Answer:
(a) 10 m/s
(b) 22.4 m/s
Explanation:
(a) Draw a free body diagram of the car when it is at the top of the loop. There are two forces: weight force mg pulling down, and normal force N pushing down.
Sum of forces in the centripetal direction (towards the center):
∑F = ma
mg + N = mv²/r
At minimum speed, the normal force is 0.
mg = mv²/r
g = v²/r
v = √(gr)
v = √(10 m/s² × 10.0 m)
v = 10 m/s
(b) Energy is conserved.
Initial kinetic energy + initial potential energy = final kinetic energy
½ mv₀² + mgh = ½ mv²
v₀² + 2gh = v²
(10 m/s)² + 2 (10 m/s²) (20.0 m) = v²
v = 22.4 m/s
Answer:
55,42 J
Explanation:
Since the height of the room is 3.45 m (distance between the floor and the ceiling) the difference between this value and the length of the rope 1.19 m; it will be equal to (3.45-1.19) =2.26 m. If we take as a reference point (Ep=0) the floor of the room, then the potential energy will be equal to Ep = M * g * h, replacing values in this equation (2.5 kg * 9.81 m/s2 * 2.26 m) will be 55,42 (N * m) or Jules.
Answer:
A,B,C,D, and F are correct
Explanation:
Answer:
T = 74°C
Explanation:
Given Mw = mass of water = 330g, Ma = mass of aluminium = 840g
Cw = 4.2gJ/g°C = specific heat capacity of water and Ca = 0.9J/g°C = specific heat capacity of aluminium
Initial temperature of water = 100°C.
Initial temperature of aluminium = 29°C
When the boiling water is poured into the aluminum pan, heat is exchanged and after a short time the water and aluminum pan both come to thermal equilibrium at a common temperature T.
Heat lost by water equal to the heat gained by aluminium pan.
Mw × Cw×(100 –T) = Ma × Ca × (T–29)
330×4.2×(100– T) = 890×0.9×(T–29)
1386(100 – T) = 801(T –29)
1386/801(100 – T) = T – 29
1.73(100 – T) = T – 29
173 –1.73T = T –29
173+29 = T + 1.73T
202 = 2.73T
T = 202/2.73
T = 74°C
