Answer:
ΔL = 3.82 10⁻⁴ m
Explanation:
This is a thermal expansion exercise
ΔL = α L₀ ΔT
ΔT = T_f - T₀
where ΔL is the change in length and ΔT is the change in temperature
Let's reduce the length to SI units
L₀ = 90.5 mm (1m / 1000 mm) = 0.0905 m
let's calculate
ΔL = 25.10⁻⁶ 0.0905 (154.6 - (14.4))
ΔL = 3.8236 10⁻⁴ m
using the criterion of three significant figures
ΔL = 3.82 10⁻⁴ m
The weights in newtowns for the given masses are
<span> masses 22.1, 33.5, 41.3, 59.2, 78
weights 216.58N 328.3N 404.74N 580.16N 764.4N
e.g, for m=22.1kg, W=22.1kgx9.8N/kg =216.58N</span>
Complete question:
A college dormitory room measures 14 ft wide by 13 ft long by 6 ft high. Weight density of air is 0.07 lbs/ft3. What is the weight of air in it under normal conditions?
Answer:
the weight of the air is 76.44 lbs
Explanation:
Given;
dimension of the dormitory, = 14 ft by 13 ft by 6 ft
density of the air, = 0.07 lbs/ft³
The volume of the air in the dormitory room = 14 ft x 13 ft x 6 ft
= 1092 ft³
The weight of the air = density x volume
= 0.07 lbs/ft³ x 1092 ft³
= 76.44 lbs
Therefore, the weight of the air is 76.44 lbs
1.
Answer:
Part a)
Part b)
Explanation:
Part a)
Length of the rod is 1.60 m
diameter = 0.550 cm
now if the current in the ammeter is given as
V = 17.0 volts
now we will have
R = 0.91 ohm
now we know that
Part b)
Now at higher temperature we have
R = 0.98 ohm
now we know that
so we will have
2.
Answer:
Part a)
Part b)
Explanation:
Part a)
As we know that current density is defined as
now we have
Now we have
so we will have
Part b)
now we have
so we have
so we have
