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2 years ago

Part H)

1 answer:

5 0

At t = 0 a truck starts from rest at x = 0 and speeds up in the positive x-direction on..., the time the truck passes the car and the coordinate the truck passes the car is mathematically given as

$t = \frac{2v_{C}}{a_{T} + a_{C}}$
$d_{T} = v_{T}t + \frac{a_{T}t^{2}}{2}$

<h3>What time does the truck pass the car and coordinate does the truck pass the car?</h3>

Generally, the equation for Distance traveled is mathematically given as

By Truck

$d_{T} = v_{T}t + \frac{a_{T}t^{2}}{2}$

By Car

$d_{C} = v_{C}t - \frac{a_{C}t^{2}}{2}$

In conclusion, when the conditions the truck passes the car

$d_{T} = d_{C}$

Therefore

$v_{t} + \frac{a_{t}t}{2} = v_{C} - \frac{a_{C}t}{2}$

$\frac{a_{T}t}{2} = v_{C} - \frac{a_{C}t}{2}$

Therefore

$t = \frac{2v_{C}}{a_{T} + a_{C}}$

differentiation of T

$d_{T} = v_{T}t + \frac{a_{T}t^{2}}{2}$

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Answer:

-2

Explanation:

Since its a compound, it's charge is 0.

So, there are 2 Fe and 3 Oxygen,

let x = oxidation number of oxygen

2(+3) + 3x =0

x =-2

Side note : even if the oxidation number is positive, make sure to put the positive sign like how you would in numbers

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3 years ago

Runner A is initially 6.0 km west of a flagpole and is running with a constant velocity of 9.0 km/h due east. Runner B is initia

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Answer:

0.176m from the flagpole, westward.

Explanation:

Let the Eastward be the positive direction. So initially runner A is at position -6km, running with velocity of 9km/h while runner B is at position 5km running at a velocity of -8km/h. We can conduct the following equation for their distances over the same time t

$s_A = -6 + 9t$

$s_B = 5 - 8t$

When A an B meets, they are at the same position and at the same time. So

$s_A = s_B$

$-6 +9t = 5 - 8t$

$17t = 5 + 6 = 11$

$t = 11/17 = 0.647 s$

$s_A = -6 + 9*0.647 = -0.176 m$

So where they meet is 0.176m from the flagpole, westward.

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3 years ago

A ball has an electric charge of +1.5 × 10-9 coulombs. At what distance from the ball's center is the electric field strength eq

dimaraw [331]

E=Fe/q
5.8x10^5N/C=Fe/(1.5x10^-9C)
Fe=(5.8x10^5N/C)(1.5x10^-9C)
Fe=8.7x10^-4N

Fe=kq1q2/r²
8.7X10^-4N=(8.99x10^9N·m²/C²)(1.5x10^-9C)(1.5x10-9C)/r²
r²=(8.99x10^9N·m²/C²)(1.5x10^-9C)(1.5x10-9C)/(8.7x10^-4N)
√r²=√0.00002325
The final answer is r=4.8x10-3m

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What can a paleontologist tell from fossil footprints of a dinosaur

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They can tell what type/species the dinosaur was

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A high voltage transmission line with a resistance of 0.51 Ω/km carries a current of 1099 A. The line is at a potential of 1300

Illusion [34]

<h2>Answer:</h2>

$\boxed{P=96.09MW}$

<h2>Explanation:</h2>

First of all, we need to figure out what is the resistance in that line. In this problem, the total resistance is not given directly, but we can calculate it because we know it in terms of 0.51 Ω/km and since the distance from the power station to the city is 156km, then:

$R_{line}=0.51 \frac{\Omega}{km}.156km \\ \\ R_{line}=79.56\Omega$

So we can calculate the power loss as:

$P=I^2R \\ \\ Where: \\ \\I=1099A \\ \\ P=(1099)^2(79.56) \\ \\ P=96092647.56W \\ \\ Remember \ that \ 1MW=10^6W \ So: \\ \\ P=96092647.56W(\frac{1M}{10^6}) \\ \\ \boxed{P=96.09MW}$

Finally, the power loss due to resistance in the line is 96.09MW

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3 years ago