Answer: d= 0.57* l
Explanation:
We need to check that before ladder slips the length of ladder the painter can climb.
So we need to satisfy the equilibrium conditions.
So for ∑Fx=0, ∑Fy=0 and ∑M=0
We have,
At the base of ladder, two components N₁ acting vertical and f₁ acting horizontal
At the top of ladder, N₂ acting horizontal
And Between somewhere we have the weight of painter acting downward equal to= mg
So, we have N₁=mg
and also mg*d*cosФ= N₂*l*sin∅
So,
d=
* tan∅
Also, we have f₁=N₂
As f₁= чN₁
So f₁= 0.357 * 69.1 * 9.8
f₁= 241.75
Putting in d equation, we have
d= 
* tan 58
d= 0.57* l
So painter can be along the 57% of length before the ladder begins to slip
Answer:
to prevent cuz now for my mind when you prevent it it will not happen to you or nothing happened to you
Answer:
vf=94.4 m/s
Explanation:
acceleration is the final velocity minus initial velocity divided by time
a = (vf-vi)/t
Given:
a= 14.2 m/s^2
vi= 0 (at rest)
t = 6.6
Solve for vf
a = (vf-vi)/t
a*t=vf-vi
(14.2)*(6.6)=vf - 0
vf=94.4 m/s
Remains the same
Explanation:
According to Gauss's law, the electric flux through a closed surface is proportional to the charge enclosed by the surface. So no matter how big or small we make the surface that encloses the charge, the electric flux remains the same because it only depends on the enclosed charge, not surface area.
Answer:
16.32 °C
Explanation:
We are given;
Mass of aluminum bowl; m_b = 0.25 kg
Mass of soup; m_s = 0.8 kg
Thus, formula to find the amount of heat energy for a temperature change of 27.6°C to 0°C is;
Q = (m_b•c_b•Δt) + (m_s•c_s•Δt)
Where;
c_b = 0.215 kcal/(kg•°C)
c_s = 1 kcal/(kg•°C)
ΔT = 27.6 - 0 = 27.6°C
Thus;
Q = (0.25 × 0.215 × 27.6) + (0.8 × 1 × 27.6)
Q = 23.5635 Kcal
Now, the energy that exits to be used to freeze the soup is;
Q' = 424 kJ - Q
Let's convert 424 KJ to Kcal
424 KJ = 424/4.184 Kcal = 101.3384 Kcal
Thus;
Q' = 101.3384 - 23.5635
Q' = 77.7749 Kcal
Amount of heat that's removed is given by;
Q_f = Q' - mL
Where;
m = m_s = 0.8 kg
L = 79.8 kcal/kg
Thus;
Q_f = 77.7749 - (0.8 × 79.8)
Q_f = 13.9349 Kcal
Then final temperature will be;
T_f = Q_f/((m_b•c_b) + (m_s•c_s))
T_f = 13.9349/((0.25 × 0.215) + (0.8 × 1))
T_f = 16.32 °C
