Answer:
If I understand correctly. Line B is parallel to the circle. Also, the angle is less than 90.
- The size of the circle determines.
- The diameter should not be fixed either.
Answer:
3A
Explanation:
Rtoal=R1+R2+R3=5+10+15=30
I=V/R 90/30
I=3
Answer:
<h2>
6.36 cm</h2>
Explanation:
Using the formula to first get the image distance
1/f = 1/u+1/v
f = focal length of the lens
u = object distance
v = image distance
Given f = 16.0 cm, u = 24.8 cm
1/v = 1/16 - 1/24.8
1/v = 0.0625-0.04032
1/v = 0.02218
v = 1/0.02218
v = 45.09 cm
To get the image height, we will us the magnification formula.
Mag = v/u = Hi/H
Hi = image height = ?
H = object height = 3.50 cm
45.09/24.8 = Hi/3.50
Hi = (45.09*3.50)/24.8
Hi = 6.36 cm
The image height is 6.36 cm
Answer:
r1 = 5*10^10 m , r2 = 6*10^12 m
v1 = 9*10^4 m/s
From conservation of energy
K1 +U1 = K2 +U2
0.5mv1^2 - GMm/r1 = 0.5mv2^2 - GMm/r2
0.5v1^2 - GM/r1 = 0.5v2^2 - GM/r2
M is mass of sun = 1.98*10^30 kg
G = 6.67*10^-11 N.m^2/kg^2
0.5*(9*10^4)^2 - (6.67*10^-11*1.98*10^30/(5*10^10)) = 0.5v2^2 - (6.67*10^-11*1.98*10^30/(6*10^12))
v2 = 5.35*10^4 m/s
Answer:
T = 20.84°C
Explanation:
From the law of conservation of energy:
Heat Lost by Copper Block = Heat Gained by Aluminum Calorimeter + Heat Gained by Water
where,
= mass of copper = 227 g
= mass of water = 844 g
= mass of aluminum = 155 g
= specific heat capacity of calorimeter = 385 J/kg.°C
= specific heat capacity of water = 4200 J/kg.°C
= specific heat capacity of aluminum = 890 J/kg.°C
= change in temperature of copper = 283°C - T
= change in temperature of water = T - 14.6°C
= change in temperature of aluminum = T - 14.6°C
T = equilibrium temperature = ?
Therefore,
<u>T = 20.84°C</u>
