What energy does a hamburger have when you eat it

Answer:

momentum = mass x velocity

Explanation:

So ultimately, find the mass of the object (be careful since this is not the same as weight. Weight is your gravitational pull to the earth. This means that your weight different amounts on different planets. But mass stays the same on all planets, it's what most people call their weight.) and then find the velocity (2 meters/second north, etc.)

Now you can apply this to your problem

Answer:

Explanation:

* For cylinders, it is unique. Since you have two circular bases, you take half the mass in the formula:

I am joyous to assist you anytime.

Answer:

778 J

Explanation:

Rotational energy is:

RE = ½ Iω²

For a solid cylinder I = ½ mr².

Rolling without slipping means ω = v/r.

RE = ½ (½ mr²) (v/r)²

RE = ¼ mv²

Plug in values:

RE = ¼ (34.5 kg) (9.5 m/s)²

RE ≈ 778 J

Round as needed.

Answer:

The pickup truck and hatchback will meet again at 440.896 m

Explanation:

Let us assume that both vehicles are at origin at the start means initial position is zero i.e. = 0. Both the vehicles will cross each other at same time so we will make equations for both and will solve for time.

Truck:

= 33.2 m/s, a = 0 (since the velocity is constant), = 0

Using

s = 33.2t .......... eq (1)

Hatchback:

, = 0 m/s (since initial velocity is zero), = 0

Using

putting in the data we will get

now putting 's' value from eq (1)

which will give,

t = 13.28 s

so both vehicles will meet up gain after 13.28 sec.

putting t = 13.28 in eq (1) will give

s = 440.896 m

So, both vehicles will meet up again at 440.896 m.

Answer:

• 440.9 m

Explanation:

initial speed of the pickup truck (Up) = 33.2 m/s

acceleration of the pickup truck (ap) = 0

initial speed of the hatchback = 0

acceleration of the hatchback (ah) = 5 m/s^{2}

how far away (s) do the cars meet up again?

from the equations of motion distance covered (s) = ut +

distance covered by the pickup = ut +

where

• u = initial speed of the pickup = 33.2 m/s
• t = time it takes
• a= acceleration of the pickup = 0
• the distance covered by the pickup (s) now becomes = 33.2t + = 33.2t ...equation 1

distance covered by the hatchback = ut +

where

• u = initial speed of the hatchback = 0 m/s
• t = time it takes
• a= acceleration of the hatchback = 5 m/s^{2}
• the distance covered by the hatchback (s) now becomes = (0)t +

        =  ......equation 2

when the cars meet, they both would have covered the same distance, therefore

• distance covered by the pickup = distance covered by the hatchback
• equation 1 = equation 2

• 33.2t =
• 33.2 = 2.5t
• t = 13.28 s

now that we have the time it takes for both cars to meet, we can put the value of the time into equation 1 or equation 2 to get the distance at which they meet

from equation 1

• distance (s) = 33.2t = 33.2 x 13.28= 440.9 m

from equation 2

• distance (s) =   = = 440.9 m