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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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A single-turn current loop, carrying a current of 4.03 A, is in the shape of a right triangle with sides 68.1, 151, and 166 cm.

SashulF [63]

Answer:

Part a)

$F = 0$

Part b)

$F = 0.25 N$

Part c)

$F = 0.25 N$

Part d)

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Explanation:

As we know that force on a current carrying wire is given as

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now we have

Part a)

current in side 166 cm and magnetic field is parallel

so we have

$F = i(\vec L \times \vec B)$

here we know that L and B is parallel to each other so

$F = 0$

Part b)

For 68.1 cm length wire we have

$F = iLB sin\theta$

here we know that

$cos\theta = \frac{68.1}{166}$

$\theta = 65.8$

so we have

$F = (4.03)(0.681)(99.3 \times 10^{-3})sin65.8$

$F = 0.25 N$

Part c)

For 151 cm length wire we have

$F = iLB sin\phi$

here we know that

$cos\phi = \frac{151}{166}$

$\theta = 24.5$

so we have

$F = (4.03)(1.51)(99.3 \times 10^{-3})sin24.5$

$F = 0.25 N$

Part d)

Net force on a closed loop in uniform magnetic field is always ZERO

$F_{net} = 0$

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Explanation:

"Acoustic resonance is a phenomenon in which an acoustic system amplifies sound waves whose frequency matches one of its own natural frequencies of vibration (its resonance frequencies)." wikipedia I hope this helps you!

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ANSWER