Which seismic waves are generally the last to arrive after an earthquake?

How to calculate moments with 3 separate weights of different amounts at different points?

Rina8888 [55]

I don't completely understand your drawing, although I can see that you certainly
did put a lot of effort into making it. But calculating the moment is easy, and we
can get along without the drawing.

Each separate weight has a 'moment'.
The moment of each weight is:

(the weight of it) x (its distance from the pivot/fulcrum) .

That's all there is to a 'moment'.

The lever (or the see-saw) is balanced when (the sum of all the moments
on one side) is equal to (the sum of the moments on the other side).

That's why when you're on the see-saw with a little kid, the little kid has to sit
farther away from the pivot than you do. The kid has less weight than you do,
so he needs more distance in order for his moment to be equal to yours.

6 0

3 years ago

A 3.00 kg object is moving in the XY plane, with its x and y coordinates given by x = 5t³ !1 and y = 3t ² + 2, where x and y are

Hatshy [7]

Answer:

The net force acting on this object is 180.89 N.

Explanation:

Given that,

Mass = 3.00 kg

Coordinate of position of $x= 5t^3+1$

Coordinate of position of $y=3t^2+2$

Time = 2.00 s

We need to calculate the acceleration

$a = \dfrac{d^2x}{dt^2}$

For x coordinates

$x=5t^3+1$

On differentiate w.r.to t

$\dfrac{dx}{dt}=15t^2+0$

On differentiate again w.r.to t

$\dfrac{d^2x}{dt^2}=30t$

The acceleration in x axis at 2 sec

$a = 60i$

For y coordinates

$y=3t^2+2$

On differentiate w.r.to t

$\dfrac{dy}{dt}=6t+0$

On differentiate again w.r.to t

$\dfrac{d^2y}{dt^2}=6$

The acceleration in y axis at 2 sec

$a = 6j$

The acceleration is

$a=60i+6j$

We need to calculate the net force

$F = ma$

$F = 3.00\times(60i+6j)$

$F=180i+18j$

The magnitude of the force

$|F|=\sqrt{(180)^2+(18)^2}$

$|F|=180.89\ N$

Hence, The net force acting on this object is 180.89 N.

3 0

3 years ago

Why is the scientific method described as cyclic?

Troyanec [42]

Because the scientific method can go around in a circle as many times as neccisary to get the results you need

8 0

3 years ago

An open pipe, 0.29 m long, vibrates in the second overtone with a frequency of 1,227 Hz. In this situation, the fundamental freq

Andru [333]

Answer:

f = 409 Hz

Explanation:

We have,

Length of the open organ pipe, l = 0.29 m

Frequency of vibration of second overtone, $f_2 = 1227 Hz$

It is required to find the fundamental frequency of the pipe. For the open organ pipe, the frequency of second overtone is given by :

$f_2=\dfrac{3v}{2l}$

v is speed of sound

Let f is the fundamental frequency. It is given by :

$f=\dfrac{v}{2l}$

The relation between f and f₂ can be written as :

$f_2=3f\\\\f=\dfrac{f_2}{3}\\\\f=\dfrac{1227}{3}\\\\f=409\ Hz$

So, the fundamental frequency of the pipe is 409 Hz.

6 0

3 years ago

Please solve the Problem.

STatiana [176]

(a) For series circuit, current in 14 ohms = current in 72 ohms = 0.698 A

(b) Power loss in each series resistor, 14 ohms = 6.82 W and 72 ohms = 35.1 W.

(c) For parallel circuit, current in 14 ohms = 4.29 A and current in 72 ohms = 0.83 A

(d) Power loss in each parallel resistor, 14 ohms = 257.4 W and 72 ohms = 49.8 W.

<h3>Current in each series resistors</h3>

The total resistance = R1 + R2

R = 14 + 72 = 86 ohms

Current = V/R = 60/86 = 0.698 A

Since the resistors are in series, current in 14 ohms = current in 72 ohms = 0.698 A

<h3>Power loss in each series resistor</h3>

P = I²R

P(14 ohms) = (0.698)² x 14 = 6.82 W

P(72 ohms) = (0.698)² x 72 = 35.1 W

<h3>Current in each resistor for parallel arrangement</h3>

Total resistance, 1/R = 1/R1 + 1/R2

1/R = 1/14 + 1/72

1/R = 0.0853

R = 1/0.0853

R = 11.72

Total current in the circuit = V/R = 60/11.72 = 5.12 A

Current in 14 ohms = 60/14 = 4.29 A

Current in 72 ohms = 60/72 = 0.83 A

<h3>Power loss in each parallel resistor</h3>

P = IV

P(14 ohms) = (4.29) x 60 = 257.4 W

P(72 ohms) = 0.83 x 60 = 49.8 W

Learn more about current here: brainly.com/question/24858512

#SPJ1

5 0

2 years ago