Answer:
T = 5.36 s
Explanation:
given,
depth of the mine shaft = 122.5 m
speed of the sound = 340 m/s
time taken = ?
time taken by the stone to reach at the bottom
using equation of motion
initial speed , u = 0 m/s
t = 5 s
time taken by the sound to travel
d =v x t
t = 0.36 s
total time taken for the sound to reach carol after dropping the stone
T = 5 + 0.36
T = 5.36 s
Answer:
Thermal conduction is the transfer of internal energy by microscopic collisions of particles and movement of electrons within a body. The colliding particles, which include molecules, atoms and electrons, transfer disorganized microscopic kinetic and potential energy, when joined known as internal energy.
Explanation:
sana makatulong
Answer:
N₂=20.05 rpm
Explanation:
Given that
R= 19 cm
I=0.13 kg.m²
N₁ = 24.2 rpm
ω₁= 2.5 rad/s
m= 173 g = 0.173 kg
v=1.2 m
Initial angular momentum L₁
L₁ = Iω₁ - m v r ( negative sign because bird coming opposite to motion of the wire motion)
Final linear momentum L₂
L₂= I₂ ω₂
I₂ = I + m r²
The is no any external torque that is why angular momentum will be conserve
L₁ = L₂
Iω₁ - m v r = I₂ ω₂
Iω₁ - m v r = ( I + m r²) ω₂
Now by putting the all values
Iω₁ - m v r = ( I + m r²) ω₂
0.13 x 2.5 - 0.173 x 1.2 x 0.19 = ( 0.13 + 0.173 x 0.19²) ω₂
0.325 - 0.0394 = 0.136 ω₂
ω₂ = 2.1 rad/s
N₂=20.05 rpm
Answer:
Explanation:
= Charge at 3000 m = 40 C
= Charge at 1000 m = -40 C
= 3000 m
= 1000 m
k = Coulomb constant = 
Electric field due to the charge at 3000 m
Electric field due to the charge at 1000 m
Electric field at the aircraft is 
.
Answer:
The correct option is A = 1960 N/m²
Explanation:
Given that,
Mass m= 20,000kg
Area A = 100m²
Pressure different between top and bottom
Assume the plane has reached a cruising altitude and is not changing elevation. Then sum the forces in the vertical direction is given as
∑Fy = Wp + FL = 0
where
Wp = is the weight of the plane, and
FL is the lift pushing up on the plane.
Let solve for FL since the mass of the plane is given:
Wp + FL = 0
FL = -Wp
FL = -mg
FL = -20,000× -9.81
FL = 196,200N
FL should be positive since it is opposing the weight of the plane.
Let Equate FL to the pressure differential multiplied by the area of the wings:
FL = (Pb −Pt)⋅A
where Pb and Pt are the static pressures on bottom and top of the wings, respectively
FL = ∆P • A
∆P = FL/A
∆P = 196,200 / 100
∆P = 1962 N/m²
∆P ≈ 1960 N/m²
The pressure difference between the top and bottom surface of each wing when the airplane is in flight at a constant altitude is approximately 1960 N/m². Option A is correct
