4200 N is the tension in the cable that pulls the elevator upwards.
The correct option is A.
<h3>What does tension ?</h3>
Tension is the force that is sent through a rope, thread, or wire whenever two opposing forces pull on it. Along the whole length of the wire, the tensile stress pulls equally on all objects at the ends. Every physical object that comes into contact with that other one exerts force on it.
<h3>Briefing:</h3>
We employ the following formula to determine the cable's tension.
Formula:
T = mg+ma............ Equation 1
Where:
T is the cable's tension.
M = Mass of the elevator and the Joey
Accelerating with a
g = Gravitational acceleration
Considering the query,
Given:
m = (300+60) = 360 kg
a = 2 m/s²
g = 9.8 m/s²
Substitute these values into equation 2
T = (360×9.8)+(360×2)
T = 3528+720
T = 4248 N
T ≈ 4200 to the nearest hundred.
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Answer:
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Answer:
1. K.E = 11.2239 kJ ≈ 11.224 kJ
2. ![C_{V} = 37.413 JK^{- 1}](https://tex.z-dn.net/?f=C_%7BV%7D%20%3D%2037.413%20JK%5E%7B-%201%7D)
3. ![Q = 10.7749 kJ](https://tex.z-dn.net/?f=Q%20%3D%2010.7749%20kJ)
Solution:
Now, the kinetic energy of an ideal gas per mole is given by:
K.E = ![\frac{3}{2}mRT](https://tex.z-dn.net/?f=%5Cfrac%7B3%7D%7B2%7DmRT)
where
m = no. of moles = 3
R = Rydberg's constant = 8.314 J/mol.K
Temperature, T = 300 K
Therefore,
K.E = ![\frac{3}{2}\times 3\times 8.314\times 300](https://tex.z-dn.net/?f=%5Cfrac%7B3%7D%7B2%7D%5Ctimes%203%5Ctimes%208.314%5Ctimes%20300)
K.E = 11223.9 J = 11.2239 kJ ≈ 11.224 kJ
Now,
The heat capacity at constant volume is:
![C_{V} = \frac{3}{2}mR](https://tex.z-dn.net/?f=C_%7BV%7D%20%3D%20%5Cfrac%7B3%7D%7B2%7DmR)
![C_{V} = \frac{3}{2}\times 3\times 8.314 = 37.413 JK^{- 1}](https://tex.z-dn.net/?f=C_%7BV%7D%20%3D%20%5Cfrac%7B3%7D%7B2%7D%5Ctimes%203%5Ctimes%208.314%20%3D%2037.413%20JK%5E%7B-%201%7D)
Now,
Required heat transfer to raise the temperature by
is:
![Q = C_{V}\Delta T](https://tex.z-dn.net/?f=Q%20%3D%20C_%7BV%7D%5CDelta%20T)
![\Delta T = 15^{\circ} = 273 + 15 = 288 K](https://tex.z-dn.net/?f=%5CDelta%20T%20%3D%2015%5E%7B%5Ccirc%7D%20%3D%20273%20%2B%2015%20%3D%20288%20K)
![Q = 37.413\times 288 = 10774.9 J = 10.7749 kJ](https://tex.z-dn.net/?f=Q%20%3D%2037.413%5Ctimes%20288%20%3D%2010774.9%20J%20%3D%2010.7749%20kJ)
Answer:
![a=4.8m/s^2](https://tex.z-dn.net/?f=a%3D4.8m%2Fs%5E2)
Explanation:
Hello,
In this case, since the acceleration in terms of position is defined as its second derivative:
![a=\frac{d^2x(t)}{dt^2}=\frac{d^2}{dt^2}(2.9+8.8t+2.4t^2)](https://tex.z-dn.net/?f=a%3D%5Cfrac%7Bd%5E2x%28t%29%7D%7Bdt%5E2%7D%3D%5Cfrac%7Bd%5E2%7D%7Bdt%5E2%7D%282.9%2B8.8t%2B2.4t%5E2%29)
The purpose here is derive x(t) twice as follows:
![a=\frac{d^2x(t)}{dt^2}=\frac{d}{dt}(8.8+2*2.4*t)\\ \\a=4.8m/s^2](https://tex.z-dn.net/?f=a%3D%5Cfrac%7Bd%5E2x%28t%29%7D%7Bdt%5E2%7D%3D%5Cfrac%7Bd%7D%7Bdt%7D%288.8%2B2%2A2.4%2At%29%5C%5C%20%5C%5Ca%3D4.8m%2Fs%5E2)
Thus, the acceleration turns out 4.8 meters per squared seconds.
Best regards.