The elasticity of a polymer is primarily due to the structure of the molecule and the cross-linking between strands. Hydrogen bonding is a contributor to the shape of the molecule, but not a major player in terms of elasticity. We would have to answer "false".
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Answer:
570 N
Explanation:
Draw a free body diagram on the rider. There are three forces: tension force 15° below the horizontal, drag force 30° above the horizontal, and weight downwards.
The rider is moving at constant speed, so acceleration is 0.
Sum of the forces in the x direction:
∑F = ma
F cos 30° - T cos 15° = 0
F = T cos 15° / cos 30°
Sum of the forces in the y direction:
∑F = ma
F sin 30° - W - T sin 15° = 0
W = F sin 30° - T sin 15°
Substituting:
W = (T cos 15° / cos 30°) sin 30° - T sin 15°
W = T cos 15° tan 30° - T sin 15°
W = T (cos 15° tan 30° - sin 15°)
Given T = 1900 N:
W = 1900 (cos 15° tan 30° - sin 15°)
W = 570 N
The rider weighs 570 N (which is about the same as 130 lb).
<h2><u>Question</u><u>:</u><u>-</u></h2>
Ryan applied a force of 10N and moved a book 30 cm in the direction of the force. How much was the work done by Ryan?
<h2><u>Answer:</u><u>-</u></h2>
<h3>Given,</h3>
=> Force applied by Ryan = 10N
=> Distance covered by the book after applying force = 30 cm
<h3>And,</h3>
30 cm = 0.3 m (distance)
<h3>So,</h3>
=> Work done = Force × Distance
=> 10 × 0.3
=> 3 Joules

Answer:4
Explanation:
Given
Frequency of beats is 3 beats per second
Frequency of first tuning fork is 
Beat frequency is difference in the frequency of tuning forks i.e.
either
or 
so 
or

so there is not enough information given to decide.
let us consider that the two charges are of opposite nature .hence they will constitute a dipole .the separation distance is given as d and magnitude of each charges is q.
the mathematical formula for potential is 
for positive charges the potential is positive and is negative for negative charges.
the formula for electric field is given as-
for positive charges,the line filed is away from it and for negative charges the filed is towards it.
we know that on equitorial line the potential is zero.hence all the points situated on the line passing through centre of the dipole and perpendicular to the dipole length is zero.
here the net electric field due to the dipole can not be zero between the two charges,but we can find the points situated on the axial line but outside of charges where the electric field is zero.
now let the two charges of same nature.let these are positively charged.
here we can not find a point between two charges and on the line joining two charges where the potential is zero.
but at the mid point of the line joining two charges the filed is zero.