Which of these is an experiment generally regarded as being first carried out by James Joule?
1 answer:
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Answer:
a) v = 0.7071 v₀, b) v= v₀, c) v = 0.577 v₀, d) v = 1.41 v₀, e) v = 0.447 v₀
Explanation:
The speed of a wave along an eta string given by the expression
v =
where T is the tension of the string and μ is linear density
a) the mass of the cable is double
m = 2m₀
let's find the new linear density
μ = m / l
iinitial density
μ₀ = m₀ / l
final density
μ = 2m₀ / lo
μ = 2 μ₀
we substitute in the equation for the velocity
initial v₀ =
with the new dough
v =
v = 1 /√2 \sqrt{ \frac{T_o}{ \mu_o} }
v = 1 /√2 v₀
v = 0.7071 v₀
b) we double the length of the cable
If the cable also increases its mass, the relationship is maintained
μ = μ₀
in this case the speed does not change
c) the cable l = l₀ and m = 3m₀
we look for the density
μ = 3m₀ / l₀
μ = 3 m₀/l₀
μ = 3 μ₀
v =
v = 1 /√3 v₀
v = 0.577 v₀
d) l = 2l₀
μ = m₀ / 2l₀
μ = μ₀/ 2
v =
v = √2 v₀
v = 1.41 v₀
e) m = 10m₀ and l = 2l₀
we look for the density
μ = 10 m₀/2l₀
μ = 5 μ₀
we look for speed
v =
v = 1 /√5 v₀
v = 0.447 v₀
Answer:
Explanation:
They traveled this distance in 2 parts, essentially. Part 1 had an average speed for a certain number of hours, part 2 had an average speed for a certain number of hours, and those 2 parts taken together took them a distance of 364 km. In equation form, that looks like this:
km/hr part 1 + km/hr part 2 = 364 km
Now we need to find each part on the left side of that equation. Part 1 first:
We traveled 65 km/hr for 2 hours, so that took us
and canceling out the hour label, we have that in part 1 we got
65(2) = 130 km. Good. Now onto the second part, where our unknown is.
We traveled 78 km/hr the second part for x hours, so that took us
and canceling out the hour label, we have that in part 2 we got
78x km. Now we can fill in the main equation (the one in bold print)
130 km + 78x km = 364 km and subtracting 130 km from both sides:
78x km = 234 km and dividing by 78 km:
x = 3 hours. Part 2 took 3 hours. Part 1 took 2 hours, so the whole trip took 5 hours.
Answer:
the final pressure of the gas is 60 kPa.
Explanation:
Given;
initial pressure of the gas, P₁ = 50 kPa = 50,000 Pa
initial temperature of the gas, T₁ = 27⁰ C = 27 + 273 = 300 k
final temperature of the gas, T₂ = 87⁰ C = 87 + 273 = 360 K
Let the final pressure of the gas = P₂
Apply pressure law;
Therefore, the final pressure of the gas is 60 kPa.