Answer:
The drift velocity is 
Explanation:
Given :
Area of metallic wire,
.
Current through wire , 
Mobile charge density , 
Charge value , 
We need to find drift velocity , 
Now, we know :

Therefore, 
Putting all given values in above equation we get,


Hence, this is the required solution.
Answer:
Explanation:
Given that
F(x) = (2.00 N/m) x +(1.00 N/m3) x3.
Which can be written as
F(x) = 2x +x³ N/m
Potential energy from x=1 to x=2
P.E is given as
P.E=∫F(x)dx
P.E=∫2x+x³ dx. From x=1 to x=2
P.E= 2x²/2 + x⁴/4 from x=1 to x=2
P.E= x² + x⁴/4 from x=1 to x=2
P.E=(2²+2⁴/4)-(1² +1⁴/4)
P.E=(4+4)-(1+1/4)
P.E, =8-5/4
P.E=27/4
P.E=6.75J
Since it is moving from lower position to upper position, then P.E should be negative
P.E = - 6.75J
2*3.5 = 7m/s
You multiply the acceleration per the time (they both are in seconds, otherwise, you should set them in the same units).
496/1127 = 0.44 = 44%
<span>sin A = 0.85/2.1. </span>
<span>A = 23.9o. </span>
<span>Fp = 1127 sin23.9 = 457 N. = Force parallel to the ramp. </span>
<span>Fn = 1127 Cos23.9 = 1,030 N. = Force </span>
<span>perpendicular to the ramp = Normal force. </span>
<span>Eff. = Fp/Fap = 457/496 = 0.92 = 92%
Correct answer: 92%</span>
Answer:
Position of object is;
s(t) = 4t³/3 + 3t + 1
Explanation:
We are told that the velocity has an expression;
v(t) = 3.00 m/s + ( 4.00 m/s³)t²
Now, to get the expression for the position(s(t)) of the object, we have to integrate the velocity expression. Thus;
s(t) = ∫3 + 4t²
s(t) = 3t + 4t³/3 + c
Now, we were told that at x = 1.00 m, time t = 0.000 s
Thus, plugging the values in;
1 = 3(0) + 4(0³/3) + c
c = 1
Thus,the expression for the position of the object is;
s(t) = 4t³/3 + 3t + 1