Thermal conductions
K= QL/ART
Aluminium T₁ = 10 + 273.15
T₂ = 283.15k
205 = 2.0 × 0.30/4× 10⁻⁴ × (T₂ - 283.15)
Copper
385 = Q × 0.70/4×10⁻⁴ ×(433.15 - T₂)
Where T₃ = 160 + 273.15
T₃ = 433.15K
From 2 to 3
205/385 = 0.30/0.70 × 433.15 - T₂/T₂ - 283.15
= 0.53T₂ -150.06 = 181.92 - 0.42 T₂
→ 0.95T₂ = 331.98 ⇒ T₂ = ₂349.45k
T₂ = 76.3°c
=77°c.
Given:
u(initial velocity)=0
a=5.54m/s^2
v(final velocity)=2 m/s
v=u +at
Where v is the final velocity.
u is the initial velocity
a is the acceleration.
t is the time
2=0+5.54t
t=2/5.54
t=0.36 sec
Responder:
20πrads ^ -1; 24πrads ^ -1; 0,1 seg; 10 Hz
Explicación:
Dado lo siguiente:
Radio (r) del círculo = 120 cm
600 revoluciones por minuto en radianes por segundo
(600 / min) * (2π rad / 1 rev) * (1min / 60seg)
(1200πrad / 60sec) = 20π rad ^ -1
Velocidad angular (w) = 20πrads ^ -1
Velocidad lineal = radio (r) * velocidad angular (w)
Velocidad lineal = (120/100) * 20πrad
Velocidad lineal = 1.2 * 20πrads ^ -1 = 24πrads ^ -1
C.) Período (T):
T = 2π / w = 2π / 20π = 0.1 seg
D.) Frecuencia (f):
f = 1 / T = 1 / 0.1
1 / 0,1 = 10 Hz
Answer:
Final velocity of electron, 
Explanation:
It is given that,
Electric field, E = 1.55 N/C
Initial velocity at point A, 
We need to find the speed of the electron when it reaches point B which is a distance of 0.395 m east of point A. It can be calculated using third equation of motion as :
........(1)
a is the acceleration, 
We know that electric force, F = qE
Use above equation in equation (1) as:
v = 647302.09 m/s
or
So, the final velocity of the electron when it reaches point B is 
. Hence, this is the required solution.
