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iVinArrow [24]
3 years ago
14

When a magnet moves above a conducting ladder, the currents induced in the ladder produces a magnetic field. This field interact

s with the magnetic field of the magnet to produce a force that A) pulls the ladder in the direction opposite to the direction of the moving magnet. B) pulls the ladder in the same direction as the moving magnet. C) brakes the ladder. D) None of the above.
Physics
1 answer:
pav-90 [236]3 years ago
4 0

Answer:

) pulls the ladder in the direction opposite

Explanation:

This is in line with lenz law that states that the magnetic field induced in a conductor act to oppose the magnetic field that produced it

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The block A and attached rod have a combined mass of 50 kg and are confined to move along the guide under the action of the 796-
Luba_88 [7]

Answer:

The bending moment is 459.16 N.m

Explanation:

From the given information;

Let's assume that the angle is 66°

Then, the free body diagram is draw and attached in the file below.

Now, the calculation of the acceleration from the first part of the free body diagram is:

\sum F_x = ma_x \\ \\ 796 - 50(9.81) sin 66=50a \\ \\ 796 - 448.094 = 50 a  \\ \\ a = \dfrac{347.906}{50} \\ \\ a  = 6.96 \ m/s^2

Bending moment M:

From the second part of the diagram:

\sum M_B = mad \\ \\ M - (15 \times 9.81) (1.5) = (25 \times 6.96)(1.5 sin 66) \\ \\ M - 220.725 = 238.435  \\ \\  M = 238.435 + 220.725 \\ \\  \mathbf{M = 459.16 \ N.m}

6 0
3 years ago
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
in volleyball if you are a server what position on the court are you? center back #6, right back #1 or left back #5? help me ple
jek_recluse [69]
In the sport of volleyball a server is in the "back left position" for standard rules.

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4 0
3 years ago
Write one general properties of non contact force​
Gekata [30.6K]

Answer:    Gravitational force between the two masses does not depend on the medium separating two masses.

3 0
3 years ago
A fan spins at 6.0 rev/s. You turn it off, and it slows at 1.0 rev/s2. What is the angular displacement before it stops
Komok [63]

Answer:

Angular displacement before it stops = 18 rev

Explanation:

Given:

Speed of fan w(i) = 6 rev/s

Speed of fan (Slow) ∝ = 1 rev/s

Final speed of fan w(f) = 0 rev/s

Find:

Angular displacement before it stops

Computation:

w(f)² = w(i) + 2∝θ

0² = 6² + 2(1)θ

0 = 36 + 2θ

2θ = -36

Angular displacement before it stops = -36 / 2

θ = -18

Angular displacement before it stops = 18 rev

4 0
3 years ago
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