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3 years ago

8

A tuning fork is struck and is vibrating at a single frequency. If a second tuning fork with the same frequency is placed near t

he first it begins to vibrate. This is an example of _____.
resonance
forced vibration
beats
Doppler effect

2 answers:

irina1246 [14]3 years ago

6 0

Forced Vibration hope this helps :P

gladu [14]3 years ago

4 0

It is not forced vibrations just turned it in and was wrong.

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A 2.0 µF capacitor is charged through a 50,000 ohm resistor. How long does it take for the capacitor to reach 90% of full charge

Nesterboy [21]

Answer:

0.23 s

Explanation:

First of all, let's find the time constant of the circuit:

$\tau=RC$

where

$R=50,000 \Omega$ is the resistance

$C=2.0\mu F=2.0\cdot 10^{-6}F$ is the capacitance

Substituting,

$\tau=(50,000 \Omega)(2.0\cdot 10^{-6}F)=0.1 s$

The charge on a charging capacitor is given by

$Q(t)=Q_0 (1-e^{-t/\tau} )$ (1)

where

$Q_0$ is the full charge

we want to find the time t at which the capacitor reaches 90% of the full charge, so the time t at which

$Q(t)=0.90 Q_0$

Substituting this into eq.(1) we find

$0.90 Q_0 = Q_0 (1-e^{-t/\tau})\\0.90=1-e^{-t/\tau}\\e^{-t/\tau}=1-0.90=0.10\\-\frac{t}{\tau}=ln(0.10)\\t=-\tau ln(0.10)=(0.1 s)ln(0.10)=0.23 s$

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3 years ago

What did classical physics predict would happen to the light given off by an object as its temperature increased?

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The correct option is C.
When the temperature of an object that is giving off light is increased, the particles in the object will move at a faster rate and there will be increased vibration of these molecules. This will makes the object to emit more light and to shine more brightly. Thus, the higher the temperature, the brighter the light that will be emitted.

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Can you answer this math homework? Please!

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$\large \mathfrak{Solution : }$

9. An object which is in circular motion (moving along a circle) is said to be accelerating because it changes it's direction constantly even if it is moving with a constant speed. cuz acceleration is change in either magnitude or direction of an object with respect to time.

therefore, it's still acceleration as change in direction with time.

10. Average speed of an object can be calculated by dividing the total distance covered by an object by time taken to cover that distance.

i.e

$\boxed{speed = \dfrac{distance}{time} }$

it can be re- arranged to find the distance as :

$\boxed{distance = speed \times time}$

$time = \dfrac{distance}{speed}$

11. speed = 20 m/s : conversion into km/h

distance covered : 4 km = 4000 m

$time = \dfrac{distance}{speed}$

$t = \dfrac{4000}{20}$

$t = 200 \: \: sec$

time taken = 200 seconds

12. let's use the first equation of motion to find the acceleration :

$v = u + at$

$50 = 80 + 120a$

$50 - 80 = 120a$

$a = \dfrac{ - 30}{120}$

$a = - \dfrac{1}{4} m/s {}^{2}$

$- 0.25 \: \: m/s {}^{2}$

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2 years ago

If 150 kcal of heat raises the temperature of 2.0 kg of a material by 400 F°, what is the specific heat capacity of the material

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it would be 0.338 kcal, which would round off to 0.34.

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In an arcade video game, a spot is programmed to move across the screen according to x = 9.00t - 0.750t³, where x is distance in

Nezavi [6.7K]

Answer:

a) The spot is at rest at t = 2.00 s.

b) The spot is at rest at x = 12 cm.

c) The acceleration at t = 2.00 s is -9.00 cm/s².

d) Before coming to rest, the spot is moving to the right.

e) After t = 2.00, the spot is moving to the left.

f) The spot reaches the edge of the screen at t = 3.46 s.

Explanation:

Hi there!

a) The velocity is defined as the variation of the position over time. If the time interval is very small and being "x" the position of the spot and "t" the time, the instantaneous velocity can be expressed as follows:

v = dx/dt

The position function is the following:

x(t) = 9.00 · t - 0.750 · t³

dx/dt is equal to the derivative of x(t):

v = dx/dt = 9.00 - 3 · 0.750 · t²

v = 9.00 - 2.25 · t²

When the spot is at rest, v = 0.

0 = 9.00 - 2.25 · t²

Solving for t:

-9.00/ -2.25 = t²

t = 2.00 s

The spot is at rest at t = 2.00 s.

b) To find the value of the position at t = 2.00, we have to evaluate the function x(t) at t = 2.00 s.

x(2.00) = 9.00 · (2.00) - 0.750 · (2.00)³

x(2.00) = 12 cm

The spot is at rest at x = 12 cm

c) The acceleration is defined as the variation of the velocity over time. If the time is very small, we get the instantaneous acceleration:

a = dv/dt

Then, we have to derivate the velocity function:

v(t) = 9.00 - 2.25 · t²

a = dv/dt = -2 · 2.25 t

a(t) = -4.50 · t

At t = 2.00 s, the acceleration will be a(2.00):

a(2.00) = -4.50 · 2.00

a(2.00) = -9.00 cm/s²

The acceleration at t = 2.00 s is -9.00 cm/s²

d and e) The velocity indicates the direction of displacement. If it is positive, the spot is moving to the right (away from the origin at x = 0) and if it is negative the spot is returning to the origin, moving to the left. Let´s evaluate the velocity function at t = 1.99 and 2.01:

v(1.99) = 9.00 - 2.25 · (1.99)²

v(1.99) = 0.09 cm/s

v(2.01) = 9.00 - 2.25 · (2.01)²

v(2.01) = -0.09 cm/s

Before coming to rest, the spot is moving to the right. After t = 2.00, the spot is moving to the left.

f) We have to find the time at which x = 0. Notice that since the acceleration is always negative and the spot comes to rest at x = 12 cm and then returns to the origin, the spot will never reach the edge at x = 15 cm. So, let´s find the time at which x = 0:

x = 9.00 · t - 0.750 · t³

At x = 0:

0 = 9.00 · t - 0.750 · t³

Solving for t:

0 = t ( 9.00 - 0.750 t²)

0 = 9.00 - 0.750 t²

-9.00 / - 0.750 = t²

t = 3.46 s

The spot reaches the edge of the screen at t = 3.46 s.

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3 years ago