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Firlakuza [10]
3 years ago
11

A wheel of diameter 8.0 cm has a cord of length 6.0 m wound around its periphery. Starting from rest, the wheel is given a const

ant angular acceleration of 3.0 rad/s2. How long will it take the cord to unwind?
Physics
1 answer:
MAVERICK [17]3 years ago
6 0

Answer: time t = 10s

Explanation:

Given;

Diameter = 8.0cm = 0.08m

Radius r = diameter/2 = 0.08/2 = 0.04m

Cord length d = 6.0m

Angular acceleration = 3.0rads/s2

Time = t

Converting the angular acceleration to linear acceleration

a = a* × r = 3.0 × 0.04 = 0.12m/s

d = vt + 1/2 (a × t^2)

Initial velocity v = 0, vt = 0 therefore;

d = 1/2 ( a × t^2)

t = √(2d/a)

t = √ [(2× 6)/0.12]

t = 10s

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Option B is correct.

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2 years ago
A stationary siren creates an 894 Hz
valentina_108 [34]

Answer:

12.3 m/s

Explanation:

The Doppler equation describes how sound frequency depends on relative velocities:

fr = fs (c + vr)/(c + vs),

where fr is the frequency heard by the receiver,

fs is the frequency emitted at the source,

c is the speed of sound,

vr is the velocity of the receiver,

and vs is the velocity of the source.

Note: vr is positive if the receiver is moving towards the source, negative if away.

Conversely, vs is positive if the receiver is moving away from the source, and negative if towards.

Given:

fs = 894 Hz

fr = 926 Hz

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5 0
3 years ago
A magnetic field is entering into a coil of wire with radius of 2(mm) and 200 turns. The direction of magnetic field makes an an
frozen [14]

Answer:

a) <em>2.278 x 10^-5 volts</em>

b) <em>1.139 x 10^-6 Ampere</em>

c) <em>2.59 x 10^-11 W</em>

Explanation:

The radius of the wire r = 2 mm = 0.002 m

the number of turns N = 200 turns

direction of the magnetic field ∅ = 25°

magnetic field strength B = 0.02 T

varying time = 2 sec

The cross sectional area of the wire = \pi r^{2}

==> A = 3.142 x 0.002^{2} = 1.257 x 10^-5 m^2

Field flux Φ = BA cos ∅ = 0.02 x 1.257 x 10^-5 x cos 25°

==> Φ = 2.278 x 10^-7 Wb

The induced EMF is given as

E = NdΦ/dt

where dΦ/dt = (2.278 x 10^-7)/2 = 1.139 x 10^-7

E = 200 x 1.139 x 10^-7 = <em>2.278 x 10^-5 volts</em>

<em></em>

<em></em>

b) If the two ends are connected to a resistor of 20 Ω, the current through the resistor is given as

I = E/R

where R is the resistor

I = (2.278 x 10^-5)/20 = <em>1.139 x 10^-6 Ampere</em>

<em></em>

<em></em>

<em> </em>c) power delivered to the resistor is given as

P = IE

P = (1.139 x 10^-6) x (2.278 x 10^-5) = <em>2.59 x 10^-11 W</em>

4 0
3 years ago
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