A block of mass M = 9 (kg) is tied to a homogeneous rope of mass m = 1 (kg) L = 1 (m) length. As shown in the figure, when the horizontal force F = 100 (N) is applied to the end of the rope, the block moves with acceleration a. What is the tensile force (N) at the midpoint of the rope? Neglect the rope hanging due to its weight. 11/20M=9 (kg) kütleli bir blok m=1 (kg) kütleli L=1 (m) uzunluklu homojen bir ipe bağlı dır. Şekil de göstenldiğigibi ipin ucuna yatay F=100 (N) kuvveti uygulandığında blok a ivmesi ile hareket etm ekte dir. Ipin ortanoktasındakı genlme kuvveti kaç (N) dur? Ipın ağırlığından dol ayı sarkm asını ihm al ediniz.

Mass of block (M) = 9 kg mass of rope(m) = 1 kg length of rope(L) = 1 m Applied force = 100 N

A block of mass M = 9 (kg) is tied to a homogeneous rope of mass m = 1 (kg) L = 1 (m) length. As shown in the figure, when the horizontal force F = 100 (N) is applied to the end of the rope, the block moves with acceleration a. What is the tensile force (N) at the midpoint of the rope? Neglect the rope hanging due to its weight.

A block of mass M = 9 (kg) is tied to a homogeneous rope of mass m = 1 (kg) L = 1 (m) length. As shown in the figure, when the horizontal force F = 100 (N) is applied to the end of the rope, the block moves with acceleration a. What is the tensile force (N) at the midpoint of the rope? Neglect the rope hanging due to its weight.

Physics for Scientists and Engineers: Foundations and Connections

1st Edition

ISBN:9781133939146

Author:Katz, Debora M.

Publisher:Katz, Debora M.

Chapter1: Getting Started

Section: Chapter Questions

Problem 11PQ: CASE STUDY On planet Betatron, mass is measured in bloobits and length in bots. You are the Earth...

See similar textbooks

Similar questions

The question: Add the following quantities: 4.05 kg + 567.95 g + 100.1 g My solution: 4.05 + 567.95 + 100.1 = 672.1 There are several things wrong with my solution. Tell me what they are.
Question 1: Complete the calculations of the table based on the data given below Test No Weight (N) H, (cm) H2 (cm) P.E. (initial) (J) P.E. (final) (J) Loss in P.E. (J) 1 1.8 60 43.0 1.08 0.774 0.306 2 2.0 60 44.4 1.2 0.888 0.312 3 2.2 60 45.7 1.32 1.005 0.315
Object Mi (kg) M2 (kg) M3 (kg) M4 (kg) M (kg) SM (kg) UM (kg) Object 1 0.01935 0.01935 0.01935 0.01935 Object 2 0.01060 0.01060 0.01060 0.01060 Object 3 0.04638 0.04638 0.04638 0.04638 M, is the average mass, Sx is standard deviation and u is uncertainty, u-Sx/ Vn, where n=number of trials
Useful constants: k=9.00×109 N.m2/C2 ε0= 8.85×10-12 C2/N.m2 µ0=4π×10-7 T.m/A mp=1.67×10-27 kg me=9.11×10-31 kg e=1.6×10-19 C g=10.0 m/s2
Sürtünmesiz, 6,=30° açılı bir eğik düzlem üzerinde bulunan m-3,0 kg kütleli bir kutu, kütlesiz bir sicim ile, açısı 0;-60° olan başka bir eğik düzlem üzerinde bulunan m-2,0 kg kütleli ikinci bir kutuya makara üzerinden bağlanıyor. Makara sürtünmesiz ve kütlesizdir. Sicimdeki gerilme ne olur? A box of mass m;=3,0 kg on a frictionless plane inclined at angle 0,-30° is connected via a cord of negligible mass to a box of mass m=2,0 kg on another frictionless plane inclined at angle 02-60° through a pulley. The pulley is frictionless and has negligible mass. What is the tension in the cord? 19,4 N 16,1 N 15,0 N 18,3 N 17,2 N
2) Find the missing values: (a) 3.0 N1.0 kg 5.0N (b) 2.0N- 120 kg 3.0 N 5.0N -5.0 N -2.0N i- 2.0 m/s? Fret =? a = 0 Ft-? 10 N -5N (d) 30 N- (e) 4000g (f) 5.0 kg 1.0N 2.0N 3.0N constant Fnet ? a-1.5 m/s? - 0.5 m/s? 7, = ? Fret = ? F, = ? -SN 20 N- (h) (i) Fz 30 18 N a = 5 m/s? Fnut = ? a = + 0.6 m/s? Frst = 1.8 × 10-2 N 10 N - constant (down] m-? m= ? Fz = ? net
1. Based on the values provided, complete the following table. Test Weight H1 H2 P.E. (initial) P.E. (final) Loss in P.E. No (N) (cm) (cm) (J) (J) (J) 1 1.4 42.5 2 1.6 65 43.8 1.8 65 45.0 4 2.0 65 47.4 2.2 65 49.7
zain IQ "ll N N N 89% Di 9:23 ぐ VY Quiz of Anal.. Misan University College of Science Department of Physics Second Stage Quiz of Analytical Mechanics Q: If Ã= fi+2f j and B= fi+2f j+3f k, f is constant, find the angle between these two vectors? Maysam 12:01p äelul 21.02.11 O O
A man weighs 470.7 kg and is 198.8 cm tall. Calculate his Body Mass Index (BMI). BMI = (Mass in kg) / (Height in metres)2 O A. 119.10 O B. 0.01 OC. 476.40 OD. 1.19 O E. 11909.99 Clear my choice
The Statue of Liberty in New York City is approximately 305 ft305 ft tall. How many U.S. dimes would be in a stack of the same height? Each dime is 1.35 mm1.35 mm thick. number of dimes: Each dime has a mass of 2.268 g.2.268 g. How much would the stack of dimes from the previous question weigh? mass: g What is the value, in dollars, of the same stack of dimes? value: dollars The 2017 U.S. gross domestic product (GDP) was valued at 19,390,604,000 dollars.19,390,604,000 dollars. How many Statue of Liberty‑height stacks of dimes are needed to match the GDP in value? number of stacks:
Math Review (a) Solve the equation 7.20 103 m/s = (4.20 103 m/s) ln (Mi/Mf) for the fraction Mi/Mf. (b) If Mi = 2.65 104 kg, calculate Mf.
A South American tribe uses a unit of mass called the zooma. 1 zooma = 9 dooka. 1 boopa = 14 uma, 1 uma 16 dooka and 7 kg = 1 boopa. What is the mass of 120 zooma in kg? 426.7 kg O 17.9 kg 33.8 kg O8640 kg O 0.7 kg