Answer:
3.67 N
Explanation:
From the question given above, the following data were obtained:
Charge of 1st object (q₁) = +15.5 μC
Charge of 2nd object (q₂) = –7.25 μC
Distance apart (r) = 0.525 m
Force (F) =?
Next, we shall convert micro coulomb (μC) to coulomb (C). This can be obtained as follow:
For the 1st object
1 μC = 1×10¯⁶ C
Therefore,
15.5 μC = 15.5 × 1×10¯⁶
15.5 μC = 15.5×10¯⁶ C
For the 2nd object:
1 μC = 1×10¯⁶ C
Therefore,
–7.25 μC = –7.25 × 1×10¯⁶
–7.25 μC = –7.25×10¯⁶ C
Finally, we shall determine the force. This can be obtained as follow:
Charge of 1st object (q₁) = +15.5×10¯⁶ C
Charge of 2nd object (q₂) = –7.25×10¯⁶ C
Distance apart (r) = 0.525 m
Electrical constant (K) = 9×10⁹ Nm²/C²
Force (F) =?
F = Kq₁q₂ / r²
F = 9×10⁹ × 15.5×10¯⁶ × 7.25×10¯⁶ / 0.525²
F = 3.67 N
Therefore, the force on the object is 3.67 N
Dr.phil can help you with that answer also the answer is 21.3
Answer:
a) At x=14 the slope will be given by:
.
b) Then, the angle between the line and the pole will be:

where
is the angle between the tangent to the catenary and the x-axis.
Explanation:
The catenary has the following general form:

a) The slope at any point will be given by the derivative of y.

At x=14:
.
b) The angle between the tangent to the catenary and the x-axis at a given point will be given by:
⇒ 
Then, the angle between the line and the pole will be:
.
Answer:
10m/s
Explanation:

Since there is no initial velocity as the object is dropped, you can write the following equation:

Now that you know how long the fall took, you can use another physics equation to find the velocity at that point.

Since there once again is no initial velocity, you can rewrite this as:

Hope this helps!