It is very important that there is only one possible explanation for experimental data because If more than one explanation is possible, a conclusion about of them cannot be reliable.
hope this helps
Answer:
The resultant vector
is given by
.
Explanation:
Let
and
, both measured in meters. The resultant vector
is calculated by sum of components. That is:
(Eq. 1)


![\vec R = 3.196\,\hat{i}-0.464\,\hat{j}\,\,\,[m]](https://tex.z-dn.net/?f=%5Cvec%20R%20%3D%203.196%5C%2C%5Chat%7Bi%7D-0.464%5C%2C%5Chat%7Bj%7D%5C%2C%5C%2C%5C%2C%5Bm%5D)
The resultant vector
is given by
.
Answer:
kinetic energy is 2, 4, 50 thirties, 2.45 into 10, raise to three Julie.
Explanation:
In the given question, we are told that what is the kinetic energy of mass M equals 0.1 kg bullet traveling at a velocity Velocity is given and 700 m/s. So we know that kinetic energy mm-hmm k equals one half m v squared. So this will be mass is given 0.1 and velocity is 700 so 700 square this is one half 0.1 in two 49 double zero, double zero. This is one-half into 49 double zero. So kinetic energy is 2, 4, 50 thirties, 2.45 into 10, raise to three Julie. This is kinetic energy. Thank you.
Explanation:
Given that,
Mass of the object, m = 7.11 kg
Spring constant of the spring, k = 61.6 N/m
Speed of the observer, 
We need to find the time period of oscillation observed by the observed. The time period of oscillation is given by :

Time period of oscillation measured by the observer is :

So, the time period of oscillation measured by the observer is 5.79 seconds.
Newton's First Law states that an object will remain at rest or in uniform motion in a straight line unless acted upon by an external force.