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2 years ago

If a bender can be used to bend 3/4-inch rmc, then it can also be used to bend__________emt.

Physics

1 answer:

USPshnik [31]2 years ago

3 0

A bender may also be used to bend 1-inch emt if it can bend 3/4-inch rmc.

<h3>A hand bender is what?</h3>

hand bender. a device with a diameter of no more than 1-1/4 in. that is used to bend EMT. To prevent crimping or flattening the conduit during bending, hand benders offer a radius that supports the conduit.

<h3>When would you utilize a hand bending device?</h3>

The user can produce very close bends with hockeys because they are 60% smaller than benders. Generally speaking, you should use a hand bender most of the time and reserve the hickey for when you need to create several really tight bends that a hand bender is unable to accomplish.

learn more about bender here

<u>brainly.com/question/9721733</u>

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2 years ago

Read 2 more answers

In a Broadway performance, an 84.5-kg actor swings from a R = 4.30-m-long cable that is horizontal when he starts. At the bottom

Arada [10]

Answer:

1.57772 m

Explanation:

M = Mass of actor = 84.5 kg

m = Mass of costar = 55 kg

v = Velocity of costar

V = Velocity of actor

$h_i$ = Intial height of actor = 4.3 m

g = Acceleration due to gravity = 9.81 m/s²

As the energy of the system is conserved

$\frac{1}{2}MV^2=Mgh_i\\\Rightarrow V=\sqrt{2gh_i}\\\Rightarrow V=\sqrt{2\times 9.81\times 4.3}\\\Rightarrow V=9.18509\ m/s$

As the linear momentum is conserved

$MV=(m+M)v\\\Rightarrow v=\frac{MV}{m+M}\\\Rightarrow V=\frac{84.5\times 9.18509}{84.5+55}\\\Rightarrow v=5.56372\ m/s$

Applying conservation of energy again

$\frac{1}{2}(m+M)v^2=(m+M)gh_f\\\Rightarrow h_f=\frac{v^2}{2g}\\\Rightarrow h_f=\frac{5.56372^2}{2\times 9.81}\\\Rightarrow h_f=1.57772\ m$

The maximum height they reach is 1.57772 m

3 0

3 years ago

The temperature of the air in a valley begins to increase after the sun comes up and heats the valley floor. What will

tester [92]

Explanation:

I think it will increase a little bit ... just image ... if the temperature is 0, the velocity will be 0 too. because the vibration of atom is so weak and the sound cant progation.

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3 years ago

Suppose a straight 1.00-mm-diameter copper wire could just "float" horizontally in air because of the force due to the Earth’s m

insens350 [35]

To solve this problem it is necessary to apply the concepts related to the Force since Newton's second law, as well as the concept of Electromagnetic Force. The relationship of the two equations will allow us to find the magnetic field through the geometric relations of density and volume.

$F_{mag}= BIL$

Where,

B = Magnetic Field

I = Current

L = Length

<em>Note: $F_{mag}$ is a direct adaptation of the vector relation $F=q \times V \times B$ </em>

From Newton's second law we know that the relation of Strength and weight is determined as

$F_g = mg$

Where,

m = Mass

g = Gravitational Acceleration

For there to be balance the two forces must be equal therefore

$F_{mag} = F_g$

$BIL = mg$

Our values are given as,

Diameter $(d) = 1.0mm = 1*10^{-3}m$

Radius $(r) = \frac{d}{2} = \frac{1*10^{-3}}{2} = 0.5*10^{-3}m$

Magnetic Field $(B) = 5.0*10^{-5} T$

From the relationship of density another way of expressing mass would be

$\rho = \frac{m}{V} \rightarrow m = \rho V$

At the same time the volume ratio for a cylinder (the shape of the wire) would be

$V = \pi r^2 L \rightarrow L =Length, r= Radius$

Replacing this two expression at our first equation we have that:

$BIL = mg$

$BIL = ( \rho V)g$

$BIL = ( \rho \pi r^2 L)g$

Re-arrange to find I

$I = \frac{( \rho \pi r^2 L)g}{BL}$

$I = \frac{( \rho \pi r^2 )g}{B}$

We have for definition that the Density of copper is $8.9*10^3 Kg/m^3$ , gravity acceleration is $9.8m/s^2$ and the values of magnetic field (B) and the radius were previously given, then:

$I = \frac{( (8.9*10^3 ) \pi (0.5*10^{-3})^2 )(9.8)}{5.0*10^{-5}}$

$I = 1370.05A$

The current is too high to be transported which would make the case not feasible.

8 0

3 years ago