The center-seeking change in velocity of an object moving in a circle is the centripetal acceleration.
So, by Newton's laws, we know that an object moving with a given velocity will remain in constant motion with a constant velocity until we apply an acceleration.
So we define acceleration as the rate of change of the velocity, also remember that velocity is a vector (has magnitude and direction), so, if there is a change the direction of the velocity, we have an acceleration that causes that.
In circular motion, the velocity vector is always perpendicular to the radius of the circle, and it can only be possible if the velocity direction is changing constantly. This will happen because of something called centripetal acceleration.
This acceleration points radially inwards (to the center of the circle) so is also perpendicular to the velocity of the moving object, and this is what causes the constant change in the direction of the velocity of the moving object.
Just to give an example, if you have a string with a mass on one end, and with your hand, you rotate the mass (from the string), the tension of the string would be the centripetal acceleration.
If you want to learn more about circular motion, you can read:
brainly.com/question/2285236
Wavelength*frequency=velocity
(331m/s)/(.6m)
Frequency = 551.666 1/s
Answer:
it is a interspecific competition
Explanation:
i just took a test with this question
Answer:
a,)3.042s
b)4.173s
c)3.281s
Explanation:
For a some pendulum the period in seconds T can be calculated using below formula
T=2π√(L/G)
Where L = length of pendulum in meters
G = gravitational acceleration = 9.8 m/s²
Then we are told to calculate
(a) What is the period of small oscillations for this pendulum if it is located in an elevator accelerating upward at 3.00 m/s2?
Since oscillations for this pendulum is located in the elevator that is accelerating upward at 3.00 then
use G = 9.8 + 3.0 = 12.8 m/s²
Period T=2π√(L/G)
T= 2π√(3/12.8)
T=3.042s
b) (b) What is the period of small oscillations for this pendulum if it is located in an elevator accelerating downward at 3.00 m/s2?
G = 9.8 – 3.0 = 6.8 m/s²
T= 2π√(3/6.8)
T=4.173s
C)(c) What is the period of this pendulum if it is placed in a truck that is accelerating horizontally at 3.00 m/s2?
Net acceleration is
g'= √(g² + a²)
=√(9² + 3²)
Then period is
T=2π√(3/11)
T=3.281s
Answer: true
(I need 20 characters to submit the answer so I wrote this not to be confused with the answer)
