It is the mitochondria of a cell that stores energy for a quick release. <span>Mitochondria break down glucose to release the energy for cells to use. Hope this answers the question. Have a nice day. Feel free to ask more questions.</span>

5 0

3 years ago

Serjik [45]

Given the distance traveled and time elapsed, the average speed of the train is approximately 26.944m/s.

<h3>What is the average speed of the train?</h3>

Speed is simply referred to as distance traveled per unit time.

Mathematically, Speed = Distance ÷ time.

Given the data in the question;

Distance traveled = 221miles
Elapsed time = 3 hours and 40 minutes

First we convert miles to meters and Hours minutes to seconds.

221 miles = ( 221 × 1609.344 )m = 355665.024 meters

3 hours and 40 minutes = ( 3×60×60)s + ( 40×60)s

= 10800s + 2400s

= 13200s

Now, determine the average speed.

Speed = Distance ÷ time

Speed = 355665.024m / 13200s

Speed = 26.944m/s

Given the distance traveled and time elapsed, the average speed of the train is approximately 26.944m/s.

Learn more about speed here: brainly.com/question/7359669

#SPJ1

4 0

2 years ago

An electron is projected with an initial speed of 5 3.2 10 / ⋅ m s directly toward a proton that is fixed in place. If the elect

Dvinal [7]

Answer:

The distance is $d =1.66*10^{-9}m$

Explanation:

From the question we are told that

The initial speed of the electron is $v_i = 3.2 *10^5 m/s$

The mass of electron is $m = 9.11*10^{-31}kg$

Let $d$ be the distance between the electron and the proton when the speed of the electron instantaneously equal to twice the initial value

Let $KE_i$ be the initial kinetic energy of the electron \

Let $KE_d$ be the kinetic energy of the electron at the distance $d$ from the proton

Considering that energy is conserved,

The energy at the initial position of the electron = The energy at the final position of the electron

i.e

$KE_i +U_1 = KE_d + U_2$

$U_1 \ and \ U_2$ are the potential energy at the initial position of the electron and at distance d of the electron to the proton

Here $U_1 = 0$

So the equation becomes

$\frac{1}{2} mv_i^2 = \frac{1}{2} mv_d^2 + \frac{kq_1 q_2}{d}$

Here $q_1 \ and \ q_2$ are the charge on the electron and the proton and their are the same since a charge on an electron is equal to charge on a proton