Answer:
Scientific Notation: 3.45 x 10^5
E Notation: 3.45e5
Answer:
2 seconds
Explanation:
The function of height is given in form of time. For maximum height, we need to use the concept of maxima and minima of differentiation.
![h(t)=-16t^{2}+64t+112](https://tex.z-dn.net/?f=h%28t%29%3D-16t%5E%7B2%7D%2B64t%2B112)
Differentiate with respect to t on both the sides, we get
![\frac{dh}{dt}=-32t+64](https://tex.z-dn.net/?f=%5Cfrac%7Bdh%7D%7Bdt%7D%3D-32t%2B64)
For maxima and minima, put the value of dh / dt is equal to zero. we get
- 32 t + 64 = 0
t = 2 second
Thus, the arrow reaches at maximum height after 2 seconds.
The current is defined as the amount of charge Q that passes through a given point of a wire in a time
![\Delta t](https://tex.z-dn.net/?f=%5CDelta%20t)
:
![I= \frac{Q}{\Delta t}](https://tex.z-dn.net/?f=I%3D%20%5Cfrac%7BQ%7D%7B%5CDelta%20t%7D%20)
Since I=500 A and the time interval is
![\Delta t=4.0 min=240 s](https://tex.z-dn.net/?f=%5CDelta%20t%3D4.0%20min%3D240%20s)
the charge is
![Q=I \Delta t=(500 A)(240 s)=1.2 \cdot 10^5 C](https://tex.z-dn.net/?f=Q%3DI%20%5CDelta%20t%3D%28500%20A%29%28240%20s%29%3D1.2%20%5Ccdot%2010%5E5%20C)
One electron has a charge of
![q=1.6 \cdot 10^{-19}C](https://tex.z-dn.net/?f=q%3D1.6%20%5Ccdot%2010%5E%7B-19%7DC)
, therefore the number of electrons that pass a point in the wire during 4 minutes is
![N= \frac{Q}{q}= \frac{1.2 \cdot 10^5 C}{1.6 \cdot 10^{-19}C}=7.5 \cdot 10^{23}](https://tex.z-dn.net/?f=N%3D%20%5Cfrac%7BQ%7D%7Bq%7D%3D%20%5Cfrac%7B1.2%20%5Ccdot%2010%5E5%20C%7D%7B1.6%20%5Ccdot%2010%5E%7B-19%7DC%7D%3D7.5%20%5Ccdot%2010%5E%7B23%7D%20%20)
electrons
Answer:
206.8965517 n
Explanation:
First, we need to see that 60:29 is 2.078965517:1. Then we need to multiply the energy put 29 cm from the fulcrum by 2.078965517, giving us the end result of our answer.
The final velocity of the red barge in the collision elastic is 0.311 m/s when it collides with blue barge pf mass 1000000 kg.
Final velocity(v3) of the red barge is calculated by following formula
m1×v1+ m2×v2= (m1+m2)v3
Substituting the value of m1= 150000 kg, v1= 0.25 m/s, m2= 1000000 kg, v2= 0.32 m/s
150000 × 0.25+ 1000000×0.32= (150000+1000000)×v3
37500+ 320000= 1150000×v3
357500= 1150000×v3
v3= 0.311 m/s
<h3>What is elastic collision velocity? </h3>
- The velocity of the target particle after a head-on elastic impact in which the projectile is significantly more massive than the target will be roughly double that of the projectile, but the projectile velocity will remain virtually unaltered.
For more information on elastic collision velocity kindly visit to
brainly.com/question/29051562
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