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Klio2033 [76]
3 years ago
10

A top box labeled X contains 2 circles with plus signs and 2 circles with minus signs. A bottom box labeled Y contains 4 circles

with minus signs and 8 circles with plus signs. An arrow Z runs from the bottom box to the top box. Which labels best complete the diagram?
X: High resistance
Y: Low resistance
Z: Flow of electrons

X: Low resistance
Y: Flow of electrons
Z: High resistance

X: Flow of electrons
Y: High potential energy
Z: Low potential energy

X: Low potential energy
Y: High potential energy
Z: Flow of electrons
Physics
2 answers:
il63 [147K]3 years ago
8 0

x:flow electrons

y:high potential energy

z:low potential energy

your welcome

Kisachek [45]3 years ago
4 0

Answer:

X: Flow of electrons

Y: High potential energy

Z: Low potential energy

Explanation:

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Answer:

C. 8.01 m/s²

Explanation:

vf²= vi² + 2 • a • d

2ad = vf² - vi²

a = (vf²- vi²)/2d

d=25.00 -5.00=20.00 m

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4 0
3 years ago
the force has to be specified both in terms of its magnitude as well as its direction true or false? ​
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Anwser

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3 0
3 years ago
Read 2 more answers
A rectangular coil with 50 turns of conducting wire and a total resistance of 10.0 Ω initially lies in the yz-plane at time t =
castortr0y [4]

Answer:

a) 43.20V

b) 2.71W/s

c) 40.25s

d) 7.77Nm

Explanation:

(a) The emf of a rotating coil with N turns is given by:

emf=NBA\omega sin(\omega t)

N: turns

B: magnitude of the magnetic field

A: area

w: angular velocity

the emf max is given by:

emf_{max}=NBA\omega=(50)(1.80T)(0.200m*0.100m)(24.0rad/s)\\\\emf_{max}=43.20V

(b) the maximum rate of change of the magnetic flux is given by:

\frac{d\Phi_B}{dt}=\frac{d(A\cdot B)}{dt}=\frac{d}{dt}(ABcos\omega t)=AB\omega sin(\omega t)\\\\\frac{d\Phi_B}{dt}_{max}=(\pi(0.200*0.100))(1.80T)(24.0rad/s)=2.71\frac{W}{s}

(c) emf(t=0.050s)=(50)(1.80T)(0.200m*0.100m)(24rad/s)sin(24.0rad/s(0.050s))\\\\emf(t=0.050s)=40.26V

(d) The torque is given by:

\tau=NABIsin\theta\\\\NAB\omega=emf_{max}\\\\\tau=\frac{emf_{max}}{\omega}\frac{emf_{max}}{R}\\\\\tau=\frac{(43.20V)^2}{(24.0rad/s)(10.0\Omega)}=7.77Nm

3 0
3 years ago
A 10kg object experiences a horizontal force which causes it to accelerate 5 m/s2 moving it a distance of 20 m horizontally. How
mr_godi [17]

Answer:

The work done by the force is 1000 J.

Explanation:

Given:

Mass of object (m) = 10 kg

Acceleration of the object (a) = 5 m/s²

Displacement of the object (S) = 20 m

Using Newton's second law, we have:

Force = Mass × Acceleration

F=ma

Plug in the given values and solve for force 'F' acting on the object. This gives,

F=(10\ kg)(5\ m/s^2)\\\\F=50\ N

Now, work done by the force 'F' is equal to the product of force 'F' and the displacement it cause in its direction which is 20 m.

Therefore, the work done by the force is given as:

Work = Force × Displacement

W=FS

Plug in the values given and solve for 'W'. This gives,

W=50\ N\times 20\ m \\\\W=1000\ J

Therefore, the work done by the force is 1000 J.

7 0
4 years ago
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