Answer:
It looks like... A machine that reads electric pulse and surge... Not sure though.
Explanation:
Answer:
1791 secs ≈ 29.85 minutes
Explanation:
( Initial temperature of slab ) T1 = 300° C
temperature of water ( Ts ) = 25°C
T2 ( final temp of slab ) = 50°C
distance between slab and water jet = 25 mm
<u>Determine how long it will take to reach T2</u>
First calculate the thermal diffusivity
∝ = 50 / ( 7800 * 480 ) = 1.34 * 10^-5 m^2/s
<u>next express Temp as a function of time </u>
T( 25 mm , t ) = 50°C
next calculate the time required for the slab to reach 50°C at a distance of 25mm
attached below is the remaining part of the detailed solution
Answer:
a) V =10¹¹*(1.5q₁ + 3q₂)
b) U = 1.34*10¹¹q₁q₂
Explanation:
Given
x₁ = 6 cm
y₁ = 0 cm
x₂ = 0 cm
y₂ = 3 cm
q₁ = unknown value in Coulomb
q₂ = unknown value in Coulomb
A) V₁ = Kq₁/r₁
where r₁ = √((6-0)²+(0-0)²)cm = 6 cm = 0.06 m
V₁ = 9*10⁹q₁/(0.06) = 1.5*10¹¹q₁
V₂ = Kq₂/r₂
where r₂ = √((0-0)²+(3-0)²)cm = 3 cm = 0.03 m
V₂ = 9*10⁹q₂/(0.03) = 3*10¹¹q₂
The electric potential due to the two charges at the origin is
V = ∑Vi = V₁ + V₂ = 1.5*10¹¹q₁ + 3*10¹¹q₂ = 10¹¹*(1.5q₁ + 3q₂)
B) The electric potential energy associated with the system, relative to their infinite initial positions, can be obtained as follows
U = Kq₁q₂/r₁₂
where
r₁₂ = √((0-6)²+(3-0)²)cm = √45 cm = 3√5 cm = (3√5/100) m
then
U = 9*10⁹q₁q₂/(3√5/100)
⇒ U = 1.34*10¹¹q₁q₂
