Answer: Option (b) is the correct answer.
Explanation:
The given data is as follows.
F =
N
g = 9.8 m/s
radius =
=
= 15 cm = 0.15 m (as 1 m = 100 cm)
Formula to calculate depth is as follows.
F = ![\rho \times g \times h \times A](https://tex.z-dn.net/?f=%5Crho%20%5Ctimes%20g%20%5Ctimes%20h%20%5Ctimes%20A)
or, h =
h =
= 751 m
Thus, we can conclude that the maximum depth in a lake to which the submarine can go without damaging the window is closest 750 m.
Assuming you are looking for the acceleration a:
1.
![m_1a = T_1 -m_1g](https://tex.z-dn.net/?f=m_1a%20%3D%20T_1%20-m_1g%20)
2.
![m_2a = m_2g - T_2](https://tex.z-dn.net/?f=m_2a%20%3D%20m_2g%20-%20T_2)
where T is the tension and a is the acceleration of the blocks. The acceleration of the two blocks and the acceleration of the pulley must be equal.
The torque on the pulley is given by:
3.
![\tau = \overrightarrow r \times \overrightarrow F = (T_2 - T_1)R = I\alpha = \frac{1}{2} MR^2 \frac{a}{R}](https://tex.z-dn.net/?f=%5Ctau%20%3D%20%5Coverrightarrow%20r%20%5Ctimes%20%5Coverrightarrow%20F%20%3D%20%28T_2%20-%20T_1%29R%20%3D%20I%5Calpha%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20MR%5E2%20%5Cfrac%7Ba%7D%7BR%7D%20)
where
![I = \frac{1}{2} mR^2](https://tex.z-dn.net/?f=I%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20mR%5E2)
and
![a = \alpha R](https://tex.z-dn.net/?f=a%20%3D%20%5Calpha%20R)
.
Combining the three equations:
Answer:
a) t₁ = 4.76 s, t₂ = 85.2 s
b) v = 209 ft/s
Explanation:
Constant acceleration equations:
x = x₀ + v₀ t + ½ at²
v = at + v₀
where x is final position,
x₀ is initial position,
v₀ is initial velocity,
a is acceleration,
and t is time.
When the engine is on and the sled is accelerating:
x₀ = 0 ft
v₀ = 0 ft/s
a = 44 ft/s²
t = t₁
So:
x = 22 t₁²
v = 44 t₁
When the engine is off and the sled is coasting:
x = 18350 ft
x₀ = 22 t₁²
v₀ = 44 t₁
a = 0 ft/s²
t = t₂
So:
18350 = 22 t₁² + (44 t₁) t₂
Given that t₁ + t₂ = 90:
18350 = 22 t₁² + (44 t₁) (90 − t₁)
Now we can solve for t₁:
18350 = 22 t₁² + 3960 t₁ − 44 t₁²
18350 = 3960 t₁ − 22 t₁²
9175 = 1980 t₁ − 11 t₁²
11 t₁² − 1980 t₁ + 9175 = 0
Using quadratic formula:
t₁ = [ 1980 ± √(1980² - 4(11)(9175)) ] / 22
t₁ = 4.76, 175
Since t₁ can't be greater than 90, t₁ = 4.76 s.
Therefore, t₂ = 85.2 s.
And v = 44 t₁ = 209 ft/s.