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garri49 [273]
3 years ago
13

A cube of metal has a mass of 11 grams and a volume of 1 cm . When fully submerged in water this metal cube hanging from an accu

rate spring scale will weigh what amount?
Physics
1 answer:
Anarel [89]3 years ago
7 0

Answer:

0.098 N

Explanation:

From the question,

Spring scale reading = W-U............... Equation 1

Where W = weight of the cube, U = upthrust.

W = mg

Where m =  mass of the cube, g = acceleration due to gravity.

Given: m = 11 g = 0.011 kg, g = 9.8 m/s².

W = 0.011(9.8)

W = 0.1078 N.

From Archimedes principle,

Upthrust = weight of water displaced.

U = (Density of water×volume of metal cube)×acceleration due to gravity.

U = (D×V)g

Given: D = 1000 kg/m², V = 1 cm³ = (1/1000000) = 1×10⁻⁶ m³, g - 9.8 m/s²

U = 1000(9.8)(10⁻⁶)

U = 0.0098 N.

Substitute the value of W and U into equation 1

Reading of the spring scale = 0.1078-0.0098

Reading of the spring scale = 0.098 N

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The Period of the resulting shm will be T=39.7

<u>Explanation:</u>

<u>Given data</u>

m=3kg

d=.06m

k=1200 N/m

Θ=3 °

T=?

we have the formulas,

I = (1/6)Md2

F = ma

F = -kx = -(mω2x)

k = mω2 τ = -d(FgsinΘ)

T=2 x 3.14/ √(m/k)

Solution for the given problem would be,

F=-Kx (where x= dsin Θ)

F=-k dsin Θ

F=-(1200)(.06)sin(3 °)

F=-10.16N

<u>By newton's second law.</u>

F = ma

a= F/m

a=(-10.16N)/3

a=3.38

<u>using the k=mω value</u>

k=mω

ω=k/m

ω=1200/3

ω=400

<u>Using F = -kx value</u>

x = F/-k

x=(-10.16)/1200

x=0.00847m

<u>Restoring the  torque value </u>

τ = -dmgsinΘ    where( τ = Iα so.).. Iα = -dmgsinΘ α = -(.06)(4)α =

α =(.06)(4)(9.81)sin(4°)

α=-1.781

<u>Rotational to linear form</u>

a = αr  

r = .1131 m

a=-1.781 x .1131 m

a=-0.2015233664

<u>Time Period</u>

T=2 x 3.14/ √(m/k)

T=6.28/√(3/1200)

T=6.28/0.158

T=39.7

6 0
3 years ago
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