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3 years ago

Calculate the period of a spring if it has a mass of 5 kg and a spring constant of 6 N/m

Physics

1 answer:

otez555 [7]3 years ago

5 0

Answer: The period of a spring if it has a mass of 5 kg and a spring constant of 6 N/m is 5.73 sec.

Explanation:

Given: Mass = 5 kg

Spring constant = 6 N/m

Formula used to calculate period is as follows.

$T = 2 \pi \sqrt\frac{m}{k}$

where,

T = period

m = mass

k = spring constant

Substitute the values into above formula as follows.

$T = 2 \pi \sqrt\frac{m}{k}\\= 2 \times 3.14 \times \sqrt\frac{5}{6}\\= 5.73 s$

Thus, we can conclude that the period of a spring if it has a mass of 5 kg and a spring constant of 6 N/m is 5.73 sec.

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Explain why the elements of the same group exhibit the same chemical behavior

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Answer:

All the elements of a period have similar chemical properties because they have the same number of valence electrons in their outermost shell. Their atoms have the same number of electrons in the highest occupied energy level

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3 years ago

An aluminum bar 600mm long, with diameter 40mm, has a hole drilled in the center of the bar. The hole is 40mm in diameter and 10

s2008m [1.1K]

Answer:

1.228 x $10^{-6}$ mm

Explanation:

diameter of aluminium bar D = 40 mm

diameter of hole d = 30 mm

compressive Load F = 180 kN = 180 x $10^{3}$ N

modulus of elasticity E = 85 GN/m^2 = 85 x $10^{9}$ Pa

length of bar L = 600 mm

length of hole = 100 mm

true length of bar = 600 - 100 = 500 mm

area of the bar A = $\frac{\pi D^{2} }{4}$ = $\frac{3.142* 40^{2} }{4}$ = 1256.8 mm^2

area of hole a = $\frac{\pi(D^{2} - d^{2}) }{4}$ = $\frac{3.142*(40^{2} - 30^{2})}{4}$ = 549.85 mm^2

Total contraction of the bar = $\frac{F*L}{AE} + \frac{Fl}{aE}$

total contraction = $\frac{F}{E} * (\frac{L}{A} +\frac{l}{a})$

==> $\frac{180*10^{3}}{85*10^{9}} *( \frac{500}{1256.8} + \frac{100}{549.85})$ = 1.228 x $10^{-6}$ mm

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3 years ago

A published hypothesis:

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Answer:

should be tested by the scientific community

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3 years ago

A 0.750kg block is attached to a spring with spring constant 13.5N/m . While the block is sitting at rest, a student hits it wit

vazorg [7]

Answer:

A)A=0.075 m

B)v= 0.21 m/s

Explanation:

Given that

m = 0.75 kg

K= 13.5 N

The natural frequency of the block given as

$\omega =\sqrt{\dfrac{K}{m}}$

The maximum speed v given as

$v=\omega A$

A=Amplitude

$v=\sqrt{\dfrac{K}{m}}\times A$

$0.32=\sqrt{\dfrac{13.5}{0.75}}\times A$

A=0.075 m

A= 0.75 cm

The speed at distance x

$v=\omega \sqrt{A^2-x^2}$

$v=\sqrt{\dfrac{K}{m}}\times \sqrt{A^2-x^2}$

$v=\sqrt{\dfrac{13.5}{0.75}}\times \sqrt{0.075^2-(0.075\times 0.75)^2}$

v= 0.21 m/s

5 0

3 years ago

If 10. joules of work must be done to move 2.0 coulombs of charge from point A to point B in an electric field, the potential di

masya89 [10]

The equation for work (W) done by an electric field is:

W = qΔV

where q is the magnitude of the charge and ΔV is the potential difference. The question gives you W and q, so plug n' play to find ΔV:

10 = 2ΔV
ΔV = 5

8 0

3 years ago