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3 years ago

How often would you push someone on a swing to create a frequency of .4 hertz

1 answer:

4 0

The unit "Hertz" means "per second".

Anything happening with a frequency of 0.4 Hz is happening 0.4 per second.

Technically, you must push the swing 0.4 times per second.
But since it's very hard to push 0.4 of one time, you have to be
able to figure out that you should push once every 2.5 seconds.

(1 push / 2.5 sec) = (1/2.5) push/sec = 0.4 per sec = 0.4 Hz .

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A 450.0 N, uniform, 1.50 m bar is suspended horizontally by two vertical cables at each end. Cable A can support a maximum tensi

bogdanovich [222]

Answer:

1) $W_{object} = 400 N$

2) x = 0.28 m from cable A.

Explanation:

1 ) Let's use the first Newton to find the , because bar is in equilibrium.

$\sum F_{Tot} = 0$

In this case we just have y-direction forces.

$\sum F_{Tot} = T_{A}+T_{B}-W_{bar}-W_{object} = 0$

Now, let's solve the equation for W(object).

$W_{object} = T_{A}+T_{B}-W_{bar} = 550 +300 - 450 = 400 N$

2 ) To find the position of the heaviest weight we need to use the torque definition.

$\sum \tau = 0$

The total torque is evaluated in the axes of the object.

Let's put the heaviest weight in a x distance from the cable A. We will call this point P for instance.

First let's find the positions from each force to the P point.

$L = 1.50 m$ ; total length of the bar.

$D_{AP} = x$ ; distance between Tension A and P point.

$D_{BP} = L-x$ ; distance between Tension B and P point.

$D_{W_{bar}P} = \frac{L}{2}-x$ ; distance between weight of the bar (middle of the bar) and P point.

Now, let's find the total torque in P point, assuming counterclockwise rotation as positive.

$\sum \tau = T_{B}(L-x)-T_{A}(x)-W_{bar}(\frac{L}{2}-x) = 0$

Finally we just need to solve it for x.

$x = \frac{T_{B}L-W_{bar}(L/2)}{T_{B}+W_{bar}+T_{A}}$

$x = 0.28 m$

So the distance is x = 0.28 m from cable A.

Hope it helps!

Have a nice day! :)

8 0

4 years ago

NEED HELP PLEASE!!!!!! 19 POINTS!!

podryga [215]

When the kinetic energy is transformed into another for of energy during the collision or impact

5 0

3 years ago

What is the power of a 12 V, 50 A light?

lora16 [44]

Answer:

42Amp

Dad's house is a message from

5 0

3 years ago

A 1.00 x 109 kg object is raised vertically at a

DerKrebs [107]

Answer:

P = 3.92 10¹⁰ W

Explanation:

The power is data by the expression

P = W / t

the work of a force is

W = F. y

the bold ones represent vectors. In this case the displacement is vertical upwards and the vertical forces upwards, therefore the angle is zero and the cos 0 = 1

W = F y

we substitute

P = F y / t

P = F v

as the body rises at constant speed the acceleration is zero and from the equilibrium condition

F -W = 0

F = mg

we substitute

P = m g v

let's calculate

P 1.00 10⁹ 9.8 4

P = 3.92 10¹⁰ W

5 0

3 years ago

Introduction Team Problem 2 You jump into the deep end of Legion Pool and swim to the bottom with a pressure gauge. The gauge at

Kaylis [27]

Answer:

Depth of the pool, h = 4.004 cm

Explanation:

Pressure at the bottom, P = 39240 N/m²

The density of water, d = 1000 kg/m³

The pressure at the bottom is given by :

P = dgh

We need to find the depth of pool. Let h is the depth of the pool. So,

$h=\dfrac{P}{dg}$

$h=\dfrac{39240\ N/m^2}{1000\ kg/m^3\times 9.8\ m/s^2}$

h = 4.004 m

So, the pool is 4.004 meters pool. Hence, this is the required solution.

5 0

4 years ago