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leonid [27]
3 years ago
10

A 0.80 kg mass attached to an ideal spring oscillates horizontally with a period of 0.50 s and amplitude of 0.30 m. What is the

spring constant of the spring?
Physics
1 answer:
sasho [114]3 years ago
8 0

Answer:

126.33 N/m

Explanation:

Angular frquency is given by

f=\frac {2\pi}{t}

Wheret is period of wave and f is angular frequency.

Given period of 0.5 s then

w=\frac {2\pi}{0.5}=4\pi rad/s\approx 12.57 rad/s

Angular frequency is also given by

w=\sqrt{\frac {k}{m}}

Making k the subject then k=mw^{2}

Where k is spring constant, m is mass

Substituting m with 0.8 then

k=0.8\times (4\pi)^{2}

K=126.330936333944 N/m

Rounded off as 126.33 N/m

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Advocard [28]

Complete question:

A fireman of mass 80 kg slides down a pole. When he reaches the bottom, 4.2 m below his starting point, his speed is 2.2 m/s. By how much has thermal energy increased during his slide?

Answer:

The thermal energy increased by 3,099.2 J

Explanation:

Given;

mass of the fireman, m = 80 kg

initial position of the fireman, hi = 4.2 m

final speed, v = 2.2 m/s

The change in the thermal energy is calculated as;

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\Delta  E_{th} +  (\frac{1}{2} mv_f^2 - \frac{1}{2} mv_i^2) + (mgh_f - mgh_i)=0\\\\initial \ velocity, \ v_i = 0\\final \ height , \ h_f = o\\\\\Delta  E_{th} + (\frac{1}{2} mv_f^2) + ( - mgh_i)=0\\\\\Delta  E_{th} +  \frac{1}{2} mv_f^2 - mgh_i = 0\\\\\Delta  E_{th} =mgh_i  - \frac{1}{2} mv_f^2\\\\\Delta  E_{th} =  80 \times 9.8 \times 4.2  \ \ - \ \ \frac{1}{2} \times 80 \times (2.2)^2  \\\\\Delta  E_{th} = 3292.8 \ J \ - \ 193.6 \ J\\\\\Delta  E_{th} = 3,099.2 \ J

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3 years ago
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Answer:

A) d_o = 20.7 cm

B) h_i = 1.014 m

Explanation:

A) To solve this, we will use the lens equation formula;

1/f = 1/d_o + 1/d_i

Where;

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d_i is image distance = 6m

1/0.2 = 1/d_o + 1/6

1/d_o = 1/0.2 - 1/6

1/d_o = 4.8333

d_o = 1/4.8333

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Where;

h_o = 3.5 cm = 0.035 m

d_i = 6 m

d_o = 20.7 cm = 0.207 m

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h_i = (6/0.207) × 0.035

h_i = 1.014 m

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