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jeka57 [31]
3 years ago
11

The electric field strength E is measured as:

Physics
1 answer:
soldi70 [24.7K]3 years ago
4 0
The correct answer is
<span>force per unit charge.

In fact, the electric field strength is defined as the electric force per unit charge experienced by a positive test charge located in the electric field. In formula:
</span>E= \frac{F}{q}
where
E is the electric field strength
F is the electric force experienced by the charge
q is the positive test charge.
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The turnbuckle is tightened until the tension in the cable AB equals 2.3 kN. Determine the vector expression for the tension T a
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a) 0.83984 i + 0.41992 j - 2.0996 k KN

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Explanation:

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Find:

a) Determine the vector expression for the tension T as a force acting on member AD.

b) Also find the magnitude of the projection of T along the line AC.

Solution:

part a)

- Find unit vector AB:

                            vector (AB) = 2 i + j - 5 k

                             mag (AB) = sqrt (2^2 + 1^2 + 5^2)

                             mag (AB) = sqrt(30)

                             unit (AB) =  ( 1 / sqrt(30) )* ( 2 i + j - 5 k )

- Find Tension vector:

                             vector (T) = unit(AB)* 2.3 KN

                                              = 0.83984 i + 0.41992 j - 2.0996 k

- The projection of T onto AC can be found from the dot product of vector T to unit vector (AC)

- For unit vector (AC)

                               vector (AC) = 2 i - 2 j - 5 k

                               mag (AC) = sqrt (2^2 + 2^2 + 5^2)

                               mag (AC) = sqrt(33)

                               unit (AC) =  ( 1 / sqrt(33) )* ( 2 i - 2 j - 5 k )

- Compute the projection:

                                T_ac = vector T . unit (AC)

T_ac = (0.83984 i + 0.41992 j - 2.0996 k)  . ( 1 / sqrt(33) )* ( 2 i - 2 j - 5 k )

                                T_ac = 0.2923947572 - 0.146973786 - 1.827467232

                                T_ac = 1.972888 KN

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