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koban [17]
3 years ago
11

A horizontal spring attached to a wall has a force constant of 850 N/m. A block of mass 1.00 kg is attached to the spring and os

cillates freely on a horizontal, frictionless surface as in Active Figure 5.20. The initial goal of this problem is to find the velocity at the equilibrium point after the block is released. (a) What objects constitute the system, and through what forces do they interact? (b) What are the two points of interest? (c) Find the energy stored in the spring when the mass is stretched 6.00 cm from equilibrium and again when the mass passes through equilibrium after being released from rest. (d) Write the conservation of energy equation for this situation and solve it for the speed of the mass as it passes equilibrium. Substitute to obtain a numerical value. (e) What is the speed at the halfway point? Why isn’t it half the speed at equilibrium
Physics
1 answer:
olga55 [171]3 years ago
3 0

Answer and Explanation:

(a)The system consists of the spring the block + Earth.

They interact through some forces which are: Gravitational form + normal

forces + spring force.

(b)The two points are:

When the spring has compressed a distance of

When the spring is no longer compressed.

(c)1.53J

0J

d. 1.75m/s

(e)1.51 m/s

The function of the final velocity is Not a linear one.

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uysha [10]
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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:
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4 0
3 years ago
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Hoochie [10]

Answer:

75 km/h

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Check the explanation

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A spring stretches 0.2 cm per newton of applied force. An object is suspended from the spring and a deflection of 3 cm is observ
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Answer:

m=1.53kg    

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