Answer:
4.3 * 10 N
Explanation:
To calculate torque, we multiply the distance from the pivot by the perpendicular (the part of the force that acts at right angles to the displacement vector) component of the force to the displacement vector from the pivot.
torque = distance from pivot * perpendicular force
170 Nm= 0.4 m * F
F = 425 N = 4.3 * 10 N rounded off to two significant figures
Answer:
The temperature of the core raises by
every second.
Explanation:
Since the average specific heat of the reactor core is 0.3349 kJ/kgC
It means that we require 0.3349 kJ of heat to raise the temperature of 1 kg of core material by 1 degree Celsius
Thus reactor core whose mass is
will require

energy to raise it's temperature by 1 degree Celsius in 1 second
Hence by the concept of proportionately we can infer 150 MW of power will increase the temperature by
Answer:
a) The shear stress is 0.012
b) The shear stress is 0.0082
c) The total friction drag is 0.329 lbf
Explanation:
Given by the problem:
Length y plate = 2 ft
Width y plate = 10 ft
p = density = 1.938 slug/ft³
v = kinematic viscosity = 1.217x10⁻⁵ft²/s
Absolute viscosity = 2.359x10⁻⁵lbfs/ft²
a) The Reynold number is equal to:

The boundary layer thickness is equal to:
ft
The shear stress is equal to:

b) If the railing edge is 2 ft, the Reynold number is:

The boundary layer is equal to:

The sear stress is equal to:

c) The drag coefficient is equal to:

The friction drag is equal to:

I think its B B)Warm water rises and cold water moves in to replace it.
<h2>distance = 523 cm</h2>
Explanation:
( a ) The rotational speed of the ladybug = 25 r.p.m = 25/60 r.p.s
= 5/12 rev/sec
( b ) The definition of frequency is the number of rotations per second .
Here the number of rotations per second is 5/12 . Thus frequency = 5/12 Hz
( c ) The tangential speed is v = angular velocity x radius of rotation
The angular velocity ω = 2π x n , where n is the number of rotations per second
Thus angular velocity = 2π x 5/12 = 5π/6 rad/sec
The linear velocity = angular velocity x distance from center of record
Thus tangential speed = 5π/6 x 10 = 25π/3 cm/sec
Angular displacement in 20 sec = ω x t = 5π/6 x 20 = 50π/3 rad
Linear displacement = angular displacement x distance from center of record
= 50π/3 x 10 = 500π/3 = 523 cm