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is plugged into the outlet of a 120−V circuit that has a 20−A circuit breaker. You plug an electric hair dryer into the same out

WINSTONCH [101]

Answer:

Given the point of maximun electric power of the hair dryer 1500 (w); The circuit breaker won´t trip at all

Explanation:

The most simplyfied relation between power, voltage and current is:

Electric power (in watts) =Voltage (in volts) * current (in ampers)

In the case of P= 1500 (w) and V= 120 (v) we have:

I = 1500/120 (a) = 12,65 (a)

This value is far away of 20 (a) (the nominal trip current

8 0

3 years ago

A 2kg box is pushed along a flat frictionless surface with an applied force of 53.91newton j+19.62 j were j is horizontal and j

tamaranim1 [39]

Since the box doesn't move vertically at all, no work is done by the vertical component of the force.

If 53.91 Newtons is the horizontal component of the force (very unclear in the question), then the work done is

Work = (force) x (distance)

Work = (53.91 N) x (10 m)

Work = 539.1 Joules

Power = (work done) / (time to do the work)

Power = (539.1 joules) / (90 sec)

Power = (539.1/90) (joule/sec)

Power = 5.99 watts

=====================================

Note: If the mass of the box is only 2 kg, and you push it along the surface with a constant force of 53⁺ Newtons, and the surface really is frictionless, then that box is gonna cover a WHOOOOOLE LOT more than 10 meters in 90 seconds. I get 109,168 meters !

5 0

3 years ago

The state of strain at a point is plane strain with εx = ε0, εy = –2ε0, γxy = 0, where ε0 is a positive constant. What is the no

Marat540 [252]

Answer:

The normal strain along an axis oriented 45° from the positive x axis in the clockwise direction is -ε₀/2

Explanation:

Given that

$\epsilon_{x}=\epsilon_{o}\\\\\epsilon_{y}=-2\epsilon_{o}\\\\\gamma_{xy}=0\\\\\theta=-45^{o}\\\\\epsilon_{x_{1}}=?$

From equation of normal strain in x direction:

$\epsilon_{x_{1}}=\epsilon_{x}cos^{2}\theta+\epsilon_{y}sin^{2}\theta+\gamma_{xy{ sin\theta cos\theta$

Substituting the values:

$\epsilon_{x_{1}}=\epsilon_{o}cos^{2}(-45)-2\epsilon_{o}sin^{2}(-45)+0\\\\\epsilon_{x_{1}}=\frac{\epsilon_{o}}{2}-2\frac{\epsilon_{o}}{2}\\\\\epsilon_{x_{1}}=-\frac{\epsilon_{o}}{2}$