Answer:
acceleration = 2.4525 m/s²
Explanation:
Data: Let m1 = 3.0 Kg, m2 = 5.0 Kg, g = 9.81 m/s²
Tension in the rope = T
Sol: m2 > m1
i) for downward motion of m2:
m2 a = m2 g - T
5 a = 5 × 9.81 m/s² - T
⇒ T = 49.05 m/s² - 5 a Eqn (a)
ii) for upward motion of m1
m a = T - m1 g
3 a = T - 3 × 9.8 m/s²
⇒ T = 3 a + 29.43 m/s² Eqn (b)
Equating Eqn (a) and(b)
49.05 m/s² - 5 a = T = 3 a + 29.43 m/s²
49.05 m/s² - 29.43 m/s² = 3 a + 5 a
19.62 m/s² = 8 a
⇒ a = 2.4525 m/s²
Answer:
10Ω
Explanation:
The 10Ω resistors are in series, so their combined resistance is:
10Ω + 10Ω = 20Ω
The 8Ω resistors are in parallel, so their combined resistance is:
(1/8Ω + 1/8Ω)⁻¹ = 4Ω
This 4Ω resistance is in series with the 16Ω resistor, so the combined is resistance is:
4Ω + 16Ω = 20Ω
Finally, this 20Ω resistance and the first 20Ω resistance are in parallel, so the total equivalent resistance is:
(1/20Ω + 1/20Ω)⁻¹ = 10Ω
Frequency has the unit hertz.
Answer:
![f = 8.8\,N](https://tex.z-dn.net/?f=f%20%3D%208.8%5C%2CN)
Explanation:
The moment equation of the toboggan is:
![m_{t}\cdot v_{o} - f\cdot \Delta t = 0](https://tex.z-dn.net/?f=m_%7Bt%7D%5Ccdot%20v_%7Bo%7D%20-%20f%5Ccdot%20%5CDelta%20t%20%3D%200)
The average friction force is:
![f=\frac{m_{t}\cdot v_{o}}{\Delta t}](https://tex.z-dn.net/?f=f%3D%5Cfrac%7Bm_%7Bt%7D%5Ccdot%20v_%7Bo%7D%7D%7B%5CDelta%20t%7D)
![f = \frac{(15.0\,kg)\cdot (4.40\,\frac{m}{s} )}{7.50\,s}](https://tex.z-dn.net/?f=f%20%3D%20%5Cfrac%7B%2815.0%5C%2Ckg%29%5Ccdot%20%284.40%5C%2C%5Cfrac%7Bm%7D%7Bs%7D%20%29%7D%7B7.50%5C%2Cs%7D)
![f = 8.8\,N](https://tex.z-dn.net/?f=f%20%3D%208.8%5C%2CN)