Answer:
The acceleration of the collar is 10 m/s²
Explanation:
Given;
mass of the collar, m = 1 kg
applied force on the bar, F = 10 N
The acceleration of the collar can be calculated by applying Newton's second law of motion;
F = ma
where;
F is the applied force
m is mass of the object
a is the acceleration
a = F / m
a = 10 / 1
a = 10 m/s²
Therefore, the acceleration of the collar is 10 m/s²
Answer:
The answer to your question is:
Explanation:
Data
mass = 4.33 kg
E = 41.7 J
v = ?
Formula
Ke = (1/2)mv²
Clear v from the equation
v = √2ke/m
Substitution
v = √2(41.7)/4.33
v = 19.26 m/s Result
Answer: 16.22 m/s^2
Explanation: g= GM/r^2 G= (6.67x 10^-11) M= 1.66(6x 10^24) r=(6400x 10^3) so
((6.67x10^-11)(1.66x 6x 10^24))/ (6400x10^3)^2 = 16.22 m/s^2
The answer to the given question above would be option B. If a topographic map included a 6,000 ft. mountain next to an area of low hills, the statement that best describe the contour lines on the map is this: <span>The contour lines around the mountain would be very close together. Hope this helps.</span>
Explanation:
An perfect mass less spring, attached at one end and with a free mass attached at the other end, will have a distinct frequency of oscillation depending on its constant spring and mass. On the other hand, a spring with mass along its length will not have a characteristic frequency of oscillation.
Alternatively, based on its spring constant and mass per length, it will now have a wave Speed. It would be possible to use all wavelengths and frequencies, as long as the component fλ= S, where S is the spring wave size. If that sounds like longitudinal waves, like solid sound waves.
