Felectric = q*E
<span> Ftranslational = m*a
</span><span> Felectric = Ftranslational
</span> <span>q*E = m*a
</span><span> Solve for a
</span><span> a = q/m*E </span>
<span> Our sign convention is "up is positive"
</span><span> q = 1.6*10^-19 C
</span><span> m = 1.67*10^-27 kg
</span><span> E = -150 N/C (- because it is down and up is positive)
</span> a =<span>
-6,4*10^5</span><span> m/s^2 (downward)
</span> answer
a = -6,4*10^5 m/s^2 (downward)
Complete Question
An aluminum "12 gauge" wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The electric field in the wire changes with time as E(t)=0.0004t2−0.0001t+0.0004 newtons per coulomb, where time is measured in seconds.
I = 1.2 A at time 5 secs.
Find the charge Q passing through a cross-section of the conductor between time 0 seconds and time 5 seconds.
Answer:
The charge is 
Explanation:
From the question we are told that
The diameter of the wire is 
The radius of the wire is 
The resistivity of aluminum is 
The electric field change is mathematically defied as

Generally the charge is mathematically represented as

Where A is the area which is mathematically represented as

So

Therefore

substituting values
![Q = 120 \int\limits^{t}_{0} { [ 0.0004t^2 - 0.0001t +0.0004] } \, dt](https://tex.z-dn.net/?f=Q%20%3D%20120%20%5Cint%5Climits%5E%7Bt%7D_%7B0%7D%20%7B%20%5B%200.0004t%5E2%20-%200.0001t%20%2B0.0004%5D%20%7D%20%5C%2C%20dt)
![Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] } \left | t} \atop {0}} \right.](https://tex.z-dn.net/?f=Q%20%3D%20120%20%5B%20%5Cfrac%7B0.0004t%5E3%20%7D%7B3%7D%20-%20%5Cfrac%7B0.0001%20t%5E2%7D%7B2%7D%20%2B0.0004t%5D%20%7D%20%20%5Cleft%20%7C%20t%7D%20%5Catop%20%7B0%7D%7D%20%5Cright.)
From the question we are told that t = 5 sec
![Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] } \left | 5} \atop {0}} \right.](https://tex.z-dn.net/?f=Q%20%3D%20120%20%5B%20%5Cfrac%7B0.0004t%5E3%20%7D%7B3%7D%20-%20%5Cfrac%7B0.0001%20t%5E2%7D%7B2%7D%20%2B0.0004t%5D%20%7D%20%20%5Cleft%20%7C%205%7D%20%5Catop%20%7B0%7D%7D%20%5Cright.)
![Q = 120 [ \frac{0.0004(5)^3 }{3} - \frac{0.0001 (5)^2}{2} +0.0004(5)] }](https://tex.z-dn.net/?f=Q%20%3D%20120%20%5B%20%5Cfrac%7B0.0004%285%29%5E3%20%7D%7B3%7D%20-%20%5Cfrac%7B0.0001%20%285%29%5E2%7D%7B2%7D%20%2B0.0004%285%29%5D%20%7D)

Answer:
= 829.69 Watt
≅ 830 Watt
Explanation:
Given that,
Velocity of air flow = 12.5m/s
Rate of flow of air = 9m³/s
Density of air = 1.18kg/m³
power by kinetic energy = 1/2(mv²)
mass = density × volume
m = 1.18 × 9
= 10.62 kg/s
power = 1/2 mV²
= 1/2 (10.62 × 12.5²)
= 829.69 Watt
≅ 830 Watt
Flow rate
u
=
9
m
3
/
s
Velocity of the air
V
=
8
m/s
Density of the air
ρ
=
1.18
kg
/
m
3
Note that
1 yd = 0.9144 m
Therefore,
The length of an American Football field is
(100 yds)*(09144 m/yd) = 91.44 m
Because the soccer field is 110 m long, its length exceeds the American Football Field by
100 - 91.44 = 8.56 m
or
(8.56/.9144) = 9.36 yd
This difference is equivalent to (8.56/91.44)*100 = 9.4%
Answer:
The Soccer Field is longer by
8.56 m, or
9.36 yd, or
9.4%
Answer:
The objects must have the same acceleration and the objects must exert the same magnitude force on each other.
Explanation:
The objects must have the same weight: FALSE. This is not needed, any two object can move together in contact no matter their mass.
The objects must have the same acceleration: TRUE. If they have different accelerations, they will separate since the distance each of them travel at a given time will be different.
The objects must have the same net force acting on them: FALSE. This is not needed, since what matters is acceleration, and a=F/m, so if both objects have different net force acting on them, they could have different masses also to compensate and result in the same acceleration.
The objects must exert the same magnitude force on each other: TRUE, this is the 3rd Newton Law, an action must follow the same reaction.