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Masja [62]
2 years ago
12

an audio CD has a diameter of 120 mm and spins at up to 540 rpm. When a CD is spinning at its maximum rate, how much time is req

uired for one revolution? Of a speck of dust rides on the outside edge of the disk,
Physics
1 answer:
Andru [333]2 years ago
6 0

Answer:

 t = 0.1111 s

Explanation:

Let's reduce the magnitudes to the SI system

    d = 120 mm (1m / 1000 mm)

    d= 0.120 m

    w = 540 rpm (2pi rad / 1 rev) (1 min / 60s)  

     w= 56.55 rad / s

When at maximum speed we can use angular kinematic relationships to find the time for a sperm revolution with zero angular acceleration

     W = θ / t

     t = θ / w

     t = 2π / 56.55

     t = 0.1111 s

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If the resistivity of copper is less than that of gold at room temperature, which of the following statements must be true? Gold
KiRa [710]

Answer:

Gold Has A Higher Resistance Than Copper. The Sample Of Gold Is Thinner Than The Sample Of Copper. Electrons In Gold Are More Likely To Be Scattered Than Electrons In Copper At Room Temperature When they are exelerated by the same electric field.

Explanation:

5 0
3 years ago
Ian walks 2 km to his best friend's house, then walks 0.5 km to the library. He then makes a 2.5 km walk home. The entire walk t
earnstyle [38]

Average speed = (total distance covered) / (time to cover the distance)

Ian's total distance covered = (2km + 0.5km + 2.5km) = 5 km.

His time to cover the distance = 3 hours.

Average speed = (5 km) / (3 hrs)

Average speed = (5/3) (km/hr)

<em>Average speed = 1.67 km/hr</em>

5 0
3 years ago
Assume: Moving to the right is positive. A(n) 7.7 g object moving to the right at 22 cm/s makes an elastic head-on collision wit
krek1111 [17]

Answer:

Explanation:

We shall apply the formula for velocity in case of elastic collision which is given below

v₁ = (m₁ - m₂)u₁ /  (m₁ +  m₂)  + 2m₂u₂ / (m₁ +  m₂)

m₁ and u₁ is mass and velocity of first object , m₂ and  u₂ is mass and velocity of second object before collision and v₁ is velocity of first velocity after collision.

Here u₁ = 22 cm /s , u₂ = - 14 cm /s . m₁ = 7.7 gm , m₂ = 18 gm

v₁ = ( 7.7 - 18 ) x 22 / ( 7.7 + 18 )  + 2 x 18 x - 14 / ( 7.7 + 18 )

= - 8.817 - 19.6

= - 28.4 cm / s

5 0
3 years ago
A 3.50-meter length of wire with a cross-sectional area of 3.14 × 10-6 meter2 is at 20° Celsius. If the wire has a resistance of
maks197457 [2]

Answer:

5.6\times 10^{-8}\ Ohm.m

Explanation:

Resistivity is given by \rho=\frac {AR}{L} where A is cross-sectional area, R is resistance, L is the length and \rho is the reistivity. Substituting 0.0625 for R, 3.14 × 10-6 for A and 3.5 m for L then the resistivity is equivalent to

\rho=\frac {3.14\times 10^{-6}\times 0.0625}{3.5}=5.60714285714285714285714285714285714285\times 10^{-8}\approx 5.6\times 10^{-8}\ Ohm.m

8 0
3 years ago
Two satellites are in circular orbits around Earth. Satellite A has speed vA. Satellite B has an orbital radius nine times that
Pani-rosa [81]

Answer:

option B

Explanation:

given,

Satellite B has an orbital radius nine times that of satellite A.

R' = 9 R

now, orbital velocity of the satellite A

........(1)

now, orbital velocity of satellite B

from equation 1

hence, the correct answer is option B

8 0
2 years ago
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