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vazorg [7]
4 years ago
9

Explain why the amplitude of the wave did not change when you increased the frequency of the stimulation. how well did the resul

ts compare with your prediction?
Physics
1 answer:
barxatty [35]4 years ago
5 0
The amplitude did not change when the recurrence was expanded on the grounds that the long headstrong time of the heart forestalls adjustment. It is the most extreme removal or separation moved by a point on a vibrating body or wave measured from its balance position. It is equivalent to the one-a large portion of the length of the vibration way.
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A CD-ROM drive in a computer spins the 12-cm-diameter disks at 9500 rpm. Find acceleration in units of g that a speck of dust on
NeTakaya
The net force on an object is:
F = ma
The only force acting on the speck of dust as it is lays on the CD is centripetal force given by:
F = (mv²)/r
Equating the two
ma = (mv²)/r
We get:
a = v²/r
v is the linear velocity in this case; however, we can calculate only the angular velocity with the given data. Therefore, we must use:
v = ωr; where
ω = 2πf; f is the rotations per second
f = 9500 / 60
f = 950/6
ω = 2π(950/6)
ω = (950π)/3
v = (950π)/3 × 0.12
v = 38π
a = (38π)²/0.12
a = 1.2 × 10⁵ m/s²
To express this in terms of gravitational acceleration, we divide by the value of gravitational acceleration, 9.81
a = 1.2 × 10⁴ g
4 0
3 years ago
If the mass of a planet is 0.231 mE and its radius is 0.528 rE, estimate the gravitational field g at the surface of the planet.
crimeas [40]

Answer:

8.1 m/s^2

Explanation:

The strength of the gravitational field at the surface of a planet is given by

g=\frac{GM}{R^2} (1)

where

G is the gravitational constant

M is the mass of the planet

R is the radius of the planet

For the Earth:

g_E = \frac{GM_E}{R_E^2}=9.8 m/s^2

For the unknown planet,

M_X = 0.231 M_E\\R_X = 0.528 R_E

Substituting into the eq.(1), we find the gravitational acceleration of planet X relative to that of the Earth:

g_X = \frac{GM_X}{R_X^2}=\frac{G(0.231M_E)}{(0.528R_E)^2}=\frac{0.231}{0.528^2}(\frac{GM_E}{R_E^2})=0.829 g_E

And substituting g = 9.8 m/s^2,

g_X = 0.829(9.8)=8.1 m/s^2

3 0
3 years ago
A(n) _____ is a gap in the geologic record where some rock layers have been lost because of erosion.
Elenna [48]
I'm not sure, I think it's option A.

Let me know if I'm wrong!
4 0
3 years ago
If a 0.15 kg ball falls and has a KE of 20 J just before striking the ground, from what height did it fall. A. 1.36m B. 3m C. 13
RUDIKE [14]
According to the conservation of mechanical energy, the kinetic energy just before the ball strikes the ground is equal to the potential energy just before it fell. 

Therefore, we can say KE = PE
We know that PE = m·g·h

Which means KE = m·g·h

We can solve for h:

h = KE / m·g
   = 20 / (0.15 · 9.8) 
   = 13.6m

The correct answer is: the ball has fallen from a height of 13.6m.

5 0
3 years ago
How much heat is needed to change 1.25 kg of steak at 100°C to water at 100°C?
cricket20 [7]

The heat required to change 1.25 kg of steak is 2825 kJ /kg.

<u>Explanation</u>:

Given, mass m = 1.25 kg,     Temperature t = 100 degree celsius

To calculate the heat required,

                                 Q = m \times L

where m represents the mass in kg,

          L represents the heat of vaporization.

When a material in the liquid state is given energy, it changes its phase from liquid to vapor and the energy absorbed in this process is called heat of the vaporization. The heat of vaporization of the water is about 2260 kJ/kg.

                                  Q = 1.25 \times 2260

                                  Q = 2825 kJ /kg.

7 0
4 years ago
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