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algol13
3 years ago
7

D= ((vf+vi)/2) t Solve for vi

Physics
2 answers:
erastovalidia [21]3 years ago
8 0
= t(Vf + Vi/2) 
<span>Vf + Vi/2 = d/t </span>
<span>Vf = (d/t) - Vi/2 </span>

<span>Answer: Vf = (d/t) - Vi/2 OR (2dt - Vit)/2t

</span>
vlada-n [284]3 years ago
7 0
D= ((vf+vi)/2) t

D = (vf +vi)*t/2

2*D = (vf +vi)*t

2D/t = vf +vi

vf +vi =  2D/t

vi = 2D/t - vf
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Explanation:

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a. Convert 21.0 cm to inches. Show your dimensional analysis setup. b. Convert 29.7 cm to inches. Show your dimensional analysis
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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
3 years ago
The voltage and power ratings of a particular light bulb, which are its normal operating values, are 110 V and 60 W. Assume the
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Answer:

The voltage operating is the 85,20563362

Explanation:

Power is the relation between Voltage and Current so knowing the resistance is going to be constant the equation of power can be just replacing in voltage terms:

1. P=I*V

using also Law OHM

2. V= I*R

I = \frac{V}{R}

Replacing 2 in the equation 1 so all the data are going to be in Voltage terms:

1. P= \frac{V}{R} *V

P= \frac{V^{2} }{R}  ⇒ R= \frac{V^{2} }{P}

R= \frac{110^{2} v }{60 w}

R= 201,6666667 Ω

So the resistance is constant so the current is going to be the same at the other Power 36 W:

P= \frac{V^{2} }{R}

V^{2} = R*P

V=\sqrt{R*P}

V=\sqrt{201,6666667*36 }

V=\sqrt{7260} ⇒V= 85,20563362

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Answer:

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Current, I = 14 kA = 14000 A

number of turns, N = 900

inner radius, r = 0.7 m

outer radius, R = 1.3 m

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B = \frac{\mu o}{4\pi}\times \frac{2 N\pi I}{R}

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(b) The magnetic field due to the outer radius is

B = 10^{-7}\times \frac{2\times 900\times 3.14\times 14000}{1.3}\\\\B = 6.09 T

5 0
3 years ago
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