Period: 1/Frequency
1/457 = 0.002188 seconds
Round it to 0.0022 seconds
The answer is D
Answer:
0.4A.
Explanation:
Current (A) = Charge (coulomb)/Time (secs)
2 coulombs/5 secs = 0.4A
Answer:
T= 1 s
Explanation:
Given that
When x= cm ,T= 1
we know that time period of spring mas system given as

T= Time period
m= mass
k=spring constant
So from above equation we can say that time period of system does not depends on the value of x.
So when x= 10 cm ,still time period will be 1 s.
T= 1 s
Answer
m/s rate of change of dispalcement per sec. ie velocity
m/s^2 is (m/s)/s ie rate of change of velocity per sec. ie accelerationplanation:
A car that experiences a deceleration of -41.62 m/s² and comes to a stop after 10.99 m has an initial velocity of 30.60 m/s.
A car experiences a deceleration (a) of -41.62 m/s² and comes to a stop (final velocity = v = 0 m/s) after 10.99 m (s).
We can calculate the initial velocity of the car (u) using the following kinematic equation.
![v^{2} = u^{2} + 2as\\\\u = \sqrt[]{v^{2}-2as} = \sqrt[]{(0m/s)^{2}-2(-42.61m/s^{2} )(10.99m)} = 30.60m/s](https://tex.z-dn.net/?f=v%5E%7B2%7D%20%3D%20u%5E%7B2%7D%20%2B%202as%5C%5C%5C%5Cu%20%3D%20%5Csqrt%5B%5D%7Bv%5E%7B2%7D-2as%7D%20%3D%20%5Csqrt%5B%5D%7B%280m%2Fs%29%5E%7B2%7D-2%28-42.61m%2Fs%5E%7B2%7D%20%29%2810.99m%29%7D%20%3D%2030.60m%2Fs)
A car that experiences a deceleration of -41.62 m/s² and comes to a stop after 10.99 m has an initial velocity of 30.60 m/s.
Learn more: brainly.com/question/14851168