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SOVA2 [1]
2 years ago
12

14. If the spring constant of a simple harmonic oscillator is doubled, by what factor will the mass of the system need to change

in order for the frequency of the motion to remain the same?
Physics
1 answer:
klio [65]2 years ago
8 0

Lets se

And

\\ \rm\Rrightarrow T=2\pi\sqrt{\dfrac{m}{k}}

\\ \rm\Rrightarrow \sqrt{k}T=2\pi\sqrt{m}

So

\\ \rm\Rrightarrow k\propto m

If spring constant is doubled mass must be doubled

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What are the damages or effects of sewage dumps into natural water sources???
masha68 [24]

Answer:

Sewage dumps pollute our waterways causing death of marine life, water pollution, and disease.

Explanation:

Sewage dumps. bring disease or death to our ecosystems

5 0
3 years ago
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A ship sets sail from Rotterdam, The Netherlands, heading due north at 8.00 m/s relative to the water. The local ocean current i
Nuetrik [128]

Answer:

The velocity of the ship relative to the earth V = 9.05 \frac{m}{s}

Explanation:

The local ocean current is  = 1.52 m/s

Direction \theta = 40°

Velocity component in X - direction V_{x} = 1.52 \cos 40°

V_{x} = 1.164 \frac{m}{s}

Velocity component in Y - direction V_{y} = 8 + 1.52 \sin 40°

V_{y} = 8.97 \frac{m}{s}

The velocity of the ship relative to the earth

V = \sqrt{V_{x}^{2} + V_{y} ^{2}  }

Put the values of V_{x} and V_{y} we get,

⇒ V = \sqrt{1.164^{2} + 8.97 ^{2}  }

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7 0
3 years ago
An ore sample weighs 17.50 N in air. When the sample
zysi [14]

Answer with Explanation:

We are given that

Weight of an ore sample=17.5 N

Tension in the cord=11.2 N

We have to find the total volume and the density of the sample.

We know that

Tension, T=W-F_b

F_b=buoyancy force

T=Tension force

W=Weight

By using the formula

11.2=17.5-F_b

F_b=17.5-11.2=6.3 N

F_b=V_{object}\times \rho_{water}\cdot g

Where

V_{object}=Volume of object

\rho_{water}=1000 kgm^{-3}=Density of water

g=9.8 ms^{-2}=Acceleration due to gravity

Substitute the values then we get

6.3=9.8\times 1000\times V_{object}

V_{object}=\frac{6.3}{9.8\times 1000}=6.43\times 10^{-4} m^3

Volume of sample=6.43\times 10^{-4} m^3

Density of sample,\rho_{object}=\frac{Mass}{volume_{object}}

Where mass of ore sample=1.79 kg

Substitute the values then, we get

\rho_{object}=\frac{1.79}{6.43\times 10^{-4}}=2.78\times 10^3 kg/m^3

Density of the sample=2.78\times 10^{3} kgm^{-3}

7 0
3 years ago
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Because different materials have different molecular organizations.
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