The time interval needed for the disk to rev from leave to 25.0 m/s
= 0.57 s
<h3>What is the time interval?</h3>
- A larger period of time can be split up into multiple shorter, equal-length segments.
- These are referred to as time periods. Consider the scenario where you wished to gauge a car's speed over an hour-long trip. You could break up an hour into ten-minute segments.
- In Hz, frequency is defined as the number of cycles per second. Simply divide 1 by the frequency to determine the time interval for a known frequency (e.g., a frequency of 100 Hz has a time interval of 1/(100 Hz) = 0.01 seconds; 500 Hz has a time interval of 1/(500Hz) = 0.002 seconds, etc.).
Given:
vo = 0
v = 25.4 m/s
r = 0.95 m
n = 1.21 rev
A.
r*ω = v
ω = v/r
= 25.4/.95
= 26.74 rad/s
B.
θ = 2πn
= 2π × (1.21)
= 7.6 m
Using equation of motion,
ωf^2 = ωi^2 + 2αθ
(26.74)^2 = 2 × α × (7.6)
α = 715.03/15.2
= 47.04 rad/s^2
C.
Using equation of motion,
θ = ωi*t + 1/2*α*t^2
θ = 0 + 1/2*α*t^2
7.6 = 47.04/2 × t^2
= sqrt(0.323)
= 0.57 s
Therefore the correct answer is 0.57 s
To learn more about time interval, refer to:
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Answer:
John Felix Anthony Cena is an American professional wrestler, actor, and television presenter. He is currently signed to WWE. Born and raised in West Newbury, Massachusetts, Cena moved to California in 1998 to pursue a career as a bodybuilder.
Explanation:
Chlorine has the smallest atomic radius since the atomic radius decreases as you travel to the right and up
Answer:
D. half as much
Explanation:
let m and M be the mass of the planets and r be the distance between them.
then: the force of attraction between them is given by,
F = G×m×M/(r^2)
if we keep one mass constant and double the other and also double the distance between them.
the force of attraction becomes:
F1 = 2G×m×M/[(2×r)^2]
= 2G×m×M/[4×(r)^2]
= (1/2)G×m×M/(r^2)
= 1/2×F
therefore, when you double one mass and keep the other mass constant and double the distance between the masses you decrease the force by a factor of 1/2.
Answer:
Acceleration, a = -31.29 m/s²
Explanation:
It is given that, Initial speed of the aircraft, u = 140 mi/h =
62.58 m/s
Finally, it stops, v = 0
Time taken, t = 2 s
Let a is the acceleration of the aircraft. We know that the rate of change of velocity is called acceleration of the object. It is given by :
a = t
0-62.58 2 a =
a = -31.29 m/s²
So, the acceleration of the aircraft is 31.29 m/s²
and the car is decelerating. Hence, this is the required solution.
(i found this answer becuase i could do the math)
