Translation of important question part (Google translation used)
Force of 6N causes a displacement of 0.03m. remove the balance and connect a body 0.5 kg to the end, pull it to move it 0.02 m, release it and see how it oscillates.
(a)Determine the spring constant.
(b)Calculate angular velocity, frequency, and oscillation period
Answer:
(a)K=200 N/m
(b) w= 20 rad/s f=3.2 Hz T=0.3125 s
Explanation:
(a)
From Hooke's law, we deduce that F=kx where F is applied force, k is spring constant and x is extension of the spring. Making k the subject of the formula then k=F/x and substituting F with 6 N and x with 0.03 m then k=6/0.03=200 N/m
(b)
Angular velocity, w is given by 
where m is the mass and k is spring constant calculated in part a above. Substituting mass with 0.5 kg and k with 200 N/m then
We know that frequency, f is given by 
and substituting 20 rad/s for w then
Finally, oscillation period, T is usually the reciprocal of frequency hence T=1/f and substituting f with 3.2 Hz then T=1/3.2=0.3125 s
The lists that arranges the Earth's compositional layers in order of increasing density is CRUST, MANTLE AND THEN CORE. The core is considered as the most dense and the lithosphere, which makes the outermost core, is considered as having the least density. Hope that this answer helps.
Answer:
Explanation:
Stefan's Boltzman Law gives the value of radiation from a black body at a particular temperature. The relation is
E = eσ ( T₂⁴ -T₁⁴ )
e is emmisivity , σ is a constant = 5.67 x 10⁻⁸ , E is energy emitted per unit time per unit area
E = .915 x 5.67 x 10⁻⁸ x ( 303⁴ -293⁴ )
= .915 x 5.67 x 10⁻⁸ x ( 84.28 -73.70 ) x 10⁸
= 55 W / m²s
Area A = 1.45 m²
time t = 14.3 min
= 858 s
Total radiation
= 55 x 858 x 1.45 W
= 68425 J
The effective height of the water for Smith's house will be 24.61m.
<h3>How to calculate the height?</h3>
Based on the information given, the volume of the water in sphere will be:
= 4/3πr³ = (5.80 × 10^5)/1000
= 4.18r³ = 580
r³ = 138.7
r = 5.18m
The effective height of the water will be:
= 18.0 + 2(5.18)
= 28.36
The gauge pressure at Faucet of Jones house will be:
= pgh
= 1000(9.8)(28.36)
= 277.9kPa
The effective height of the water for Smith's house will be:
= 18.0 + 2(5.18) - 3.75
= 24.61m
The gauge pressure at Faucet of Jones house will be:
= 1000 × 9.8 × 24.61
= 241.2kPa
Learn more about height on:
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This problem we can represent as right triangle where we can use basic trygonometric functions.
in this case we have largest side (hypothenuse) of the triangle and we have one angle. The angle we got is the one between side of triangle that is horizontal with the ground and the hypothenuse, meaning that we need sin function to connect relation between angle, known side lenght and unknown side lenght.
sin(alpha) = (side of triangle oposite from observed angle)/(hypothenuse)
sin(41) = x/1503
x = 1503*sin41 = 986.06 meters
Answer is 986.06 meters
