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Eddi Din [679]
2 years ago
5

A 200 g air-track glider is attached to a spring. The glider is pushed in 9.8 cm against the spring, then released. A student wi

th a stopwatch finds that 12 oscillations take 15.0 s. Part A What is the spring constant
Physics
1 answer:
Likurg_2 [28]2 years ago
6 0

Answer:

k =  5.05 N/m

Explanation:

In order to calculate the spring mass of the system, you use the following formula:

T=2\pi \sqrt{\frac{m}{k}}     (1)

T: period of oscillation of the system

m: mass of the air-track glider = 200g = 0.200 kg

k: spring constant = ?

You first calculate the period of oscillation:

T=\frac{1}{f}=\frac{1}{12/15.0s}=1.25s

Next, you solve the equation (1) for k, and then you replace the values of the other parmateres:

k=4\pi^2 \frac{m}{T^2}\\\\k=4\pi^2 \frac{0.200kg}{(1.25s)^2}=5.05\frac{N}{m}

The spring constant of the spring is 5.05 N/m

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Any help would be great! Thank you x<br><br><br> Giving brainliest answer xoxo
garri49 [273]

Answer:

A vacuum would have been created. I hope this helps have a great day

3 0
3 years ago
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
valina [46]
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
6 0
3 years ago
A man stands still on a moving walkway that is going at a speed of 0.2 m/s to
miskamm [114]

Answer:

0.2 m/s east

Explanation:

mark as brainlyest

7 0
2 years ago
active and passive secruity measures are employed to identify, detect, classify and analyze possible threats inside of which zon
Juli2301 [7.4K]

Answer:

Assessment zone

Explanation:

It is the assessment zone in various security zones where active and passive security measures are employed to identify, detect, classify and analyze possible threats inside the assessment zones.

8 0
3 years ago
Invader Zim’s spaceship is sitting at rest in outer space. The ship then accelerates at a uniform rate of 12 m/s2 for 10 seconds
melisa1 [442]

Answer:

120 m/s

Explanation:

Given:

v₀ = 0 m/s

a = 12 m/s²

t = 10 s

Find: v

v = at + v₀

v = (12 m/s²) (10 s) + 0 m/s

v = 120 m/s

6 0
3 years ago
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