Answer:
Explanation:
Given that,
The spring constant of spring 1, 
The motion of the object on spring 1 has twice the amplitude as the motion of the object on spring 2, 
As the magnitude of the maximum velocity is the same in each case, it means the maximum kinetic energy is same in each case. In other words, the total energy is same.
So, the spring constant of spring 2 is 920 N/m. Hence, this is the required solution.
Answer:
k=320N/m
Explanation:
Step one:
given data
Let the initial/equilibrum position be x
mass m1= 0.2kg
F1= 0.2*10= 2N
elongation e= 9.5cm= 0.095m
mass m2=1kg
F2=1*10= 10N
elongation e= 12cm= 0.12m
Step two:
From Hooke's law, which states that provided the elastic limits of a material is not exceeded the extention e is proportional to applied Force F
F=ke
2=k(0.095-a)
2=0.095k-ka----------1
10=k(0.12-a)
10=0.12k-ka----------2
solving equation 1 and 2 simultaneously
10=0.12k-ka----------2
- 2=0.095k-ka----------1
8=0.025k-0
divide both side by 0.025
k=8/0.025
k=320N/m
Answer;
- This statement about matter and its behavior is best classified as a Law.
-It is the Law of Universal Gravitation.
Explanation;
-The Law of Universal Gravitation states that every point mass attracts every other point mass in the universe by a force pointing in a straight line between the centers-of-mass of both points, and this force is proportional to the masses of the objects and inversely proportional to their separation.This attractive force always points inward, from one point to the other.
-The Law applies to all objects with masses, big or small. Two big objects can be considered as point-like masses, if the distance between them is very large compared to their sizes or if they are spherically symmetric.
A, something to remember is that if the numbers are even the answer too will be even. Every time hope this helps :>
Answer:
E
Explanation:
There is not enough information in the question. A beat is the observed differences between two frequencies.
Hence, 3 beat represents a difference of 3-hertz between the two frequencies. So the frequency of the piano string could be 1056+3 = 1059-hertz or 1056-3= 1053-hertz.
In order to ascertain the actual value i.e if it is an increase or a decrease in frequency observed, we need more information.
