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egoroff_w [7]
3 years ago
5

How do you think the weight and the lift should not be one of the same magnitude​

Physics
1 answer:
Rus_ich [418]3 years ago
3 0

because weight disappear and doesn't show the actual weight in the lift

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A mass of 4kg suspended by a light string 2m long and at rest is projected horizontally with a velocity of 1.5 m/s. find the ang
Dafna11 [192]

Answer:

19.5°

Explanation:

The energy of the mass must be conserved. The energy is given by:

1) E=\frac{1}{2}mv^2+mgh

where m is the mass, v is the velocity and h is the hight of the mass.

Let the height at the lowest point of the be h=0, the energy of the mass will be:

2) E=\frac{1}{2}mv^2

The energy when the mass comes to a stop will be:

3) E=mgh

Setting equations 2 and 3 equal and solving for height h will give:

4) h=\frac{v^2}{2g}

The angle ∅ of the string with the vertical with the mass at the highest point will be given by:

5) cos\phi=\frac{l-h}{l}

where l is the lenght of the string.

Combining equations 4 and 5 and solving for ∅:

6) \phi={cos}^{-1}(\frac{l-h}{l})={cos}^{-1}(1-\frac{h}{l})={cos}^{-1}(1-\frac{v^2}{2gl})

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3 years ago
In a ballistics test at the police station an officer
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You times the 6 by the 350 duvided 1.8
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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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Answer:

a) 0.83984 i + 0.41992 j - 2.0996 k KN

b) T_ac = 1.972888 KN

Explanation:

Given:

- The tension in cable AB = 2.3 KN

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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