Answer:
μ = 0.408
Explanation:
given,
speed of the automobile (u)= 20 m/s
distance = 50 m
final velocity (v) = 0 m/s
kinetic friction = ?
we know that,
v² = u² + 2 a s
0 = 20² + 2 × a × 50

a = 4 m/s²
We know
F = ma = μN
ma = μ mg
a = μ g


μ = 0.408
hence, Kinetic friction require to stop the automobile before it hit barrier is 0.408
Answer:
If R₂=25.78 ohm, then R₁=10.58 ohm
If R₂=10.57 then R₁=25.79 ohm
Explanation:
R₁ = Resistance of first resistor
R₂ = Resistance of second resistor
V = Voltage of battery = 12 V
I = Current = 0.33 A (series)
I = Current = 1.6 A (parallel)
In series

In parallel


Solving the above quadratic equation


∴ If R₂=25.78 ohm, then R₁=10.58 ohm
If R₂=10.57 then R₁=25.79 ohm
Prior to touching the bar magnet, the magnetic domains in the nail were pointing in random directions. When Taylor touched the nail to the bar magnet the magnetic fields of the magnetic domains aligned and it became a temporary magnet.
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)
Answer:
Part a)

Part b)

Explanation:
Part a)
Electric field due to large sheet is given as







now the electric field is given as


Part b)
Now since the electric field is required at same distance on other side
so the field will remain same on other side of the plate
