Electric field due to a charged rod along its axis is given by
here we know that
L = 14 cm
r = distance from end of rod
r = 36 - 7 = 29 cm
Q = 222 mC
now we will have
Answer:
d = 0.343 m
Explanation:
Given that,
The speed of a longitudinal wave, v = 343 m/s
Your ear is capable of differentiating sounds that arrive at the ear just 1.00 milliseconds apart.
We need to find the minimum distance between two speakers that produce sounds that arrive at noticeably different times.
Let the distance be d. So,
So, the required distance is equal to 0.343 m.
To solve this problem it is necessary to apply the definition of Young's Module which states that
Where,
F = Force
A = Cross sectional Area
L = Length
= Initial Length
We need to find the ratio between the two values when the another values are constant, that is
Re-arrange to find
Therefore the bone stretch around
Answer:
0.795 kg
Explanation:
Assuming the complete question:
A person with compromised pinch strength in their fingers can
only exert a normal force of 6.0 N to either side of a pinch-held object. What is the mass of the heaviest book this person can
hold onto vertically before it slips out of his or her fingers? The coefficient of static friction of the surface between the fingers and the book cover is 0.60
SOLUTION:
The maximum weight of the book will equal the maximal friction force that can be produced:
m g = 2 f
Note that there are two sides of the book, so the friction force equals 2 times the friction force on one side (hence the factor 2).
So the maximum mass of the book is
m = 2 f / g
m = 20.65 6.0N / (9.81N/kg)
m = 0.795 kg
Answer:
k = 44000 N/m
Explanation:
Given the following data;
Maximum gravitational potential energy = 770 J
Elastic potential energy = 14% of 770 J = 14/100 * 770 = 107.8 J
Extension, x = 42 - 35 = 7cm to meters = 7/100 = 0.07 m
To find the spring constant, k;
The elastic potential energy of an object is given by the formula;
Substituting into the equation, we have;
Cross-multiplying, we have;
k = 44000 N/m