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kirill115 [55]
1 year ago
5

The amplitude of a lightly damped harmonic oscillator decreases from 60.0 cm to 40.0 cm in 10.0 s. What will be the amplitude of

the harmonic oscillator after another 10.0 s passes?

Physics
1 answer:
Svetllana [295]1 year ago
3 0
This looks like exponential decay starting at 60cm when t is set to zero. See enclosed ...

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_____ has a longer wavelength than _____. _____ has a longer wavelength than _____. Blue ... green Green ... yellow Red ... gree
nadya68 [22]

Answer:

red has a longer wavelength than yellow. Yellow has a longer wavelength than green.

Explanation

in the visible spectrum a color with high frequency  contains shorter wavelength.In this case red color has high wavelength followed by yellow followed by green,blue and violet.

3 0
2 years ago
Read 2 more answers
The total charge that an automobile battery can supply without being recharged is given in terms of ampere-hours. A typical 12 V
Lelechka [254]

Answer:

7.894 Hours.

Explanation:

Based on information number hours that this battery will last with give load  has mathematical relation of.

t = \frac{60Ah}{load in amperes.}

with load 60A t =  1h, 30A t = 2h so on and forth.

two head lights draw total current of 2x3.8A = 7.6A.

putting this in above relation gives.

t = \frac{60Ah}{7.6A}=7.894 h.

That is how long will it be before battery is dead.

6 0
1 year ago
Consider the waveform expression. y (x, t) = ym sin (0.333x + 5.36 + 585t) The transverse displacement (y) of a wave is given as
Sonja [21]

Explanation:

The waveform expression is given by :

y(x,t)=y_m\ sin(0.333x+5.36+585t)...........(1)

Where

y is the position

t is the time in seconds

The general waveform equation is given by :

y(x,t)=y_m\ sin(kx+\phi+\omega t)..........(2)

Where

k=\dfrac{2\pi}{\lambda}

\omega=2\pi f

On comparing equation (1) and (2) we get :

0.333=\dfrac{2\pi}{\lambda}

\lambda=18.86\ m

585=2\pi f

f = 93.10 Hz

Time period, T=\dfrac{1}{f}

T=\dfrac{1}{0.010}

T = 0.010 s

Phase constant, \phi=5.36\ radian

Hence, this is the required solution.

8 0
2 years ago
A 3.0-kg mass and a 5.0-kg mass hang vertically at the opposite ends of a very light rope that goes over an ideal pulley. If the
AleksAgata [21]

Answer:

acceleration = 2.4525‬ m/s²

Explanation:

Data: Let m1 = 3.0 Kg, m2 = 5.0 Kg, g = 9.81 m/s²

Tension in the rope = T

Sol: m2 > m1

i) for downward motion of m2:

m2 a = m2 g - T

5 a = 5 × 9.81 m/s² - T  

⇒ T = 49.05‬ m/s² - 5 a     Eqn (a)‬

ii) for upward motion of m1

m a = T - m1 g

3 a = T - 3 × 9.8 m/s²

⇒ T =  3 a + 29.43‬ m/s²   Eqn (b)

Equating Eqn (a) and(b)

49.05‬ m/s² - 5 a = T =  3 a + 29.43‬ m/s²

49.05‬ m/s² - 29.43‬ m/s² = 3 a + 5 a

19.62 m/s² = 8 a

⇒ a = 2.4525‬ m/s²

5 0
1 year ago
You want to move a heavy box with mass 30.0 kg across a carpeted floor. You pull hard on one of the edges of the box at an angle
charle [14.2K]

Answer:

a=5.54m/s^{2}

Explanation:

The net force, F_{net} of the box is expressed as a product of acceleration and mass hence

F_{net}=ma where m is mass and a is acceleration

Making a the subject, a= \frac {F_{net}}{m}

From the attached sketch,  

∑ F_{net}=Fcos\theta-F_{f} where F_{f} is frictional force and \theta is horizontal angle

Substituting ∑ F_{net} as F_{net} in the equation where we made a the subject

a= \frac {Fcos\theta-F_{f}}{m}

Since we’re given the value of F as 240N, F_{f} as 41.5N, \theta as 30^{o} and mass m as 30kg

a= \frac {240cos30-41.5}{30.0}=\frac {166.346}{30.0}=5.54m/s^{2}

6 0
2 years ago
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