Answer:
0.546 
Explanation:
From the given information:
The force on a given current-carrying conductor is:

where the length usually in negative (x) direction can be computed as

Now, taking the integral of the force between x = 1.0 m and x = 3.0 m to get the value of the force, we have:



![F = I (9.0) \bigg [\dfrac{x^3}{3} \bigg ] ^3_1 \hat k](https://tex.z-dn.net/?f=F%20%3D%20I%20%20%289.0%29%20%5Cbigg%20%5B%5Cdfrac%7Bx%5E3%7D%7B3%7D%20%5Cbigg%20%5D%20%5E3_1%20%5Chat%20k)
![F = I (9.0) \bigg [\dfrac{3^3}{3} - \dfrac{1^3}{3} \bigg ] \hat k](https://tex.z-dn.net/?f=F%20%3D%20I%20%20%289.0%29%20%5Cbigg%20%5B%5Cdfrac%7B3%5E3%7D%7B3%7D%20-%20%5Cdfrac%7B1%5E3%7D%7B3%7D%20%5Cbigg%20%5D%20%20%5Chat%20k)
where;
current I = 7.0 A
![F = (7.0 \ A) (9.0) \bigg [\dfrac{27}{3} - \dfrac{1}{3} \bigg ] \hat k](https://tex.z-dn.net/?f=F%20%3D%20%287.0%20%5C%20A%29%20%20%289.0%29%20%5Cbigg%20%5B%5Cdfrac%7B27%7D%7B3%7D%20-%20%5Cdfrac%7B1%7D%7B3%7D%20%5Cbigg%20%5D%20%20%5Chat%20k)
![F = (7.0 \ A) (9.0) \bigg [\dfrac{26}{3} \bigg ] \hat k](https://tex.z-dn.net/?f=F%20%3D%20%287.0%20%5C%20A%29%20%20%289.0%29%20%5Cbigg%20%5B%5Cdfrac%7B26%7D%7B3%7D%20%5Cbigg%20%5D%20%20%5Chat%20k)
F = 546 × 10⁻³ T/mT 
F = 0.546 
- Mass of the elevator (m) = 570 Kg
- Acceleration = 1.5 m/s^2
- Distance (s) = 13 m
- Let the force be F.
- We know, F = ma,
- Therefore, F = (570 × 1.5) N = 855 N
- Angle between distance and force (θ) = 0°
- We know, work done = F s Cos θ
- Therefore, work done by the cable during this part
- = (855 × 13 × Cos 0°) J
- = (855 × 13 × 1) J
- = 11115 J
<u>Answer</u><u>:</u>
<u>1</u><u>1</u><u>1</u><u>1</u><u>5</u><u> </u><u>J</u>
Hope you could get an idea from here.
Doubt clarification - use comment section.
Answer: FR=2.330kN
Explanation:
Write down x and y components.
Fx= FSin30°
Fy= FCos30°
Choose the forces acting up and right as positive.
∑(FR) =∑(Fx )
(FR) x= 5-Fsin30°= 5-0.5F
(FR) y= Fcos30°-4= 0.8660-F
Use Pythagoras theorem
F2R= √F2-11.93F+41
Differentiate both sides
2FRdFR/dF= 2F- 11.93
Set dFR/dF to 0
2F= 11.93
F= 5.964kN
Substitute value back into FR
FR= √F2(F square) - 11.93F + 41
FR=√(5.964)(5.964)-11.93(5.964)+41
FR= 2.330kN
The minimum force is 2.330kN
If the comb and the paper are attracted to each other the charge on the paper, is D)must be positive
<h3>
Laws of electrical attraction</h3>
This states that
- Like charges attract
- Unlike charges repel
Now, given that a negatively charged plastic comb is brought close to, but does not touch, a small piece of paper. If the comb and the paper are attracted to each other the charge on the paper, this implies that both the negatively charged plastic comb and the paper have opposite charges.
Since the charge on the plastic comb is negative, this means that the charge on the paper must be positive
So, if the comb and the paper are attracted to each other the charge on the paper, is D)must be positive
Learn more about electric charge here:
brainly.com/question/2373424
Answer:
43.41 mm
Explanation:
Given:
thickness of sheet, t = 6 mm
Force exerted by punch, F = 45 KN
Average shearing stress, T = 55 MPa
From average shearing stress T = Force F / Area A
Hence area = force/stress =45000/ 55 =
From area = pi*diameter*thickness
diameter = area/(pi* thickness)
= 818.18/(3.142*6)
= 43.41 mm