20. The systems shown in Figure P5.20 are in equilibrium. Ifthe spring scales are calibrated in newtons, what do theyread? Ignore the masses of the pulleys and strings andassume the pulleys and the incline in Figure P5.20d arefrictionless.5.00 kg5.00 kg5.00 kgb5.00 kg30.00Creating archive Westworld.S01 .part075. rar5.00 kg5.00 kgсra

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Asked Dec 2, 2019
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20. The systems shown in Figure P5.20 are in equilibrium. If
the spring scales are calibrated in newtons, what do they
read? Ignore the masses of the pulleys and strings and
assume the pulleys and the incline in Figure P5.20d are
frictionless.
5.00 kg
5.00 kg
5.00 kg
b
5.00 kg
30.00
Creating archive Westworld.S01 .part075. rar
5.00 kg
5.00 kg
с
ra
help_outline

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20. The systems shown in Figure P5.20 are in equilibrium. If the spring scales are calibrated in newtons, what do they read? Ignore the masses of the pulleys and strings and assume the pulleys and the incline in Figure P5.20d are frictionless. 5.00 kg 5.00 kg 5.00 kg b 5.00 kg 30.00 Creating archive Westworld.S01 .part075. rar 5.00 kg 5.00 kg с ra

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Expert Answer

Step 1

part A

Given info: The system is in equilibrium and two pulleys are connected
with a spring and two masses are freely hanged on both sides with
5.00 kg
From the given figure, the tension on the spring is,
T =mg
Here
T is the tension on the spring.
m is mass of objects
g is the acceleration due to gravity
Substitute 5.00kg for m and 9.81 m/s2 for gin above equation
T (5 kg) (9.81 m/s)
= 49 N
Conclusion
Therefore, the tension in spring in part (a) of given figure is 49 N.
help_outline

Image Transcriptionclose

Given info: The system is in equilibrium and two pulleys are connected with a spring and two masses are freely hanged on both sides with 5.00 kg From the given figure, the tension on the spring is, T =mg Here T is the tension on the spring. m is mass of objects g is the acceleration due to gravity Substitute 5.00kg for m and 9.81 m/s2 for gin above equation T (5 kg) (9.81 m/s) = 49 N Conclusion Therefore, the tension in spring in part (a) of given figure is 49 N.

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Step 2

Part B

Given info: The system is in equilibrium and one pulley is connected
with mass with 5.00 kg and a spring connected to the wall
From the given figure, the tension on the spring is,
T = mg
Substitute 5.00kg for m and 9.81 m/s for gin above equation
T (5 kg) (9.81 m/s
= 49 N
Conclusion:
Therefore, the tension in spring in part (b) of given figure is 49 N.
help_outline

Image Transcriptionclose

Given info: The system is in equilibrium and one pulley is connected with mass with 5.00 kg and a spring connected to the wall From the given figure, the tension on the spring is, T = mg Substitute 5.00kg for m and 9.81 m/s for gin above equation T (5 kg) (9.81 m/s = 49 N Conclusion: Therefore, the tension in spring in part (b) of given figure is 49 N.

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Step 3

Part C :...

Given info: The system is in equilibrium and two masses of 5.00 kg are
hanged freely and connected to pulley of both sides and pulley is hanged
by spring.
From the given figure, the tension on the spring is,
T (m1m2) g
Here
mi, m2 are two masses.
Substitute 5.00 kg for mi and m2 both and 9.81 m/s for gin above
equation
T-(5 kg + 5kg) (9.81 m/s?)
10kg x 9.81 m/
98 N
Conclusion:
Therefore, the tension in spring in part (c) of given figure is 98 N.
help_outline

Image Transcriptionclose

Given info: The system is in equilibrium and two masses of 5.00 kg are hanged freely and connected to pulley of both sides and pulley is hanged by spring. From the given figure, the tension on the spring is, T (m1m2) g Here mi, m2 are two masses. Substitute 5.00 kg for mi and m2 both and 9.81 m/s for gin above equation T-(5 kg + 5kg) (9.81 m/s?) 10kg x 9.81 m/ 98 N Conclusion: Therefore, the tension in spring in part (c) of given figure is 98 N.

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