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3 years ago

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A plane cruising at 233 m/s accelerates at 17 m/s 2 for 4.8 s. What is its final velocity? Answer in units of m/s. 013 (part 2 o

f 2) 10.0 points How far will it have traveled in that time? Answer in units of m.

1 answer:

Volgvan3 years ago

4 0

Answer:

Final velocity will be 314.6 m/sec

Distance traveled = 1314.24 m

Explanation:

We have given initial velocity u = 233 m/sec

Acceleration $a=17m/sec^2$

Time t = 4.8 sec

From first equation of motion $v=u+at$ , here v is final velocity, u is initial velocity and t is time

So $v=233+17\times 4.8=314.6m/sec$

Now we have to find distance traveled

From second equation of motion

$S=ut+\frac{1}{2}at^2=233\times 4.8+\frac{1}{2}\times 17\times 4.8^2=1314.24m$

So distance traveled in given time will be 1314.24 m

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A solenoidal coil with 26 turns of wire is wound tightly around another coil with 350 turns. The inner solenoid is 20.0 cm long

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Answer:

Part a)

$\phi = 2.76 \times 10^{-7} T m^2$

Part B)

$M = 5.52 \times 10^{-5} H$

Part C)

$EMF = 0.1 V/s$

Explanation:

Part a)

Magnetic field due to a long ideal solenoid is given by

$B = \mu_0 n i$

n = number of turns per unit length

$n = \frac{N}{L}$

$n = \frac{350}{0.20}$

$n = 1750 turn/m$

now we know that magnetic field due to solenoid is

$B = (4\pi \times 10^{-7})(1750)(0.100)$

$B = 2.2 \times 10^{-4} T$

Now magnetic flux due to this magnetic field is given by

$\phi = B.A$

$\phi = (2.2 \times 10^{-4})(\pi r^2)$

$\phi = (2.2 \times 10^{-4})(\pi(0.02)^2)$

$\phi = 2.76 \times 10^{-7} T m^2$

Part B)

Now for mutual inductance we know that

$\phi_{total} = M i$

$\phi_{total} = N\phi$

$\phi_{total} = 20(2.76 \times 10^{-4})$

$\phi_{total} = 5.52 \times 10^{-6}$

now we have

$M = \frac{5.52 \times 10^{-6}}{0.100}$

$M = 5.52 \times 10^{-5} H$

Part C)

As we know that induced EMF is given as

$EMF = M \frac{di}{dt}$

$EMF = 5.52 \times 10^{-5} (1800)$

$EMF = 0.1 V/s$

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3 years ago

Consider the points below. P(1, 0, 1), Q(−2, 1, 4), R(6, 2, 7) (a) Find a nonzero vector orthogonal to the plane through the poi

kozerog [31]

Answer:

a) (0, -33, 12)

b) area of the triangle : 17.55 units of area

Explanation:

<h2>a) </h2>

We know that the cross product of linearly independent vectors $\vec{A}$ and $\vec{B}$ gives us a nonzero, orthogonal to both, vector. So, if we can find two linearly independent vectors on the plane through the points P, Q, and R, we can use the cross product to obtain the answer to point a.

Luckily for us, we know that vectors $\vec{A} = \vec{P}-\vec{Q}$ and $\vec{B} = \vec{R} - \vec{Q}$ are living in the plane through the points P, Q, and R, and are linearly independent.

We know that they are linearly independent, cause to have one, and only one, plane through points P Q and R, this points must be linearly independent (as the dimension of a plane subspace is 3).

If they weren't linearly independent, we will obtain vector zero as the result of the cross product.

So, for our problem:

$\vec{A} = \vec{P} - \vec{Q} \\\\\vec{A} = (1,0,1) - (-2,1,4)\\\\\vec{A} = (1 +2,0-1,1-4)\\\\\vec{A} = (3,-1,-3)$

$\vec{B} = \vec{R} - \vec{Q} \\\\\vec{B} = (6,2,7) - (-2,1,4)\\\\\vec{B} = (6 +2,2-1,7-4)\\\\\vec{B} = (8,1,3)$

$\vec{A} \times \vec{B} = (A_y B_z - B_y A_z) \ \hat{i} - ( A_x B_z-B_xA_z) \ \hat{j} + (A_x B_y - B_x A_y ) \ \hat{k}$

$\vec{A} \times \vec{B} = ( (-1) * 3 - 1 * (-3) ) \ \hat{i} - ( 3 * 3 - 8 * (-3)) \ \hat{j} + (3 * 1 - 8 * (-1) ) \ \hat{k}$

$\vec{A} \times \vec{B} = ( - 3 + 3 ) \ \hat{i} - ( 9 + 24 ) \ \hat{j} + (3 + 8 ) \ \hat{k}$

$\vec{A} \times \vec{B} = 0 \ \hat{i} - 33 \ \hat{j} + 12 \ \hat{k}$

$\vec{A} \times \vec{B} =(0, -33, 12)$

<h2>B)</h2>

We know that $\vec{A}$ and $\vec{B}$ are two sides of the triangle, and we also know that we can use the magnitude of the cross product to find the area of the triangle:

$|\vec{A} \times \vec{B} | = 2 * area_{triangle}$

so:

$\sqrt{(-33)^2 + (12)^2} = 2 * area_{triangle}$

$\sqrt{1233} = 2 * area_{triangle}$

$35.114= 2 * area_{triangle}$

$17.55 \ units \ of \ area = area_{triangle}$

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3 years ago