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sveta [45]
3 years ago
5

Indicate the result of each of the following unit vector cross products (unit vector hat-symbols not shown): . kxi = . jxi= -jxk

= • ixi = • ur Xue =
Physics
1 answer:
Reil [10]3 years ago
4 0

Answer:

Explanation:

The cross product of two vectors is given by

\overrightarrow{A}\times \overrightarrow{B}=\left | \overrightarrow{A} \right |\left | \overrightarrow{B} \right |Sin\theta \widehat{n}

Where, θ be the angle between the two vectors and \widehat{n} be the unit vector along the direction of cross product of two vectors.

Here, K x i = - j

As K is the unit vector along Z axis, i is the unit vector along X axis and j be the unit vector along  axis.

The direction of cross product of two vectors is given by the right hand palm rule.

So, k x i = j

j x i = - k

- j x k = - i

i x i = 0

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A force of 10 lb is required to hold a spring stretched 4 in. beyond its natural length. How much work W is done in stretching i
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114.44 J

Explanation:

From Hook's Law,

F = ke................. Equation 1

Where F = Force required to stretch the spring, k = spring constant, e = extension.

make k the subject of the equation

k = F/e.............. Equation 2

Given: F = 10 lb = (10×4.45) N = 44.5 N, e = 4 in = (4×0.254) = 1.016 m.

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Work done in stretching the 9 in beyond its natural length

W = 1/2ke²................. Equation 3

Given: e = 9 in = (9×0.254) = 2.286 m, k = 43.799 N/m

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W = 1/2×43.799×2.286²

W = 114.44 J

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Suppose you wanted to break a meter stick over your knee. The cross-section of a meter stick is rectangular. Will it be easier t
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It will be easier to break the meter rule with the long side against my knee.

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a solid metal sphere of radius 3.00m carries a total charge of -5.50. what is the magnitude of the electric field at a distance
aivan3 [116]

Answer:

(a) Electric field at 0.250 m is zero.

(b)  Electric field at 2.90 m is zero.

(c) Electric field at 3.10 m is - 5.15 x 10³ V/m.

(d) Electric field at 8.00 m is - 0.77 x 10³ V/m.

Explanation:

Let Q and R are the charge and radius of the solid metal sphere. The solid metal sphere behave as conductor, so total charge Q is on the surface of the sphere.

Electric field inside and outside the metal sphere is :

E = 0 for r ≤ R ( inside )

  = \frac{KQ}{r^{2} } for r > R ( outside )

Here K is electric constant and r is the distance from the center of the metal sphere.

(a) Electric field at 0.250 m is zero as r < R i.e. 0.250 m < 3 m from the above equation.

(b)  Electric field at 2.90 m is zero as r < R i.e. 2.90 m < 3 m from the above equation.

(c) Electric field at 3.10 m is given by the relation as r > R :

E = \frac{KQ}{r^{2} }

Substitute 9 x 10⁹ N m²/C² for K, -5.50 μC for Q and 3.10 m for r in the above equation.

E = - \frac{9\times10^{9}\times5.50\times10^{-6}  }{3.10^{2} }

E = - 5.15 x 10³ V/m

(d) Electric field at 8.00 m is given by the relation as r > R :

E = \frac{KQ}{r^{2} }

Substitute 9 x 10⁹ N m²/C² for K, -5.50 μC for Q and 8.00 m for r in the above equation.

E = - \frac{9\times10^{9}\times5.50\times10^{-6}  }{8^{2} }

E = - 0.77 x 10³ V/m

8 0
3 years ago
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