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liberstina [14]

3 years ago

10

A vertical scale on a spring balance reads from 0 to 200 \rm{N}. The scale has a length of 10.0 \rm{cm} from the 0 to 200 \rm{N}

reading. A fish hanging from the bottom of the spring oscillates vertically at a frequency of 2.05 \rm{Hz}.
Ignoring the mass of the spring, what is the mass m of the fish?

1 answer:

pav-90 [236]3 years ago

8 0

Answer:

m = 12.05 kg

Explanation:

Spring constant in K, N/m

K = 200/10* 100

K = 2000 N/m

Angular Frequency = sqrt (Spring constant / (Mass )

ω = 2 π f

ω = 2π* 2.05 Hz = 12.8805 rad/s

ω^2 = Spring constant / Mass

Mass= Spring constant / ω^2

ω^2 = 165.907 rad^2/s^2

m = 2000 (N/m)/165.907 (rad^2/s^2)

m = 12.05 kg

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7. A mother pushes her 9.5 kg baby in her 5kg baby carriage over the grass with a force of 110N @ an angle

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We have to look into the FBD of the carriage.

Horizontal forces and Vertical forces separately.

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To calculate normal force we have to look upon the vertical component of forces, as Normal force is acting vertically.We have weight which is a downward force along with $F_x$ , force of $110\ N$ acting vertically downward.Both are downward and Normal is upward so Normal force $=Summation\ of\ both\ forces$

Normal force (N) $= Fsin(\theta)+W=110sin(30) + 142.1 =197.1\ N$

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To calculate acceleration we will use Newtons second law.

That is Force is product of mass and acceleration.

We can see in the diagram that $F_y=Horizontal$ and $F_x=Vertical$ component of forces.

So Fnet = Fy(Horizontal) - f(friction) $= m\times a$

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2 years ago

At the surface, atmospheric pressure is 1.013 × 10^5 Pa. People can normally snorkel down to a depth of roughly one meter. What

natulia [17]

Answer:

1.01 × 10⁵ Pa

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At the surface, atmospheric pressure is 1.013 × 10⁵ Pa.

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How does a bimetallic strip work in a sinple alarm system

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Read 2 more answers

How far apart (in mm) must two point charges of 70.0 nC (typical of static electricity) be to have a force of 1.30 N between the

Sever21 [200]

Answer:r=5.824 mm

Explanation:

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Charge $q_1=70 nC$

$q_2=70 nC$

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Electrostatic Force is given by

$F=\frac{kq_1q_2}{r^2}$

where $q_1$ and $q_2$ are the charge on the Particles

r=distance between them

$k=Coulomb\ constant =9\times 10^9 N-m^2/C^2$

$1.3=\frac{9\times 10^9\times 70\times 70\times 10^{-18}}{r^2}$

$r^2=\frac{9\times 10^9\times 70\times 70\times 10^{-18}}{1.3}$

$r=\sqrt{33.923\times 10^{-6}}$

$r=5.824\times 10^{-3}$

$r=5.824 mm$

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3 years ago