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Natasha2012 [34]
2 years ago
6

A Foucault pendulum consists of a brass sphere with a diameter of 32.0 cm suspended from a steel cable 10.0 m long (both measure

ments made at 20.0∘C). Due to a design oversight, the swinging sphere clears the floor by a distance of only 3.00 mm when the temperature is 20.0∘C.c. at what temperature will the sphere begin to brush the floor?
Physics
1 answer:
Sergio039 [100]2 years ago
8 0

Answer:

T = 44.35 °C

Explanation:

d = 32cm

R = 16 cm

Lsteel = 10m

T1 = 20° C

Space = 0.3cm

The space between the sphere and the floor is represented by δL(total) after the temperature increases.

As the temperature increases, both will expand.

So,

0.3 x 10^(-2) = δL(steel) + δR(brass)

= {L(o) x α(steel) x δT} + {R(o) x {α(brass) x δT}

= {10 x 1.2 x 10^(-5) x (T-20)} + {0.16 x 2 x 10^(-5) x (T-20)}

= 12.32 x 10^(-5) x(T-20)

Therefore (T-20) = (0.3 x 10^(-2)) / {12.32 x 10^(-5)}

T = 20 + 24. 35 = 44.35 °C

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Two carts (m= 0.4 kg each) are placed on an aluminum track. The first cart pushed with the initial velocity of 1.5 m/s towards t
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here two cars are placed on an aluminium track

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now if there is no external force so momentum is conserved

m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f]

here

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now plug in all values in it

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divide whole equation by mass 0.4

v_{1f} + v_{2f} = 1.5

also be the equation of coefficient of restitution

e = 1 = \frac{v_{2f} - v_{1f}}{v_{1i} - v_{2i}}

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v_{2f} - v_{1f} = 1.5

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2 years ago
How does a galvanometer use a magnetic field to indicate the strength of an electric current?
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A circular loop of wire with radius 0.0410 m and resistance 0.169 Ω is in a region of spatially uniform magnetic field, as shown
ArbitrLikvidat [17]

(a) The induced current in the loop will be counterclockwise.

(b) The rate at which electrical energy is being dissipated by the resistance of the loop is 0.012 W.

<h3>Direction of the current</h3>

The induced current in the loop will be counterclockwise to the direction of magnetic field.

<h3>Emf induced in the loop</h3>

emf = -NdФ/dt

emf = -NBA/dt

where;

A is area of the loop

A = πr² = π(0.041)² = 5.28 x 10⁻³ m²

emf = -(-0.605 - 7.78) x 5.28 x 10⁻³

emf = 8.385 x 5.28 x 10⁻³

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<h3>Rate of energy dissipation</h3>

P = emf²/R

P = (0.0442)²/0.169

P = 0.012 W

Thus, the rate at which electrical energy is being dissipated by the resistance of the loop is 0.012 W.

Learn more about power dissipation here: brainly.com/question/15015986

#SPJ1

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