The angular velocity of the wheel at the bottom of the incline is 4.429 rad/sec
The angular velocity (ω) of an object is the rate at which the object's angle position is changing in relation to time.
For a wheel attached to an incline angle, the angular velocity can be computed by considering the conservation of energy theorem.
As such the total kinetic energy (K.E) and rotational kinetic energy (R.K.E) at a point is equal to the total potential energy (P.E) at the other point.
i.e.
P.E = K.E + R.K.E
Therefore, we can conclude that the angular velocity of the wheel at the bottom of the incline is 4.429 rad/sec
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Answer:
240 V
Explanation:
number of turns in primary coil, Np = 10
Number of loops in secondary coil, Ns = 20
Voltage in primary coil, Vp = 120 V
Let the voltage in secondary coil is Vs.
So, Vs / Vp = Ns / Np
Vs / 120 = 20 / 10
Vs / 120 = 2
Vs = 240 V
Thus, the voltage in secondary coil is 240 Volt.
We have to choose the correct name for the zone along the southern margins of the Sahara. Laterite is a soil in hot and wet tropical areas. Savanna is the tropical grassland. It has tropical savanna climate. Veldt is name for the areas in the South Africa. Finally, the Sahel is the zone along the south margins of the Sahara. It has a semi-arid climate. The Arabic word "sahel" means "coast". Answer: C. Sahel. <span> </span>
Answer: q = 2.781e-9C = 2.781nC
E=200C
Explanation:
E = Qd/(2πEor^3)
Where
E=Electric field intensity
Q=Charge
d=distance between the dipole=0.008m
Eo=permitivitty
400 N/C = Q(0.80e-2 m)/(2πε*(10e-2 m)^3)
Q= (400* 2* 3.142 * 8.85 x 10-12 * 0.1^3)/0.008
q = 2.781e-9C = 2.781nC
b)
Though the dipole are two separate charges. And since the point is on the x-axis, the electric field strengths are equivalent. The magnitude of the vector sum is:
E = kq*2sin θ/r^2
= 2(8.99e9 N*m^2/C^2)(2.781e-9 C)*sin(arctan(.4/10))/(10e-2 m)^2
= 2(8.99e9) * (2.781e-9) * sin(2.290)/(10e-2 m)^2
=200 C
let's use first equation of motion to solve this ;
Velocity after 30 seconds = 170 m/s
