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Solnce55 [7]
3 years ago
12

An object attached to a horizontal spring is oscillating back and forth along a frictionless surface. The maximum speed of the o

bject is 1.15 m/s, and its maximum acceleration is 6.52 m/s2. How much time elapses between an instant when the objects speed is at a maximum and the next instant when its acceleration is at a maximum?
Physics
1 answer:
Readme [11.4K]3 years ago
5 0
<span>Object's speed is max when passing through 0 displacement.
Object's acceleration is max at either end of its oscillation, and these two times are 1/4 period apart.
y = A sin (wt)
v= Aw cos(wt)
a = - A (w^2) sin (wt)
 
v max = Aw
 a max= -A w^2
 These relations can give you w= 2 pi/ T where T = period of oscillation.</span>
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The speed of sound is greater in ice (4000 m/s), then in water (1500 m/s), then in air (340 m/s). The explanation for this is the differente state of the matter in the three cases.

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3 years ago
Read 2 more answers
You charge an initially uncharged 65.7-mf capacitor through a 39.1-Ï resistor by means of a 9.00-v battery having negligible int
uysha [10]
In a RC-circuit, with the capacitor initially uncharged,  when we connect the battery to the circuit the charge on the capacitor starts to increase following the law:
Q(t) = Q_0 (1-e^{-t/\tau})
where t is the time, Q_0 = CV is the maximum charge on the capacitor at voltage V, and \tau = RC is the time constant of the circuit.
Using this law, we can answer all the three questions of the problem.

1) Using R=39.1 \Omega and C= 65.7 mF=65.7\cdot 10^{-3}F, the time constant of the circuit is:
\tau = RC=(39.1 \Omega)(65.7 \cdot 10^{-3}F)=2.57 s

2) To find the charge on the capacitor at time t=1.95 \tau, we must find before the maximum charge on the capacitor, which is
Q_0 = CV=(65.7 \cdot 10^{-3}F)(9 V)=0.59 C
And then, the charge at time t=1.95 \tau is equal to
Q(1.95 \tau) = Q_0 (1-e^{-t/\tau})=(0.59 C)(1-e^{-1.95})=0.51 C

3) After a long time (let's say much larger than the time constant of the circuit), the capacitor will be fully charged, this means its charge will be Q_0 = 0.59 C. We can see this also from the previous formule, by using t=\infty:
Q(t) = Q_0 (1-e^{-\infty})=Q_0(1-0) = 0.59 C

4 0
3 years ago
Two postal delivery workers have different routes. They both travel from the post office, to neighborhoods to deliver mail, and
SOVA2 [1]

they travel the same distance

7 0
3 years ago
You throw a ball upward with an initial speed of 4.3 m/s. When it returns to your hand 0.88 s later, it has the same speed in th
djverab [1.8K]

Answer:

The acceleration is -9.8 m/s²

Explanation:

Hi there!!

When you throw a ball upward, there is a downward acceleration that makes the ball return to your hand. This acceleration is produced by gravity.

The average acceleration is calculated as the variation of the speed over time. In this case, we know the time and the initial and final speed. Then:

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4 0
3 years ago
A car travels 40km/hr for 2 hours and 55km/hr for 2 hours how far has the car traveled what’s its average velocity?
vfiekz [6]

Answer;

Average speed  = 47.5 km/hr

Explanation and solution;

Average Speed = Total distance /Total time

Total distance;

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(40 × 2) = 80 km

-In next two hours distance covered is 110 km. (55 ×2)

Total distance = 110 + 80 = 190 km

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                         = 47.5 km/hr

7 0
3 years ago
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