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riadik2000 [5.3K]

3 years ago

7

The equation of damped oscillations is given in the form x=0.05e^-0.25sin½πt (m). Find the velocity of an oscillating point at t

he moments of time: 0, T, 2T, 3T and 4T.

2 answers:

zhannawk [14.2K]3 years ago

8 0

Explanation:

The logarithmic damping decrement of a mathematical pendulum is DeltaT=0.5. How will the amplitude of oscillations decrease during one full oscillation of the pendulum

Salsk061 [2.6K]3 years ago

6 0

Answer:

Explanation:

ω = π/2 = 2π/T

T = 2π/(π/2) = 4 s

velocity is the derivative of position

v(t) = (-1/80)(e^(-t/4)[sin(πt/2) - 2πcos(πt/2)]

0T = v(0) = (-1/80)(e^(-0/4)[sin(π0/2) - 2πcos(π0/2)] = 0.078540

T = v(4) = (-1/80)(e^(-4/4)[sin(π4/2) - 2πcos(π4/2)] = 0.028893

2T = v(8) = (-1/80)(e^(-8/4)[sin(π8/2) - 2πcos(π8/2)] = 0.010629

3T = v(12) = (-1/80)(e^(-12/4)[sin(π12/2) - 2πcos(π12/2)] = 0.003910

4T = v(16) = (-1/80)(e^(-16/4)[sin(π16/2) - 2πcos(π16/2)] = 0.001439

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A pendulum is made by letting a 4 kg mass swing at the end of a string that has a length of 1.5 meter. The maximum angle that th

olga nikolaevna [1]

Answer:

Approximately $7.8\; \rm J$ .

Explanation:

The change in the gravitational potential energy of the pendulum is directly related to the change in its height.

Refer to the sketch attached. The pendulum is initially at $\rm P_2$ . Its highest point is at $P_1$ . The length of segment $\rm BP_2$ gives the change in its height.

The lengths of $\rm AP_1$ and $\rm AP_2$ are simply the length of the string, $1.5\; \rm m$ . To find the length of $\rm BP_2$ , start by calculating the length of $\rm AB$ .

$\rm AB$ forms a leg in the right triangle $\rm \triangle AP_1B$ . Besides, it is adjacent to the $30^\circ$ angle $\rm P_1\hat{A}B$ . Its length would be:

$\rm AB = 1.5 \times \cos(30^\circ) \approx 1.30\; \rm m$ .

The length of $\rm BP_2$ would thus be

$\rm BP_2 = AP_2 - AB = 1.5 - 1.30 \approx 0.20\; \rm m$ .

The change in gravitational potential energy can be found with the equation

$\Delta \mathrm{GPE} = m \cdot g \cdot \Delta h$ . In this equation,

$m$ is the mass of the object,
$g \approx 9.81\; \rm N \cdot kg^{-1}$ near the surface of the earth, and
$\Delta h$ is the change in the object's height.

In this case, $m = 4\; \rm kg$ and $\Delta h \approx 0.20\; \rm m$ . Therefore:

$\Delta \mathrm{GPE} = 4 \times 9.81 \times 0.20 \approx 7.8\; \rm J$ .

6 0

3 years ago

A go-cart and rider have a mass of 14 kg. If the cart accelerates at 6m/s^2 during a 40 m sprint in 100 seconds, how much power

lara [203]

Answer:

P = 33.6 [N]

Explanation:

To solve this problem we must use Newton's second law, which tells us that the sum of forces on a body is equal to the product of mass by acceleration.

∑F = m*a

where:

F = forces [N]

m = mass = 14 [kg]

a = acceleration = 6 [m/s²]

$F = 14*6\\F = 84 [N]$

In the second part of this problem we must find the work done, where the work in physics is known as the product of force by distance, it is important to make it clear that force must be applied in the direction of movement.

$W = F*d$

where:

W = work [J]

F = force = 84 [N]

d = displaciment = 40 [m]

$W = 84*40\\W = 3360 [J]$

Finally, the power can be calculated by the relationship between the work performed in a given time interval.

$P=W/t\\$

where:

P = power [W]

W = work = 3360 [J]

t = time = 100 [s]

Now replacing:

$P=3360/100\\P=33.6[W]$

The power is given in watts

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3 years ago

A dancer starts 3 meters from the curtain then moves 8 meters from the curtain in 15 seconds. What was his velocity?

myrzilka [38]

I am pretty sure it is either 45 or 0.2

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3 years ago

Friends Burt and Ernie stand at opposite ends of a uniform log that is floating in a lake. The log is 3.0 m long and has mass 20

Taya2010 [7]

Answer:

The distance the log has moved by the time Ernie reaches Bur is 1.33 m.

Explanation:

give information:

The log is 3.0 m long and has mass 20.0 kg.

Burt has mass 30.0 kg; Ernie has mass 40.0 kg

Ernie has mass 40.0 kg.

to find the distance, first, we have to calculate the center of mass

X = ∑ m x /∑m

= (20 x (3/2)) + (30 x 0) + (40 x 3)/ (20+30+40)

= 150/90

= 5/3

when Ernie walk, the center of the mass is

X = (70 x 0) + (20 x (3/2))/(70 + 20)

= 30/90

= 1/3

the distance of log moved = 5/3 - 4/3 = 1.33 m

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