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ziro4ka [17]
3 years ago
12

(a) What is the angular speed ω about the polar axis of a point on Earth's surface at a latitude of 55° N? (Earth rotates about

that axis.) (b) What is the linear speed v of the point? What are (c) ω and (d) v for a point at the equator? (Note: Earth radius equals 6370 km and let one day be 24 hours)
Physics
1 answer:
Vaselesa [24]3 years ago
6 0

Answer

given,

radius of earth = 6370 km

earths latitude = 55° N

time  = 24 hour = 86400 s

angular speed = ω

a) \omega = \dfrac{2\pi}{T}

\omega = \dfrac{2\pi}{86400}

\omega = 7.27 \times 10^{-5}rad/s

b) v = r\omega cos\theta

v = 6370\times 7.27 \times 10^{-5}\times cos 55^0

v = 266 m/s

c) ω at equator

\omega = \dfrac{2\pi}{T}

\omega = \dfrac{2\pi}{86400}

\omega = 7.27 \times 10^{-5}rad/s

d) v = r\omega cos\theta

v = 6370\times 7.27 \times 10^{-5}\times cos 0^0

v = 463 m/s

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3 years ago
Two long, parallel wires separated by 2.00 cm carry currents in opposite directions. The current in one wire is 1.75 A, and the
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The force per unit length between the two wires is 6.0\cdot 10^{-5} N/m

Explanation:

The magnitude of the force per unit length exerted between two current-carrying wires is given by

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I_1, I_2 are the currents in the two wires

r is the separation between the two wires

For the wires in this problem, we have

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Substituting into the equation, we find

\frac{F}{L}=\frac{(4\pi \cdot 10^{-7})(1.75)(3.45)}{2\pi (0.02)}=6.0\cdot 10^{-5} N/m

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An object releasing energy is evidence of which type of chemical change?
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A truck has shock absorbers with a spring constant of 24200 N/m. When it hits a bump, it oscillates at 0.429 Hz. What is the mas
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Answer:

3331.5 kg

Explanation:

Given:

Spring constant of the spring (k) = 24200 N/m

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Let the mass be 'm' kg.

Now, we know that, a spring-mass system undergoes Simple Harmonic Motion (SHM). The frequency of oscillation of SHM is given as:

f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}

Rewrite the above equation in terms of 'm'. This gives,

2\pi f=\sqrt{\frac{k}{m}}\\\\Squaring\ both\ sides,\ we\ get:\\\\(2\pi f)^2=\frac{k}{m}\\\\m=\frac{k}{4\pi^2 f^2}

Now, plug in the given values and solve for 'm'. This gives,

m=\frac{24200\ N/m}{4\pi^2\times (0.429\ Hz)^2 }\\\\m=\frac{24200\ N/m}{4\pi^2\times 0.184\ Hz^2}\\\\m\approx3331.5\ kg

Therefore, the mass of the truck is 3331.5 kg.

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