An extension cord made of two wires of diameter 0.129 cmcm (no. 16 copper wire) and of length 2.7 mm (9 ftft ) is connected to a
1 answer:
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Answer:
The answer is given here would be a simplified equation, seeing as there are some missing variables in the question.
<u>F1 = T- 46, 674.656 gm/s² </u>
Explanation:
<em>Note: Once we have the mass of the second object and/or acceleration of the cord, we can solve for the force of the ground acting on the box.</em>
To calculate the force caused by gravity on the basic pulley system we use the following equation:
F2 = M2 x g; where g= gravitational acceleration (a constant equal to 9.8 m/s²). The mass M2 = 10.5 lb = 4762.72g
∴ F2 = 4762.72g x 9.8 m/s²
= 46, 674.656 gm/s² or 46, 674.656 N
But since this F2 is acting in a downlowrd direction, it would be negative.
Tension of the cord, T = Mass, x × acceleration. ( x is in the pulley diagram)
⇒ F1 = T - F2
<u>F1 = T- 46, 674.656 gm/s² </u>
Answer:
A+B; 5√5 units, 341.57°
A-B; 5√5 units, 198.43°
B-A; 5√5 units, 18.43°
Explanation:
Given A = 5 units
By vector notation and the axis of A, it is represented as -5j
B = 3 × 5 = 15 units
Using the vector notations and the axis, B is +15i. The following vectors ate taking as the coordinates of A and B
(a) A + B = -5j + 15i
A+B = 15i -5j
|A+B| = √(15)²+(5)²
= 5√5 units
∆ = arctan(5/15) = 18.43°
The angle ∆ is generally used in the diagrams
∆= 18.43°
The direction of A+B is 341.57° based in the condition given (see attachment for diagrams
(b) A - B = -5j -15i
A-B = -15i -5j
|A-B|= √(15)²+(-5)²
|A-B| = √125
|A-B| = 5√5 units
The direction is 180+18.43°= 198.43°
See attachment for diagrams
(c) B-A = 15i -( -5j) = 15i + 5j
|B-A| = 5√5 units
The direction is 18.43°
See attachment for diagram
Answer:
The fraction of kinetic energy to the total energy is .
Explanation:
displacement is one third of the amplitude.
Let the amplitude is A.
x= A/3
The kinetic energy of the body executing SHM is
The total energy is
Divide (1) by (2)