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spayn [35]
3 years ago
9

oscillating spring mass systems can be used to experimentally determine an unknown mass without using a mass balance. a student

observes that a particular spring-mass system has a frequency of oscillation of 10 Hz. the spring constant of the spring is 250 N/m. what is the mass?​
Physics
1 answer:
puteri [66]3 years ago
3 0

Answer:

Mass, m = 6.18 kg

Explanation:

Given the following data;

Frequency, F = 10 Hz

Spring constant, k = 250 N/m

We know that pie, π = 22/7

To find the mass, we would use the following formula;

F = 1/2π√(k/m)

Where;

F is the frequency of oscillation.

k is the spring constant.

m is the mass of the spring.

Substituting into the formula, we have;

10 = 1/2 * 22/7 * √250/m

10 = 22/14 * √250/m

Cross-multiplying, we have;

140 = 22 * √250/m

Dividing both sides by 22, we have;

140/22 = √250/m

6.36 = √250/m

Taking the square of both sides, we have;

6.36² = (√250/m)²

40.45 = 250/m

Cross-multiplying, we have;

40.45m = 250

Mass, m = 250/40.45

Mass, m = 6.18 kg

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Answer:

a. T₂ = 901.43 ° K

b. T₂ = 843.85 ° K

Explanation:

Given

η = 0.165 mol , Q = 690 J , V = 280 cm³ , P = 3.40 x 10 ⁶ Pa , T₁ = 700 ° K

Using the equation that describe the experiment of heat

Q = η * Cv * ΔT

a.

Nitrogen   Cv =  20.76 J / mol ° K  

ΔT = T₂ - T₁

T₂ = [ Q / ( η * Cv) ] + T₁

T₂ = [ 690 J / ( 0.165 mol * 20.76 J / mol ° K ) ] + 700 ° K

T₂ = 901.43 ° K

b.

Nitrogen   Cp = 29.07 J / mol ° K

T₂ = [ Q / ( η * Cv) ] + T₁

T₂ = [ 690 J / ( 0.165 mol * 29.07 J / mol ° K ) ] + 700 ° K

T₂ = 843.85 ° K

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3 years ago
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8 0
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A hollow sphere of radius 0.200 m, with rotational inertia I = 0.0484 kg·m2 about a line through its center of mass, rolls witho
d1i1m1o1n [39]

Answer:

Part a)

KE_r = 8 J

Part b)

v = 3.64 m/s

Part c)

KE_f = 12.7 J

Part d)

v = 2.9 m/s

Explanation:

As we know that moment of inertia of hollow sphere is given as

I = \frac{2}{3}mR^2

here we know that

I = 0.0484 kg m^2

R = 0.200 m

now we have

0.0484 = \frac{2}{3}m(0.200)^2

m = 1.815 kg

now we know that total Kinetic energy is given as

KE = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2

KE = \frac{1}{2}mv^2 + \frac{1}{2}I(\frac{v}{R})^2

20 = \frac{1}{2}(1.815)v^2 + \frac{1}{2}(0.0484)(\frac{v}{0.200})^2

20 = 1.5125 v^2

v = 3.64 m/s

Part a)

Now initial rotational kinetic energy is given as

KE_r = \frac{1}{2}I(\frac{v}{R})^2

KE_r = \frac{1}{2}(0.0484)(\frac{3.64}{0.200})^2

KE_r = 8 J

Part b)

speed of the sphere is given as

v = 3.64 m/s

Part c)

By energy conservation of the rolling sphere we can say

mgh = (KE_i) - KE_f

1.815(9.8)(0.900sin27.1) = 20- KE_f

7.30 = 20 - KE_f

KE_f = 12.7 J

Part d)

Now we know that

\frac{1}{2}mv^2 + \frac{1}{2}I(\frac{v}{r})^2 = 12.7

\frac{1}{2}(1.815) v^2 + \frac{1}{2}(0.0484)(\frac{v}{0.200})^2 = 12.7

1.5125 v^2 = 12.7

v = 2.9 m/s

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From the diagram The value of cos C × sin A = \frac{3}{4}

<h3>Determine the numerical value of cos C × sin A</h3>

First step : determine the values of cos C and sin A

cos C = adjacent / hypotenuse

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Therefore the numerical value of cos C * sin A = \frac{3}{4}

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