The fundamental frequency of a musical note is 330Hz. Which harmonic is this?

What is the set of processes used to encode, store, and retrieve information over different periods of time?

kakasveta [241]

The set of processes used to encode, store, and retrieve information over different periods of time is called memory. Memory<span> makes us.</span><span> If we couldn't recall the who, what, where, and when of our everyday lives, we would never be able to manage.</span>

8 0

4 years ago

ASK YOUR TEACHER A meter stick is found to balance at the 49.7-cm mark when placed on a fulcrum. When a 51.5-gram mass is attach

Simora [160]

Answer:

0.114 kg or 114 g

Explanation:

From the diagram attaches,

Taking the moment about the fulcrum,

sum of clockwise moment = sum of anticlockwise moment.

Wd = W'd'

Where W = weight of the mass, W' = weight of the meter rule, d = distance of the mass from the fulcrum, d' = distance of the meter rule.

make W' the subject of the equation

W' = Wd/d'................ Equation 1

Given: W = mg = 0.0515(9.8) = 0.5047 N, d = (39.2-16) = 23.2 cm, d' = (49.7-39.2) = 10.5 cm

Substitute these values into equation 1

W' = 0.5047(23.2)/10.5

W' = 1.115 N.

But,

m' = W'/g

m' = 1.115/9.8

m' = 0.114 kg

m' = 114 g

6 0

3 years ago

The hour and minute hands of a tower clock like Big Ben in London are 2.79 m and 4.44 m long and have masses of 58.2 kg and 90 k

LuckyWell [14K]

Answer: 895.85 x 10^-6 J or 8.96 x 10^-4 J

Explanation:

Angular kinetic energy E in Joules

E = ½Iw^2

W is angular velocity in radians/sec

1 radian/sec = 9.55 rev/min

I is moment of inertia in kgm^2

I = cMR^2

M is mass (kg), R is radius (meters)

c = 1/3 for a rod around its end, R = length

For minute hand

I = (1/3)(90)(4.44)^2 = 0.33 x 90 x 19.7136 = 585.49

w= 1 rev/hour = 1 rev/3600sec = 2pi/ 3600 = pi/1800 rad/s

KE = 1/2(1/3)(90)(4.44)^2(pi/1800)^2 = 0.5 x 0.33 x 90 x 19.7136 x 3.05 x 10^-6

KE = 0.00089 J

For hour hand

I = (1/3)(58.2)(2.79)^2 = 0.33 x 58.2 x 2.79^2 = 149.5

w = 1 rev/12hour = 1 rev/(12x3600sec) = 2pi/ 12x3600 = pi/21600 rad/s

KE = 1/2(1/3)(90)(4.44)^2(pi/21600)^2 = 0.5 x 0.33 x 90 x 19.7136 x 2.12 x 10^-8

KE = 5.85 x 10^-6 J

Therefore total kinetic energy = 895.85 x 10^-6 J

4 0

4 years ago

A concave mirror has a radius of curvature of 10 cm. Find the location an height of the image if the distance of the object is 8

puteri [66]

R=10
F=R/2
F=10/2=5
F=-5(CONCAVE MIRROR)
U=-8(CONCAVE MIRROR)
HEIGHT OF OBJECT=1.5
V=?
HEIGHT OF IMAGE=?
I/F=1/U+1/V
-I/5=-1/8-1/V
-1/V=-1/5+1/8
-1/V=-8+5/40
-1/V=-3/40
1/V=3/40
V=40/3

HEIGHT OF IMAGE/HEIGHT OF OBJECT =-V/U
HEIGHT OF IMAGE=40/3*1/-8*15/10
=-20/8
=-2.5

7 0

3 years ago

You are driving up an inclined road. After 3.0 km you notice a roadside sign that indicates that your elevation has increased by

shutvik [7]

Answer:

~3°

Explanation:

x^2 + 0.16^2 = 3^2

x^2 + 0.0256 = 9

x^2 = 8.9744

x = sqrt(8.9744) = 2.99... = ~3

8 0

3 years ago