Answer:
A Normal tape is very weak under tensile force and when a small fracture is caused, it will affect the whole tape and the tape will fail easily. while a fiber-reinforced tape will not fail easily under same stress and this is the major advantage of a fiber-reinforced tape has over a normal tape
Explanation:
The fracture behavior would be different for a fiber-reinforced tape in the following way :
* It's behavior during tensile stress and its fracture behavior.
A Normal tape is very weak under tensile force and when a small fracture is caused, it will affect the whole tape and the tape will fail easily. while a fiber-reinforced tape will not fail easily under same stress and this is the major advantage of a fiber-reinforced tape has over a normal tape
Match the use of the magnetic field to its respective description.
oooExplanation:
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1. Consider a solid cube of dimensions 1ft x 1ft x 1ft (=0.305m x 0.305m x 0.305m). Its top surface is 10
ft (=3.05 m) below the surface of the water. The density of water is pf=1000 kg/m3.
Consider two cases:
a) The cube is made of cork (pB=160.2 kg/m3)
b) The cube is made of steel (pB=7849 kg/m3)
In what direction does the body tend to move?
Answer:
a) up
b) down
Explanation:
When the cube is less dense than water, it will tend to float (move upward). When it is more dense, it will sink (move downward).
a) 160.2 kg/m^3 < 1000 kg/m^3. The cube will move up.
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b) 7849 kg/m^3 > 1000 kg/m^3. The cube will move down.
A differential amplifier is to have a voltage gain of 100. What will be the feedback resistance required if the input resistances are both 1 kΩ?
Answer:
required feedback resistance ( R2 ) = 100 k Ω
Explanation:
Given data :
Voltage gain = 100
input resistance ( R1 ) = 1 k ohms
calculate feedback resistance required
voltage gain of differential amplifier
[tex]\frac{Vout}{V2 - V1 } = \frac{R2}{R1}[/tex]
= Voltage gain = R2/R1
= 100 = R2/1
hence required feedback resistance ( R2 ) = 100 k Ω
A 550 kJ of heat quantity needed to increase water temperature from 32°C to 80°C. Calculate the mass
of the water when the specific heat capacity of water is 4200 J/kg °C.
Answer:
2.728 kg
Explanation:
The units help you keep the calculation straight.
[tex]\dfrac{550\text{ kJ}}{(80^\circ\text{C}-32^\circ\text{C})(4.200\text{ kJ/kg\,$^\circ$C})}=\dfrac{550}{48\cdot4.2}\text{ kg}\approx\boxed{2.728\text{ kg}}[/tex]
Quadrilateral ABCD is a rectangle.
If m ZADB = 7k + 60 and mZCDB = -5k + 40, find mZCBD.
Hope this helps...........