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3 years ago

12

During a care on level ground, Andra runs with an average velocity of 6.02 m/s to the East. What distance does Andra cover in 13

7 seconds?

1 answer:

garik1379 [7]3 years ago

4 0

Answer:

The distance covered is: 824.74 meters

Explanation:

Use the formula for velocity (v) as the distance (d) covered over the time (t) it took:

v = d / t

in our case:

6.02 m/s = d / 137 s

d = 6.02 * 137 = 824.74 meters

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Two large conducting parallel plates A and B are separated by 2.4 m. A uniform field of 1500 V/m, in the positive x-direction, i

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Answer:

a. 1.027 x 10^7 m/s b. 3600 V c. 0 V and d. 1.08 MeV

Explanation:

a. KE =1/2 (MV^2) where the M is mass of electron

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3 years ago

When non-metric units were used in the United Kingdom, a unit of mass called the pound-mass (lbm) was employed, where 1lbm=0.453

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Answer:

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Percent uncertainty is 0.022%

b) For 1 kg uncertainty mass in kg would be

$\frac{1}{0.022}\times {100}=4545.5\ kg$

Mass in pounds would be

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3 years ago

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spin [16.1K]

Answer:

Explanation:

I got everything but i. Don't know why but it's eluding me. So let's do everything but that.

a. PE = mgh so

PE = (2.5)(98)(14) and

PE = 340 J

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$KE=\frac{1}{2}(2.5)(14)^2$ and

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TE = 340 + 250 and

TE = 590 J

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PE = (2.5)(9.8)(8.7) and

PE = 210 J

e. The KE at the same height:

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KE = 380 J

f. The velocity at that height:

$380=\frac{1}{2}(2.5)v^2$ and

$v=\sqrt{\frac{2(380)}{2.5} }$ so

v = 17 m/s

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590 = KE + PE and

PE = (2.5)(9.8)(11.6) so

PE = 280 then

590 = KE + 280 so

KE = 310 then

$310=\frac{1}{2}(2.5)v^2$ and

$v=\sqrt{\frac{2(310)}{2.5} }$ so

v = 16 m/s

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$v=v_0+at$ and

26 = 0 + 9.8t and

26 = 9.8t so the time at 26 m/s is

t = 2.7 seconds. Now we use that in the equation for displacement:

Δx = $v_0t+\frac{1}{2}at^2$ and filling in the time the object was at 26 m/s:

Δx = 0t + $\frac{1}{2}(-9.8)2.7)^2$ so

Δx = 36 m

i. ??? In order to find the velocity at which the object hits the ground we would need to know the initial height so we could find the time it takes to hit the ground, and then from there, sub all that in to find final velocity. In my estimations, we have 2 unknowns and I can't seem to see my way around that connundrum.

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2 years ago