The study of science involves the study of the natural world.
<h2>Answer:</h2><h2>The depth of barge float=
3 cm</h2><h2>
Explanation:</h2>
Length of rectangular barge=5.2 m
Width of rectangular barge=2.4m
Mass of crate=410 kg
Let h be the height of barge float
Volume of barge float=
Density of water=
Weight of water displaced by barge=Buoyant force=-Weight of horse
1 m=100 cm
cm
Hence, the depth of barge float=3 cm
<h2 />
Answer:
Balanced.
Explanation:
A Balanced Chemical equation is a scientific term that describes a chemical equation that has the same number of atoms on each side of the equation.
Hence, when the number of atoms on the right side of a chemical equation matches the number of atoms on the left side of a chemical equation, it is said to be BALANCED.
Explanation:
12) q = mCΔT
125,600 J = (500 g) (4.184 J/g/K) (T − 22°C)
T = 82.0°C
13) Solving for ΔT:
ΔT = q / (mC)
a) ΔT = 1 kJ / (0.4 kg × 0.45 kJ/kg/K) = 5.56°C
b) ΔT = 2 kJ / (0.4 kg × 0.45 kJ/kg/K) = 11.1°C
c) ΔT = 2 kJ / (0.8 kg × 0.45 kJ/kg/K) = 5.56°C
d) ΔT = 1 kJ / (0.4 kg × 0.90 kJ/kg/K) = 2.78°C
e) ΔT = 2 kJ / (0.4 kg × 0.90 kJ/kg/K) = 5.56°C
f) ΔT = 2 kJ / (0.8 kg × 0.90 kJ/kg/K) = 2.78°C
14) q = mCΔT
q = (2000 mL × 1 g/mL) (4.184 J/g/K) (80°C − 20°C)
q = 502,000 J
20) q = mCΔT
q = (2000 g) (4.184 J/g/K) (100°C − 15°C) + (400 g) (0.9 J/g/K) (100°C − 15°C)
q = 742,000 J
24) q = mCΔT
q = (0.10 g) (0.14 J/g/K) (8.5°C − 15°C)
q = -0.091 J
Answer:
The force required to move the cart is divided by three or 33.33 N.
Explanation:
