The working equation to be used here is written below:
Q = kA(T₁ - T₂)/Δx
where
Q is the rate of heat transfer
k is the heat transfer coefficient
A is the cross-sectional area of the wall
T₁ - T₂ is the temperature difference between the sides of the wall
Δx is the thickness of the wall
The solution is as follows:
Q = (0.69 W/m²·°C)(5 m × 6 m)(50°C - 20°C)/(30 cm * 1 m/100 cm)
Q = 2,070 W/m
Answer:
0.975 m
Explanation:
The center of mass y = ∑m₁y₁/∑m₁ where m₁ = the individual masses and y₁ = the y - coordinates of the individual masses
So,
y = (2.02 kg × 3.03 m + 2.92 kg × 2.43 m + 2.53 kg × 0 m + 4.02 kg × -0.501 m)/(2.02 kg + 2.92 kg + 2.53 kg + 4.02 kg)
y = (6.1206 kgm + 7.0956 kgm + 0 kgm - 2.01402 kgm)/11.49 kg
y = 11.20218 kgm/11.49 kg
y = 0.975 m
So, the y- component of the center of mass of these objects is 0.975 m
Answer:
42.5 m/s
Explanation:
Given:
x₀ = 0 m
x = 62 m
y₀ = 80 m
y = 0 m
v₀ᵧ = 0 m/s
aₓ = 0 m/s²
aᵧ = -9.8 m/s²
Find: v
First, find the time it takes to land.
y = y₀ + v₀ᵧ t + ½ aᵧ t²
(0 m) = (80 m) + (0 m/s) t + ½ (-9.8 m/s²) t²
t = 4.04 s
Find the horizontal component vₓ:
x = x₀ + vₓ t − ½ aₓ t²
(62 m) = (0 m) + vₓ (4.04 s) − ½ (0 m/s²) (4.04 s)²
vₓ = 15.3 m/s
Find the vertical component vᵧ:
vᵧ = aᵧ t + v₀ᵧ
vᵧ = (-9.8 m/s²) (4.04 s) + (0 m/s)
vᵧ = -39.6 m/s
Find the speed using Pythagorean theorem:
v = √(vₓ² + vᵧ²)
v = √((15.3 m/s)² + (-39.6)²)
v = 42.5 m/s
Sorry!
This cannot be answered. We don't have weight, height, etc.
