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tangare [24]
3 years ago
14

Una barra conductora de L = 0,9 m se mueve sin fricción sobre dos rieles conductores horizontales, Unos extremos de los rieles s

e conectan a una batería de 15 V y un interruptor.
La barra se mueve en una región donde existe un campo magnético uniforme
B = −1,5 ˆz T. La barra tiene una masa de 0,5 kg y una resistencia de 8 Ω.
a) Encuentre la aceleración de la barra, en t = 0 s, justo cuando se cierra el interruptor.
b) Encuentre la aceleración de la barra cuando se mueve hacia la derecha con rapidez 2 m/s.
Physics
1 answer:
MAVERICK [17]3 years ago
8 0

Answer:

sorry I don't understand

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lidiya [134]

Answer:

hi

Explanation:

hi

4 0
3 years ago
(ANSWER NOW PLS) A diver with a mass of 90kg is at a height of 10m, and he has not jumped off of the board yet (v=0m/s) what's h
Korolek [52]

Answer: there is zero kinetic energy but there is Gravitational Potential Energy (GPE) and GPE = 8826.3 J

Explanation:

3 0
3 years ago
A thick steel sheet of area 100 in.2 is exposed to air near the ocean. After a one-year period it was found to experience a weig
vladimir1956 [14]

Answer:

Therefore the rate of corrosion 37.4 mpy and 0.952 mm/yr.

Explanation:

The corrosion rate is the rate of material remove.The formula for calculating CPR or corrosion penetration rate is

CPR=\frac{KW}{DAT}

K= constant depends on the system of units used.

W= weight =485 g

D= density =7.9 g/cm³

A = exposed specimen area =100 in² =6.452 cm²

K=534 to give CPR in mpy

K=87.6  to give CPR in mm/yr

mpy

CPR=\frac{KW}{DAT}

        =\frac{534\times( 485g)\times( 10^3mg/g)}{(7.9g/cm^3) \times (100in^2)\times (24h/day)\times (365day/yr)\times 1yr}

        =37.4mpy

mm/yr

CPR=\frac{KW}{DAT}

        =\frac{87.6\times (485g)\times (10^3 mg/g)}{(7.9g/cm^3)\times (100in^2)\times(2.54cm/in)^2\times (24h/day)\times (365day/yr)\times 1yr}

       =0.952 mm/yr

Therefore the rate of corrosion 37.4 mpy and 0.952 mm/yr.

3 0
4 years ago
A 0.18-kg turntable of radius 0.32 m spins about a vertical axis through its center. A constant rotational acceleration causes t
borishaifa [10]

Answer:

Angular acceleration will be 18.84rad/sec^2

Explanation:

We have given that mass m = 0.18 kg

Radius r = 0.32 m

Initial angular velocity \omega _i=0rev/sec

And final angular velocity \omega _f=24rev/sec

Time is given as t = 8 sec

From equation of motion

We know that \omega _f=\omega _i+\alpha t

24=0+\alpha \times 8

\alpha =3rev/sec^2=3\times 2\times \pi rad/sec^2=18.84rad/sec^2

So angular acceleration will be 18.84rad/sec^2

4 0
4 years ago
A penny is placed on a rotating turntable. where on the turntable does the penny require the largest centripetal force to remain
Mama L [17]

m = mass of the penny

r = distance of the penny from the center of the turntable or axis of rotation

w = angular speed of rotation of turntable

F = centripetal force experienced by the penny

centripetal force "F" experienced by the penny of "m" at distance "r" from axis of rotation is given as

F = m r w²

in the above equation , mass of penny "m"  and angular speed "w" of the turntable is same at all places. hence the centripetal force directly depends on the radius .

hence greater the distance from center , greater will be the centripetal force to remain in place.  

So at the edge of the turntable , the penny experiences largest centripetal force to remain in place.

4 0
3 years ago
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