Answer:
1b 2c 4f 5 a 6y
Explanation: i did this earlier
Answer: The answer is 19,600 newtons.
Explanation:
Answer:
335°C
Explanation:
Heat gained or lost is:
q = m C ΔT
where m is the mass, C is the specific heat capacity, and ΔT is the change in temperature.
Heat gained by the water = heat lost by the copper
mw Cw ΔTw = mc Cc ΔTc
The water and copper reach the same final temperature, so:
mw Cw (T - Tw) = mc Cc (Tc - T)
Given:
mw = 390 g
Cw = 4.186 J/g/°C
Tw = 22.6°C
mc = 248 g
Cc = 0.386 J/g/°C
T = 39.9°C
Find: Tc
(390) (4.186) (39.9 - 22.6) = (248) (0.386) (Tc - 39.9)
Tc = 335
Answer:
v₂ = 79.69 m/s
Explanation:
The initial diameter of the hose, d₁ = 3.0 cm = 0.03 m
Initial Cross Sectional Area, A₁ = πd₁²/4
A₁ = (π* 0.03²)/4
A₁ = 0.00071 m²
The initial speed of water from the hose, v₁ = 2.2 m/s
The diameter of the hose after blocking the end, d₂ = 0.50 cm = 0.005 m
Cross Sectional Area of the hose after blocking the end, A₂ = πd₂²/4
A₂ = (π* 0.005²)/4
A₂ = 0.0000196 m²
To get the speed, v₂, at which the water spray from the hose after blocking the end, we will use the continuity equation:
A₁v₁ = A₂v₂
0.00071 * 2.2 = 0.0000196 v₂
0.001562 = 0.0000196 v₂
v₂ = 0.001562/0.0000196
v₂ = 79.69 m/s