Respuesta:
0,0560 cal / gºC.
Explicación:
Cantidad de calor; (Q)
Q = mcΔt; Δt = t2 - t1
m = masa, c = capacidad calorífica específica; Δt = cambio de temperatura
c de agua = 1 cal / gºC
c de aluminio = 0,22 cal / gºC
QTotal = Q de agua + Q de aluminio
Q de agua = 450 * 1 * (26 - 23) = 1350 cal
Q de aluminio = 60 * 0.22 * (26 - 23) = 39.6 cal
QTotal = 1350 + 39,6 = 1389,6 cal
Calor perdido = calor ganado
QTotal = calor perdido
- 1389,6 = 335,2 * c * (26 - 100)
-1389,6 = −24804,8 * c
c = 1389,6 / 24804,8
c = 0,056021 cal / gºC.
Capacidad calorífica específica de la plata = 0,0560 cal / gºC.
Answer:

Explanation:
As per mechanical energy conservation we can say that here since friction is present in the barrel so we will have
Work done by friction force = Loss in mechanical energy
so we will have

here we know that



Initial compression in the spring is given as



now from above equation




Answer:
W= F × d
W= 2kn × 3.6
W= 7.2 J
Work is measured in Joules!
Answer:
3.33 N
Explanation:
First, find the acceleration.
Given:
Δx = 3 m
v₀ = 0 m/s
t = 3 s
Find: a
Δx = v₀ t + ½ at²
3 m = (0 m/s) (3 s) + ½ a (3 s)²
a = ⅔ m/s²
Use Newton's second law to find the force.
F = ma
F = (5 kg) (⅔ m/s²)
F ≈ 3.33 N