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4 years ago

Find the length of a pendulum that oscillates with a frequency of 0.15 Hz.

Physics

1 answer:

tester [92]4 years ago

5 0

The 'period' of a simple pendulum is

Time = 2 π · √ (pendulum's length) / (acceleration of gravity)

Time = 1/frequency = 6.67 seconds

g = 9.81 m/s²

Stuff those numbers into the formula, and you get

<em>Pendulum's length = 11.04 meters</em>

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How can astronomers infer wich elements are found in a star

Alchen [17]

By previous elements found in other stars, accurate guesses and lots of studying

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3 years ago

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An electron microscope is usually used for microorganisms that are _____. too small to be seen with the unaided eye too small to

GaryK [48]

<h3><u>Answer;</u></h3>

<em>too small to be seen with an optical microscope </em>

<h3><u>Explanation;</u></h3>

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<em><u>Electron microscope has very high resolution and magnification as compared to other optical microscope </u></em>hence can be observed in the study of micro-organisms such as viruses which would be difficult to study their features using optical microscopes.

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3 years ago

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Which is a unit of volume?<br><br> A) centimeter <br> B) meter <br> C) milligram <br> D) milliliter

german

Answer:

I think milligram is the correct answer.

6 0

3 years ago

A block of mass 0.1 kg is attached to a spring of spring constant 21 N/m on a frictionless track. The block moves in simple harm

bogdanovich [222]

Answer:

A) 2.75 m/s B) 0.1911 m C) 0.109 s

Explanation:

mass of block = M =0.1 kg

spring constant = k = 21 N/m

amplitude = A = 0.19 m

mass of bullet = m = 1.45 g = 0.00145 kg

velocity of bullet = vᵇ = 68 m/s

as we know:

Angular frequency of S.H.M = ω₀ = $\sqrt\frac{k}{M}$

= $\sqrt\frac{21}{0.1}$

= 14.49 rad/sec

<h3>A) Speed of the block immediately before the collision:</h3>

displacement of Simple Harmonic Motion is given as:

$x = A sin (\omega t + \phi)\\$

Differentiating this to find speed of the block immediately before the collision:

$v=\frac{dx}{dt}= A\omega_{o} cos (\omega_{o}t =\phi}\\$

As bullet strikes at equilibrium position so,

φ = 0

t= 2nπ

⇒ cos (ω₀t + φ) = 1

⇒ $v= A\omega_{o}$

$v=(.19)(14.49)\\v= 2.75 ms^{-1}$

<h3>B) If the simple harmonic motion after the collision is described by x = B sin(ωt + φ), new amplitude B:</h3>

S.H.M after collision is given as :

$x= Bsin(\omega t + \phi)$

To find B, consider law of conservation of energy

$K.E = P.E\\K.E= \frac{1}{2}(m+M)v^{2} \\P.E = \frac{1}{2} kB^{2}$

$\frac{m+M}{k} v^{2} = B^{2} \\B =\sqrt\frac{m+M}{k} v\\B = \sqrt\frac{.00145+0.1}{21} (2.75)\\B = .1911m$

<h3>C) Time taken by the block to reach maximum amplitude after the collision:</h3>

Time period S.H.M is given as:

$T=2\pi \sqrt\frac{m}{k}\\ for given case\\m= m=M\\then\\T=2\pi \sqrt\frac{m+M}{k}$

Collision occurred at equilibrium position so time taken by block to reach maximum amplitude is equal to one fourth of total time period

$T=\frac{\pi }{2}\sqrt\frac{m+M}{k} \\T=0.109 sec$

5 0

4 years ago

A pair of eyeglass frames is made of epoxy plastic. At room temperature (20.0°C), the frames have circular lens holes 2.50 cm in

nignag [31]

Answer:

T₂ = 114 °C

Explanation:

Area or superficial expansion: can be defined as an increase in area, per unit area per degree rise in temperature. It unit is (1/K) or (1/°C).

β = ΔA/(A₁ΔT) ............................................. equation 1

β = 2α ......................................................... equation 2

Area of circle = πr².................................... equation 3

Where β = Area expansivity, α = linear expansivity, ΔA = increase in area = (A₂ - A₁), ΔT = change in temperature, A₁ = initial area, A₂ = Final area r = radius,

from the question, The coefficient of linear expansion for epoxy = 1.3 × 10⁻⁴ °C⁻¹,

∴ Coefficient of area expansion for epoxy = 2 × 1.3 × 10⁻⁴ =

β = 2.6 × 10⁻⁴ °C⁻¹,

Using equation 3 to calculate for area, and taking (π = 3.143)

r₁ = 2.5 cm ∴ A₁ = πr₁² = 3.143 × 2.5² =19.64 cm².

r₂ = 2.53 cm ∴ A₂ = πr₂² = 3.143 × 2.53² =20.12 cm².

ΔA = A₂ - A₁ = 20.12 - 19.64 = 0.48 cm².

Making ΔT the subject of the relation in equation 1.

ΔT = ΔA/(βA₁) ...................................... equation 4

Substituting the values above into equation 4,

ΔT = 0.48/(2.6 × 10⁻⁴ × 19.64)

ΔT = 0.48/(51.064 × 10⁻⁴)

ΔT =( 0.48/51.064) × 10000

ΔT = 0.0094 × 10000 = 94 °C

But, ΔT = T₂ -T₁,

Then, T₂ = ΔT + T₁

Where T₁ = 20 °C, ΔT =94 °C

∴ T₂ = 94 + 20 = 114 °C.

Therefore, temperature at which the frame must be heated if the the lenses 2.53 cm are to be inserted in them = 114 °C

3 0

3 years ago