This problem is about the rate of the current. It's important to know that refers to the quotient between the electric charge and the time, that's the current rate.
Where Q = 2.0×10^−4 C and t = 2.0×10^−6 s. Let's use these values to find I.
<em>As you can observe above, the division of the powers was solved by just subtracting their exponents.</em>
<em />
<h2>Therefore, the rate of the current flow is 1.0×10^2 A.</h2>
<span>Organelles which are very important
in giving nutrients. During cellular respiration, the food molecules such as
glucose, are oxidized to carbon dioxide (CO2) and water (H2O) and trapped in
ATP (Adenosine triphosphate) form for further us of cell’s activities. ATP’s
are formed at mitochondria – the cell’s powerhouse. This type of organelle
takes and breaks nutrients absorbed by the cell and creates energy afterward.
The energy from ATP is then used by the body in kinetic activities like running
& walking or involuntary activities like breathing, blood circulation,
stimulus-responding, etc.</span>
Answer:
mass consumed by 235U each day = 2 kg
Explanation:
electrical power produced = 1 GW = 1 × 10⁹ × (6.24151 × 10¹⁸ ) eV
= 6.24151× 10²¹ MeV/s
thermal energy = 0.420 * 250 = 105 MeV
= 5.94 × 10¹⁹ fission/second
=5.94 × 10¹⁹× 24 × 60 ×60)
= 5.13 × 10²⁴ fission/day
mu = 235.04393 × 1.660× 10 ⁻²⁷ = 390.1729× 10⁻²⁷ Kg
M = mu ×5.13 × 10²⁴
= 390.1729× 10⁻²⁷ ×5.13 × 10²⁴
M = 2 kg(approx.)
mass consumed by 235U each day = 2 kg
Answer:
<h2> 27m/s</h2>
Explanation:
Given data
initital velocity u=15m/s
deceleration a=3m/s^2
time t= 4 seconds
final velocity v= ?
Applying the expression
v=u+at------1
substituting our data into the expression we have
v=15+3*4
v=15+12
v=27m/s
The velocity after 4 seconds is 27m/s
