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aniked [119]
3 years ago
13

A horizontal rope is tied to a 57.0 kg kg box on frictionless ice. What is the tension in the rope if: You may want to review (P

ages 135 - 137) . Part A The box is at rest? Express your answer as an integer and include the appropriate units. T T = nothing nothing SubmitRequest Answer Part B The box moves at a steady v x vxv_x = 4.20 m/s m/s ? Express your answer as an integer and include the appropriate units. T T = nothing nothing SubmitRequest Answer Part C The box v x vxv_x = 4.20 m/s m/s and a x axa_x = 5.80 m/ s 2 m/s2 ?
Physics
1 answer:
VARVARA [1.3K]3 years ago
8 0

Answer:

Explanation:

A ) The surface is frictionless . If the box is at rest , there will be no tension in the rope.

B ) In this case , the box is moving with steady rate of 4.20 m /s . In this case also no force is acting on the box so tension in the rope will be nil.

C ) In this case the box is moving with acceleration of 5.8 m /s² so force on the box = mass x acceleration

= 57 x 5.8

= 330.6 N .

So tension in the rope will be equal to force acting on the rope . Hence tension = 330.6 N .

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Answer:

d = 39.7 km

Explanation:

initial position of the boat is 45 km away at an angle of 15 degree East of North

so we will have

r_1 = 45 sin15 \hat i + 45 cos15 \hat j

r_1 = 11.64 \hat i + 43.46\hat j

after some time the final position of the boat is found at 30 km at 15 Degree North of East

so we have

r_2 = 30 cos15\hat i + 30 sin15 \hat j

r_2 = 28.98\hat i + 7.76 \hat j

now the displacement of the boat is given as

d = r_2 - r_1

d = (28.98\hat i + 7.76 \hat j) - (11.64 \hat i + 43.46\hat j)

d = 17.34 \hat i - 35.7 \hat j

so the magnitude is given as

d = \sqrt{17.34^2 + 35.7^2}

d = 39.7 km

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3 years ago
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Do you have the options? I would say swerve?

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