<span>Let Q be the charge, thus Q = -20.0 µC.</span>
Define D as the distance between the center of the rod and the specified point. Therefore,
D=0.32 - 0.12 = 0.2 m
<span>L = 0.12 m, which represents the length of the rod
</span><span>To find the magnitude and direction of the electric field along the axis of the rod at a point 32.0 cm from its center, use the formula:
</span><span>E = K·Q/r²
</span>or<span>E = kQ/D(D+L), where k</span> is a constant equal to 8.99 x 10<span>9</span> N m
2/C2.<span>Consequently,[TAG_21]]E=(</span>8.99 x 109 N m2/C2.* (-20.0 µC))/(<span>0.2 m*0.32m)</span><span>
</span>
Answer:
19.62 ms
Explanation:
t = Time taken = 2 s
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration due to gravity = 9.81 m/s² (we take downward direction as positive)
Using the equations of motion
The pebble's speed upon contact with the water is 19.62 ms
The object's density is 8000 kg/m^3. The object's weight in air is 7.84 N while it measures 6.86 N when submerged in water, where the density of water is 1000 kg/m^3. According to Archimedes' principle, an immersed object experiences an upward buoyant force equivalent to its loss of weight in the fluid. By calculating the weight difference (7.84 - 6.86 = 0.98 N) and employing the standard equations relating density and volume, we find that 10^-4 m^3 corresponds to a density of 8000 kg/m^3.
Answer:
consult a teacher
Explanation:
go to your school
locate the teacher
request assistance from him/her
complete the question and you're done:)