Answer:
The magnitude of the electric force on a protein with this charge is 
Explanation:
Given that,
Electric field = 1500 N/C
Charge = 30 e
We need to calculate the magnitude of the electric force on a protein with this charge
Using formula of electrostatic force
Where, F = force
E = electric field
q = charge
Put the value into the formula
Hence, The magnitude of the electric force on a protein with this charge is
Answer:
-8.56V
Explanation:
Our values are given by,
e = 6.04 V
Φ = 30.3
VC = 5.32
We can calculate the voltage across the circuit with the emf formula, that is,
Now, Using Kirchoff Voltage Law,
Finally we have the potential difference across the inductor.
Answer:
The rise from A to B is 0.887
Solution:
As per the question:
The following reading of an inverted staff is given as:
A = 2.915
B = -2.028
Here, for inverted staff, the greater reading shows greater elevation and lesser reading shows lower elevation.
Thus
The rise from A to B is given as:
A - B = 2.915 - 2.028 = 0.887
We use the work formula to solve for the unknown in the problem. The formula for work is expressed as the product of the net force and the distance traveled by the object. We were given both the force and the distance so we can solve work directly.
Work = 250 N x 50 m = 12500 J
Thus, the answer is C.
