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3 years ago

8

Where is the independent variable usually located in a scientific table? First column Second column Third through fifth columns

First and second columns

1 answer:

Reika [66]3 years ago

4 0

Explanation:

On the horizontal side left side

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On isolated ground receptacles, the metal yoke ____ integrally bonded to the equipment grounding terminal of the receptacle.

Ede4ka [16]

On isolated ground receptacles, the metal yoke is not allowed to be integrally bonded to the equipment grounding terminal of the receptacle.

Any device with two distinct switches or receptacles is a duplex device. It can be shaped to fit a Decora opening or a typical duplex plate opening. It should be noted that they can be combination devices with a switch/outlet, switch/pilot light, etc.

Because of grounding connection removal and receptacle, it is utterly undesirable to connect the two bare equipment grounding conductors together directly.

The equipment grounding conductor associated with those circuits must be connected to the box when circuit conductors are terminated on equipment inside a metal box to prevent unneeded current discharge.

Learn more about grounding conductors here brainly.com/question/14886979

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2 years ago

Why do lithospheric plates move constantly?

Charra [1.4K]

I think the correct answer from the choices listed above is option C. Lithospheric plates move constantly due to the density differences. Lithospheric plates<span> are another name for tectonic plates, which are part of the Earth's uppermost layer, the crust.</span>

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3 years ago

Suppose a 2662 kg giant seal is placed on a scale and produces a 20.0 cm compres- sion. If the seal and spring system are set in

Yuki888 [10]

Answer:

0.8976 seconds

Explanation:

The period of oscillation for the simple harmonic motion can be found using the formula ...

T = 2π√(d/g)

where d is the displacement of the spring due to the attached weight, and g is the acceleration due to gravity.

__

For d = 0.20 meters, the period is ...

T = 2π√(0.20/9.8) ≈ 0.8976 . . . . seconds

_____

<em>Additional comment</em>

The formula for the oscillator period is usually seen as ...

T = 2π√(m/k)

where m is the mass in the system and k is the spring constant. The value of the spring constant is calculated from ...

k = mg/d

Using that in the formula, we find it simplifies to ...

$T=2\pi\sqrt{\dfrac{m}{k}}=2\pi\sqrt{\dfrac{m}{\left(\dfrac{mg}{d}\right)}}=2\pi\sqrt{\dfrac{d}{g}}$

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2 years ago

agasfer [191]

Answer:

For a velocity versus time graph how do you know what the velocity is at a certain time?

Ans: By drawing a line parallel to the y axis (Velocity axis) and perpendicular to the co-ordinate of the Time on the x axis (Time Axis). The point on the slope of the graph where this line intersects, will be the desired velocity at the certain time.

_____________________________________________________

How do you know the acceleration at a certain time?

$Ans: We\ know\ that\ acceleration = \frac{Final\ velocity-Initial\ Velocity}{Time\ taken}$

Hence,

By dividing the difference of the Final and Initial Velocity by the Time Taken, we could find the acceleration.

_________________________________________________________

How do you know the Displacement at a certain time?

Ans: As Displacement equals to the area enclosed by the slope of the Velocity-Time Graph, By finding the area under the slope till the perpendicular at the desired time, we find the Displacement.

_________________________________________________________

4 0

3 years ago

A rocket, initially at rest on the ground, accelerates straight upward from rest with constant acceleration 34.3 m/s^2 . The acc

Eddi Din [679]

Answer:

The maximum height reached by the rocket is 1.94 × 10³ m.

Explanation:

The height of the rocket can be calculated using the following equations:

y = y0 + v0 · t + 1/2 · a · t² (when the rocket is accelerated upward).

y = y0 + v0 · t + 1/2 · g · t² (after the rocket runs out of fuel).

Where:

y = height at time t.

y0 = initial height.

v0 = initial velocity.

t = time.

a = acceleration due to engines of the rocket.

g = acceleration due to gravity.

In the same way, the velocity of the rocket can be calculated as follows:

v = v0 + a · t (when the rocket has fuel)

v = v0 + g · t (when the rocket runs out of fuel)

Where "v" is the velocity at time "t"

First, let´s find the height reached until the rocket runs out of fuel.

y = y0 + v0 · t + 1/2 · a · t²

y = 0 m + 0 m/s · t + 1/2 · 34.3 m/s² · (5.00 s)²

y = 429 m

And now, let´s find the velocity reached in that time of upward acceleration:

v = v0 + a · t

v = 0 m/s + 34.3 m/s² · 5.00 s

v = 172 m/s

When the rocket runs out of fuel, it is accelerated downward due to gravity. But, since the rocket has initially an upward velocity (172 m/s), it will not fall immediately and will continue to go up until the velocity becomes 0. In that instant, the rocket is at its maximum height and thereafter it will start to fall with negative velocity.

Then, using the equation for velocity, we can calculate the time it takes the rocket to reach its maximum height:

v = v0 + g · t

0 = 172 m/s - 9.80 m/s² · t

-172 m/s / -9.80 m/s² = t

t = 17.6 s

With this time, we can now calcualte the maximum height. Notice that the initial velocity and height are the ones reached during the upward acceleration phase:

y = y0 + v0 · t + 1/2 · g · t²

ymax = 429 m + 172 m/s · 17.6 s - 1/2 · 9.80 m/s² · (17.6 s)²

ymax = 1.94 × 10³ m

4 0

4 years ago