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Damm [24]
3 years ago
5

A 1.00 kg object is attached to a horizontal spring. the spring is initially stretched by 0.500 m, and the object is released fr

om rest there. it proceeds to move without friction. the next time the speed of the object is zero is 0.100 s later. what is the maximum speed of the object?
Physics
1 answer:
valina [46]3 years ago
6 0
The  spring is initially stretched, and the mass released from rest (v=0). The next time the speed becomes zero again is when the spring is fully compressed, and the mass is on the opposite side of the spring with respect to its equilibrium position, after a time t=0.100 s. This corresponds to half oscillation of the system. Therefore, the period of a full oscillation of the system is
T=2 t = 2 \cdot 0.100 s = 0.200 s
Which means that the frequency is
f= \frac{1}{T}= \frac{1}{0.200 s}=5 Hz
and the angular frequency is
\omega=2 \pi f = 2 \pi (5 Hz)=31.4 rad/s

In a spring-mass system, the maximum velocity of the object is given by
v_{max} = A \omega
where A is the amplitude of the oscillation. In our problem, the amplitude of the motion corresponds to the initial displacement of the object (A=0.500 m), therefore the maximum velocity is
v_{max} = A \omega = (0.500 m)(31.4 rad/s)= 15.7 m/s
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Answer:

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Q_{coffee} = - Q_{water}\\c_{c} \times m_{c} \times \Delta T_{c} = -c_{w} \times m_{w} \times \Delta T_{w}\\m_{c} \times \Delta T_{c} = -m_{w} \times \Delta T_{w}\\m_{w} = \frac{m_{c} \times \Delta T_{c}}{-\Delta T_{w}} \\m_{w}=\frac{162g \times (62.5 \°C - 94.6 \°C ) }{-(62.5 \°C - 25.0 \°C)} \\m_{w} = 139 g

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A thermometer containing 0.10 g of mercury is cooled from 15 degrees celsius to 8.5 degrees celcius. How much energy left the me
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To solve this exercise we will use the concept related to heat loss which is mathematically given as

Q = mC_p \Delta T

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