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3 years ago

6

Motion equation of object is

1 answer:

Ganezh [65]3 years ago

5 0

Take the derivative to find the velocity of the object:

$v=\dfrac{\mathrm dx}{\mathrm dt}=-12+6t$

The object stops when $v=0$ :

$-12+6t=0\implies6t=12\implies t=2$

so the answer is E.

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A car slid off an icy 10m bridge and landed 12m away from the bridge. How much time was the car in the air? (Hunt: Projectile)

Contact [7]

You will use the height of the bridge from the ground.

Solution:

Formula to be used is y=Viy(t)+g(t^2)/2

Where:

Vi=initial velocity which is 0 m/s

y=10 m

Gravitational acceleration or g =9.8m/s^2

T= time you need

Substitute all the given to the formula

10m=(0m/s)(t)+(9.8m/s^2)(t^2)/2

10mx2=9.8m/s^2(t^2)

Now isolate the variable you want to find which is T or time

10mx2/9.8m/s^2=t^2

20m/9.8m/s^2=t^2

Square root of 2.04= square root of t^2

T=1.43 secs

The answer is 1.43 seconds

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3 years ago

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You are a support technician working in a data closet in a remote office. You suspect that a connectivity problem is related to

Nostrana [21]

Answer:

crimping tool

Explanation:

This is a tool employed in affixing a connector to the end of a network cable.

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3 years ago

Monochromatic light of wavelength 385 nm is incident on a narrow slit. On a screen 3.00 m away, the distance between the second

LiRa [457]

To solve this problem it is necessary to apply the concepts related to the concept of overlap and constructive interference.

For this purpose we have that the constructive interference in waves can be expressed under the function

$a sin\theta = m\lambda$

Where

a = Width of the slit

d = Distance of slit to screen

m = Number of order which represent the number of repetition of the spectrum

$\theta =$ Angle between incident rays and scatter planes

At the same time the distance on the screen from the central point, would be

$sin\theta = \frac{y}{d}$

Where y = Represents the distance on the screen from the central point

PART A ) From the previous equation if we arrange to find the angle we have that

$\theta = sin^{-1}(\frac{y}{d})$

$\theta = sin^{-1}(\frac{1.4*10^{-2}}{3})$

$\theta = 0.2673\°$

PART B) Equation both equations we have

$a sin\theta = m\lambda$

$a \frac{y}{d} = m\lambda$

Re-arrange to find a,

$a = \frac{(2)(385*10^{-9})(3)}{(1.4*10^{-2})}$

$a = 1.65*10^{-4}m$

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3 years ago

A 0.106-A current is charging a capacitor that has square plates 6.00 cm on each side. The plate separation is 4.00 mm. (a) Find

FrozenT [24]

Answer:

The time rate of change of flux is $1.34 \times 10^{10}$ $\frac{V}{s}$

Explanation:

Given :

Current $I = 0.106$ A

Area of plate $A = 36 \times 10^{-4}$ $m^{2}$

Plate separation $d = 4 \times 10^{-3}$ m

(A)

First find the capacitance of capacitor,

$C = \frac{\epsilon _{o} A }{d}$

Where $\epsilon _{o} = 8.85 \times 10^{-12}$

$C = \frac{8.85 \times 10^{-12 } \times 36 \times 10^{-4} }{4 \times 10^{-3} }$

$C = 7.9 \times 10^{-12}$ F

But $C = \frac{Q}{V}$

Where $Q = It$

$C = \frac{It}{V}$

$V = \frac{It}{C}$

Now differentiate above equation wrt. time,

$\frac{dV}{dt} = \frac{I}{C}$

$= \frac{0.106}{7.9 \times 10^{-12} }$

$= 1.34 \times 10^{10}$ $\frac{V}{s}$

Therefore, the time rate of change of flux is $1.34 \times 10^{10}$ $\frac{V}{s}$

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3 years ago

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Semenov [28]

"Ionization energy" is the one among the following choices given in the question that <span>decreases with increasing atomic number in Group 2A. The correct option among all the options that are given in the question is the third option or option "C". I hope that the answer has helped you.</span>

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