W = ∫ (x from 0.1 to +oo) F dx
= ∫ (x from 0.1 to +oo) A e^(-kx) dx
= A/k x [ - e^(-kx) ](between 0.1 and +oo)
= A/k x [ 0 + e^(-k * 0.1) ]
<span>
= A/k x e^(-k/10) </span>
Answer:
Explanation:
Given that,
The volume of the balloon is
V = 440 × 10³ m³
Buoyant force F?
Given the density of the surrounding to be 2.58 kg/m³
ρ = 2.58 kg/m³
The buoyant force is the weight of water displaced and it is calculated using
F_b = ρVg
Where
F_b is buoyant force
ρ is density
V is the volume of the liquid displace.
g is the acceleration due to gravity
Then,
F_b = ρVg
F_b = 2.58 × 440 × 10³ × 9.81
F_b = 1.1 × 10^7 N
Answer:
The datapoint 9.0 ppm is outlier at the 90% confidence level.
Explanation:
The old data has following values
mean=10.5 mm
standard deviation 0.2 mm
Now the mean of new values is calculated as following

So the value as 9.0 ppm can be considered easily as outlier in this regard.
Mass = 0.201kg
Energy = 15J
temperature change = 10C
Energy(E) = mass(m) × specific heat capacity(c) × temperature change(θ)
we can rearrange this to make specific heat capacity the subject
c =

c =

c =7.46268657