Complete question:
The coordinate of a particle in meters is given by x(t)=1 6t- 3.0t³ , where the time tis in seconds. The
particle is momentarily at rest at t is:
Select one:
a. 9.3s
b. 1.3s
C. 0.75s
d.5.3s
e. 7.3s
Answer:
b. 1.3 s
Explanation:
Given;
position of the particle, x(t)=1 6t- 3.0t³
when the particle is at rest, the velocity is zero.
velocity = dx/dt
dx /dt = 16 - 9t²
16 - 9t² = 0
9t² = 16
t² = 16 /9
t = √(16 / 9)
t = 4/3
t = 1.3 s
Therefore, the particle is momentarily at rest at t = 1.3 s
Answer:
400000
Explanation:
So first solve one part:
(3.25 * 10^5)
(3.25 * 100,000)
= 325000
Then solve the next part:
(7.5 * 10^4)
(7.5 * 10000)
= 75000
Now lastly, add the two answers:
325000 + 75000 = 400000
Therefore,
(3.25 x 10^5) + (7.5 x 10^4) = 400000
Answer:
a. 2.1 s
b.0.48 Hz
c. A=24cm
d. 72cm/s
Explanation:
An air-track glider attached to a spring oscillates between the 10.0 cm mark and the 57.0 cm mark on the track. The glider completes 15.0 oscillations in 31.0 s.What are the (a) period, (b) frequency, (c) amplitude, and (d) maximum speed of the glider?
What are the period,
period is the time taken for a wave particle to make one complete oscillation
a) 31 / 15 = 2.066 seconds
= 2.1 s
(b) frequency
: this the number of oscillation made in one seconds.
it is also the inverse of the period.
= oscillations / time
= 15/31= 0.48 Hz
(c) amplitude
: maximum displacement from the origin
amplitude = 1/2 of the difference of oscillation marks
= 1/2(57-10) = 47/2cm
23.5cm
A=24cm
(d) maximum speed of the glider?
V=ωA
angular frequency *Amplitude
V=a*pi*f*amplitude
2π x frequency x amplitude = maximum speed
= 2π x .48 x 24
=72.38 cm/s
72cm/s
The answer is 2) 1.0c. Light will always propagate through a vacuum at the speed of light “c”; even when moving at a significant fraction of the speed of light, observers will still measure this as the speed of light and the difference is resultant of time dilation.
