=m_{3}v_{3}$

$v_{3}=\frac{m_{1}v_{1}+m_{2}v_{2}}{m_{3} }=\frac{(102kg)(5.79m/s)+(103kg)(18.5m/s)}{(205kg)}$

$v_{3}=12.17m/s$

B. The change in velocity of both vehicles are:

For the car

$v_{car}=v_{f}-v_{o}=12.17m/s-5.79m/s=6.38m/s$

For the truck

$v_{truck}=12.17m/s-18.5m/s=-6.3m/s$

C. The change in kinetic energy is:

ΔK= $K_{2}-K_{1} =\frac{1}{2}m_{3}v_{3}^{2}-(\frac{1}{2}m_{1}v_{1}^{2}+\frac{1}{2}m_{2}v_{2}^{2})$

ΔK= $\frac{1}{2}(205)(12.17)^{2}-(\frac{1}{2}(102)(5.79)^{2}+\frac{1}{2}(103)(18.5)^{2})=-4.13x10^{3}J$

ΔK $=-4.13x10^{3}J$

6 0

3 years ago

The image shows the displacement of a motorboat. The data table shows the magnitudes of the components of each displacement vect

Diano4ka-milaya [45]

Rx= 3.5 km

Ry= 2.9 km

4 0

3 years ago

Read 2 more answers

When must scientific theories be changed

madreJ [45]

Answer:

when new information disproving the current theory becomes available.

Explanation:

hope this helps..

4 0

2 years ago

tamaranim1 [39]

the friction force provided by the brakes is 30000 N.

<h3>What is friction force?</h3>

Friction force is the force that opposes the motion between two bodies in contact.

To calculate the average friction force provided by the brakes, we apply the formula below.

Formula:

K.E = F'd............. Equation 1

Where:

K.E = Kinetic energy of the train
F' = Friction force provided by the brakes
d = distance

Make F' the subject of the equation

F' = K.E/d............ Equation 2

From the question,

Given:

K.E = 8.1×10⁶
d = 270 m

Substitute these values into equation 2

F' = (8.1 ×10⁶)/270
F' = 30000 N

Hence, the friction force provided by the brakes is 30000 N

Learn more about friction force here: brainly.com/question/13680415

8 0

2 years ago

You pull two blocks connected by a lightweight string across a horizontal table. Block 1 has a mass of 1.00 kg, and block 2 has

coldgirl [10]

Use google
-Just sayi’n don’t get mad

7 0

3 years ago