Given:
The force of attraction is F = 48.1 N
The separation between the charges is

Also, the magnitude of charge q1 = q2 = q.
To find the magnitude of charge.
Explanation:
The magnitude of charge can be calculated by the formula

Here, k is the Coulomb's constant whose value is

On substituting the values, the magnitude of charge will be

Thus, the magnitude of each charge is 9.91 x 10^(-4) micro Coulombs.
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Answer:
94.67 N
Explanation:
Consider a free body diagram with force, F of 41 N applied at an angle of 37 degrees while the weight acts downwards. Resolving the force into vertical and horizontal components, we obtain a free body diagram attached.
At equilibrium, normal reaction is equal to the sum of the weight and the vertical component of the force applied. Applying the condition of equilibrium along the vertical direction.

Substituting 70 N for W, 41 N for F and
for 37 degrees
N=70+41sin37=94.67441595 N and rounding off to 2 decimal places
N=94.67 N
The density of the object is approximately 1.91 kg per m³.
42 kg is a measure of mass, and 22 m³ is a measure of volume. Knowing this, you can use the relationship

to solve for the object's density.
42 kg

22 m³

1.91 kg per m³.