Эт 4. A supertanker of mass 1.5 x 108 kg is being towed by two tugboats. The tension in the towing cables are T₁ and T2 at equal angles 30°. In addition, the tanker?s engine produces a forward drive force D = 75 x 10³ N, while the water applies an opposing force R = 40 x 10³ N. The tanker moves forward with an acceleration of 20 x 10-3 m/s. Find the magnitude of the tension T. Note that T₁ = T2= T the same cord. Hint: Newton's 2nd Law equation along the x-direction will be sufficient to help you solve the question. M=15x108 kg a=20x10-³ A) B) E) 104 R 40X103 N T= ? EF=ma T-R + pcoso=ma 85,000 N 1.5 x 105 N 1.7 x 106 N 2.5 x 107 N none of the above 0.020 a = 20x 10²³m/s 30.0° T₁ D COSO 30.0⁰ T₂ T=ma+R-DCOSO (1.5x10²) (20.10³) + 40x10³ - 75X10³ (ose sino 1 Sine

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Ff
2
+x
4. A supertanker of mass 1.5 x 108 kg is being towed by two tugboats. The tension in the towing cables are Ti
and T2 at equal angles 30°. In addition, the tanker?s engine produces a forward drive force D = 75 x 10³ N,
while the water applies an opposing force R = 40 x 103 N. The tanker moves forward with an acceleration of
20 x 10-³ m/s. Find the magnitude of the tension T. Note that T₁ = T2= T the same cord. Hint: Newton's 2nd
Law equation along the x-direction will be sufficient to help you solve the question.
M = 1,5x108 kg
a=20x10-³
R
40X10³
A)
B)
N
85,000 N
1.5 x 105 N
1.7 x 106 N
2.5 x 107 N
E) none of the above
0.020
EF=ma T-R + Dcoso=ma
T= ?
9
a = 20x10 ³ m/s
F
30.0°
T₁
D COSO
30.0⁰
000000000
T₂
T=ma+R-Droso
(1.5x10³) (20.10²³) + 40x10³ - 75X10³ (ose
sine
Sine
Transcribed Image Text:Ff 2 +x 4. A supertanker of mass 1.5 x 108 kg is being towed by two tugboats. The tension in the towing cables are Ti and T2 at equal angles 30°. In addition, the tanker?s engine produces a forward drive force D = 75 x 10³ N, while the water applies an opposing force R = 40 x 103 N. The tanker moves forward with an acceleration of 20 x 10-³ m/s. Find the magnitude of the tension T. Note that T₁ = T2= T the same cord. Hint: Newton's 2nd Law equation along the x-direction will be sufficient to help you solve the question. M = 1,5x108 kg a=20x10-³ R 40X10³ A) B) N 85,000 N 1.5 x 105 N 1.7 x 106 N 2.5 x 107 N E) none of the above 0.020 EF=ma T-R + Dcoso=ma T= ? 9 a = 20x10 ³ m/s F 30.0° T₁ D COSO 30.0⁰ 000000000 T₂ T=ma+R-Droso (1.5x10³) (20.10²³) + 40x10³ - 75X10³ (ose sine Sine
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