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3 years ago

What is the technical name for super glue type adhesives

Physics

2 answers:

Murrr4er [49]3 years ago

5 0

The technical name for super glue type adhesive is Cyanoacrylate glue

Bumek [7]3 years ago

5 0

Cyanoacrylate adhesives are the technical name for super glue type adhesives.

<u>Explanation:</u>

Cyanoacrylate adhesives are generally known as instant glues.
Glues are generally used for pasting in the books and the broken items .
They are also known as power glues or super glues .
The term cyanoacrylate called as CA is commonly used for industrial grade cyanoacrylate.
Cyanoacrylate adhesives describes quick bonding super glues .
They are also called as Crazy glues .
They can be used to transform metal to plastic and human skin also .

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An Object, Start from rest w Confront Aiceleration 8m/s2 along a

shutvik [7]

Answer:

A) v = 40 m / s, B) v_average = 20 m / s

Explanation:

For this exercise we will use the kinematics relations

A) the final velocity for t = 5 s and since the body starts from rest its initial velocity is zero

v = vo + a t

v = 0 + 8 5

v = 40 m / s

B) the average velocity can be found with the relation

v_average = vf + vo / 2

v-average = 0+ 40/2

v_average = 20 m / s

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3 years ago

Suppose the magnitude of the gravitational force between two spherical objects is 2000 n when they are 100 km apart. what is the

Allushta [10]

=> use the inverse square relationship

Fg = 889 N

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4 years ago

7. What type of mineral can scratch glass, but cannot scratch steel?

zimovet [89]

The answer is (D) Fluorite

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3 years ago

You have a 160-Ω resistor and a 0.430-H inductor. Suppose you take the resistor and inductor and make a series circuit with a vo

S_A_V [24]

Answer:

A. Z = 185.87Ω

B. I = 0.16A

C. V = 1mV

D. VL = 68.8V

E. Ф = 30.59°

Explanation:

A. The impedance of a RL circuit is given by the following formula:

$Z=\sqrt{R^2+\omega^2L^2}$ (1)

R: resistance of the circuit = 160-Ω

w: angular frequency = 220 rad/s

L: inductance of the circuit = 0.430H

You replace in the equation (1):

$Z=\sqrt{(160\Omega)^2+(220rad/s)^2(0.430H)^2}=185.87\Omega$

The impedance of the circuit is 185.87Ω

B. The current amplitude is:

$I=\frac{V}{Z}$ (2)

V: voltage amplitude = 30.0V

$I=\frac{30.0V}{185.87\Omega}=0.16A$

The current amplitude is 0.16A

C. The current I is the same for each component of the circuit. Then, the voltage in the resistor is:

$V=\frac{I}{R}=\frac{0.16A}{160\Omega}=1*10^{-3}V=1mV$ (3)

D. The voltage across the inductor is:

$V_L=L\frac{dI}{dt}=L\frac{d(Icos(\omega t))}{dt}=-LIsin(\omega t)\\\\V_L=-(0.430H)(160\Omega)sin(220 t)=68.8sin(220t)\\\\V_L_{max}=68.8V$

E. The phase difference is given by:

$\phi=tan^{-1}(\frac{\omega L}{R})=tan^{-1}(\frac{(220rad/s)(0.430H)}{160\Omega})\\\\\phi=30.59\°$

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3 years ago

If you swim with the current in a river, your speed is increased by the speed of the water; if you swim against the current, you

Blababa [14]

Answer:

11.23%

Explanation:

Lets take

Speed of man in still water =u= 1.73 m/s

Speed of flow of water = v=0.52 m/s

When swims in downward direction then speed of man = u + v

When swims in upward direction then speed of man = u - v

Lets time taken by man when he swims in downward direction is $t_1$ and when he swims in downward direction is $t_2$

Lets distance is d and it will be remain constant in both the case

$d=(u+v)t_1$

$d=(u-v)t_2$

$(1.73+0.52)t_1=(1.73-0.52)t_2$

$t_2=1.85t_1$

Time taken in still water

2 d= t x 1.73

t=1.15 x d sec

$t_1=0.44d\ sec$

$t_2=0.82d\ sec$

total time in current = 0.82 +0.44 d=1.26 d sec

So the percentage time

$percentage\ time =\dfrac{1.28-1.15}{1.15}$

Percentage time =11.32%

So it will take 11.32% more time as compare to still current.

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3 years ago