Which trophic level usually has the LEAST energy available to it?

OlgaM077 [116]

The correct answer is (c.) resists a change in motion of an object. Inertia basically resists any physical changes in an object in terms of its state. As stated in the Newton's first law of motion or also known as law of inertia, an object that is at rest will stay at rest and an object that is in motion will stay in motion. In short, an object will keep doing what it is already doing UNLESS net force is acted upon it,

4 0

3 years ago

Assume that when you stretch your torso vertically as much as you can, your center of mass is 1.0 m above the floor. The maximum

Elenna [48]

1) 0.77 m

2) 0.23 m

Explanation:

1)

Here we want to find the time elapsed for crouching in order to jump and reach a height of 2.0 m above the floor, starting from 1.0 m above the floor.

First of all, we start by calculating the speed required to jump up to a height of 2.0 m. Since the total energy is conserved, the initial kinetic energy is converted into gravitational potential energy, so:

$\frac{1}{2}mv^2 = mgh$

where

m is the mass of the man

v is the speed after jumping

$g=9.8 m/s^2$ is the acceleration due to gravity

h = 2.0 - 1.0 = 1.0 m is the change in height

Solving for v,

$v=\sqrt{2gh}=\sqrt{2(9.8)(1.0)}=4.43 m/s$

In the acceleration phase, we know that the initial velocity is

$u=0$

And the force exerted on the floor is 2.3 times the gravitational force, so

$F=2.3 mg$

This means the net force on you is

$F_{net} = F-mg=2.3mg-mg=1.3 mg$

because we have to consider the force of gravity acting downward.

So the acceleration of the man is

$a=\frac{F_{net}}{m}=\frac{1.3mg}{m}=1.3g$

Now we can use the following suvat equation to find the displacement in the acceleration phase, which is how low the man has to crouch in order to jump:

$v^2-u^2=2as$

where s is the quantity we want to find. Solving for s,

$s=\frac{v^2-u^2}{2a}=\frac{4.43^2-0}{2(1.3g)}=0.77 m$

2)

At the beginning, we are told that the height of the center of mass above the floor is

h = 1.0 m

During the acceleration phase and the crouch, the height of the center of mass of the body decreases by

$\Delta h = -0.77 m$

This means that the lowest point reached by the center of mass above the floor during the crouch is

$h'=h+\Delta h = 1.0 - 0.77 = 0.23 m$

This value seems unpractical, since it is not really easy to crouch until having the center of mass 0.23 m above the ground.

3 0

3 years ago

Acceleration is often measured in what unit? A. newtons B. kilograms C. meters per second D. meters per second squared

spayn [35]

D. meters per second squared

3 0

3 years ago

Read 2 more answers

The change in motion (acceleration) of an object depends on

7nadin3 [17]

Answer:

BOTH the size of the force AND the mass of the object

Explanation:

Acceleration of an object is the rate of change of its velocity.

The relation between force, mass and acceleration is given by the formula as follows :

F = ma

m is mass

a is acceleration

It would mean that the change in motion or the acceleration of an object depends on both the size of the force and the mass of the object. Hence, the correct option is (c).

8 0

3 years ago

A 72.0 kg stuntman jumps from a moving car to a 2.50 kg skateboard at rest. If the velocity of the car is 15.0 m/s to the east w

yawa3891 [41]

Answer:

B 14.5 m/s to the east

Explanation:

We can solve this problem by using the law of conservation of momentum.

In fact, if the system is isolated, the total momentum of the system must be conserved.

Here the total momentum before the stuntman reaches the skateboard is:

$p_i = Mv$

where

M = 72.0 kg is the mass of the stuntman

v = 15.0 m/s is his initial velocity (to the east)

The total momentum after the stuntmen reaches the skateboard is:

$p_f = (M+m)v'$

where

m = 2.50 kg is the mass of the skateboard

v' is the final velocity of the stuntman and the skateboard

Since momentum must be conserved, we have

$p_i = p_f\\Mv=(M+m)v'$

And solvign for v',

$v'=\frac{Mv}{M+m}=\frac{(72.0)(15.0)}{(72.0+2.50)}=14.5 m/s$

And since the sign is the same as v, the direction is the same (to the east).

7 0

3 years ago