Answer:
Explanation:
First we need to state our assumptions:
Thermal properties of ice and water are constant, heat transfer to the glass is negligible, Heat of ice 
Mass of water,
.
Energy balance for the ice-water system is defined as
a.The mass of ice at 
is defined as:
![[mc(0-T_1)+mh_i_f+mc(T_2-0)]_i_c_e+[mc(T_2-T_1)]_w=0\\\\m_i_c_e[0+333.7+418\times(5-0)]+0.3\times4.18\times(5-20)=0\\m_i_c_e=0.0546Kg=54.6g](https://tex.z-dn.net/?f=%5Bmc%280-T_1%29%2Bmh_i_f%2Bmc%28T_2-0%29%5D_i_c_e%2B%5Bmc%28T_2-T_1%29%5D_w%3D0%5C%5C%5C%5Cm_i_c_e%5B0%2B333.7%2B418%5Ctimes%285-0%29%5D%2B0.3%5Ctimes4.18%5Ctimes%285-20%29%3D0%5C%5Cm_i_c_e%3D0.0546Kg%3D54.6g)
b.Mass of ice at 
is defined as:
![[mc(0-T_1)+mh_i_f+mc(T_2-0)]_i_c_e+[mc(T_2-T_1)]_w=0\\\\m_i_c_e[2.11\times(0-(20))+333.7+4.18\times(5-0)]+0.3\times4.18\times(5-20)=0\\\\m_i_c_e=0.0487Kg=48.7g](https://tex.z-dn.net/?f=%5Bmc%280-T_1%29%2Bmh_i_f%2Bmc%28T_2-0%29%5D_i_c_e%2B%5Bmc%28T_2-T_1%29%5D_w%3D0%5C%5C%5C%5Cm_i_c_e%5B2.11%5Ctimes%280-%2820%29%29%2B333.7%2B4.18%5Ctimes%285-0%29%5D%2B0.3%5Ctimes4.18%5Ctimes%285-20%29%3D0%5C%5C%5C%5Cm_i_c_e%3D0.0487Kg%3D48.7g)
c.Mass of cooled water at 
![[mc(T_2-T_1)]_c_w+[mc(T_2-T_1)]_w=0\\m_c_w\times4.18\times(5-0)+0.3\times4.18\times(5-20)\\m_c_w=0.9kg=900g](https://tex.z-dn.net/?f=%5Bmc%28T_2-T_1%29%5D_c_w%2B%5Bmc%28T_2-T_1%29%5D_w%3D0%5C%5Cm_c_w%5Ctimes4.18%5Ctimes%285-0%29%2B0.3%5Ctimes4.18%5Ctimes%285-20%29%5C%5Cm_c_w%3D0.9kg%3D900g)
Answer:
D. It is upright.
Because convex mirror produce the real image, and the plane mirror produce it's upright.
Answer:
5 .07 s .
Explanation:
The child will move on a circle of radius r
r = 1.5 m
Let the velocity of rotation = v
radial acceleration = v² / r
v² / r = 2.3
v² = 2.3 r = 2.3 x 1.5
= 3.45
v = 1.857 m /s
Time of revolution = 2π r / v
= 2 x 3.14 x 1.5 / 1.857
= 5 .07 s .
Answer:
Explanation:
Let T be the tension
For linear motion of hoop downwards
mg -T = ma , m is mass of the hoop . a is linear acceleration of CG of hoop .
For rotational motion of hoop
Torque by tension
T x R , R is radius of hoop.
Angular acceleration be α,
Linear acceleration a = α R
So TR = I α
= I a / R
a = TR² / I
Putting this value in earlier relation
mg -T = m TR² / I
mg = T ( 1 + m R² / I )
T = mg / ( 1 + m R² / I )
mg / ( 1 + R² / k² )
Tension is less than mg or weight because denominator of the expression is more than 1.
First, calculate the initial velocity of the dog given with the vertical height and the acceleration due to gravity which is calculated through the equation,
2ad = Vo²
Substituting the known values,
2(9.8 m/s²)(1.2 m) = V₀²
V₀ = 4.85 m/s
The kinetic energy is solved through the equation,
KE = 0.5mv²
Substituting the known values to the latest equation,
KE = 0.5 (7.2 kg)(4.85 m/s)²
KE = 17.46 J
Thus, the kinetic energy is 17.46 J.
