This is somewhat misleading, and I encountered the same question in my homework. An electric field strength of 1*10^5 N/C is provided, along with a drag force of 7.25*10^-11 N, and the critical detail is that it maintains a constant velocity, indicating that the particle is in equilibrium and not accelerating.
<span>To solve, utilize F=(K*Q1*Q2)/r^2 </span>
<span>You'll want to equate F with the drag force, where the electric field strength translates to (K*Q2)/r^2; substituting the values results in </span>
<span>(7.25*10^-11 N) = (1*10^5 N/C)*Q1 ---> Q1 = 7.25*10^-16 C </span>
The answer is 9938.8 km. Explanation: 1 pound-force = 4.48 N. Hence, 30.0 pounds-force = 134.4 N. The gravitational force between Earth and an object on its surface is defined by: Where M denotes Earth’s mass, m is the object's mass, and R represents the Earth's radius (6371 km). To determine height (h) above Earth's surface, we compare ratios. Ultimately, Pete's weight would be 30 pounds at a height of 9938.8 km from the Earth's surface.
Explanation:
Attached is a document that contains the solution.
Answer:
force = 6.53× 
N
Explanation:
Provided data
downward force = 0.60 m
mass m = kg
distance h = 0.40 m
to determine
magnitude of the downward force
solution
we know here mg is apply 0.4 m away from support and
thus applied force is d = 0.6 m from support
therefore
by balancing torque we can compute force
as
force = mass × g × h / d
substituting the values
force = mass × g × h / d
force = ( × 9.81 × 0.4 ) / 0.6
force = 6.53× N
