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igomit [66]
4 years ago
11

Oscilloscope channel addition problem. Read the followingoscilloscope settings carefully. Suppose you display a 1 kHz, 2 Vsine w

ave in channel A and a DC, 1V signal in channel B withsensitivities set on 1V/div. You select 'Add' so that the twosignals are combined and you readjust the position so the trace isin the middle of the screen. If you switch the signal in channel Bfrom DC to AC, what will happen? (Draw a picture if it helps.)A.The trace jumps down 2 divisions.B.The trace jumps down 1 divisionC.The trace jumps up 1 division.D.Nothing happens
Physics
1 answer:
Novay_Z [31]4 years ago
4 0

Answer: B.  The trace jumps down 1 division

Explanation:

From the above data, the two signals are combined and adjusted in the middle of the screen. If you switch the signal in channel B from DC to AC, then the screen jumps down 1 division.

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A person stands on a scale in an elevator. As the elevator starts, the scale has a constant reading of 592 N. As the elevator la
gladu [14]

Answer:

<h2>a) 496N</h2><h2>b) 50.56kg</h2><h2>c) 1.90m/s²</h2>

Explanation:

According to newton's secomd law, ∑F = ma

∑F is the summation of the force acting on the body

m is the mass of the body

a is the acceleration

Given the normal force when the elevator starts N1 = 592N

Normal force after the elevator stopped N2 = 400N

When the elevator starts, its moves upward, the sum of force ∑F = Normal (N)force on the elevator - weight of the person( Fg)

When moving up;

N1 - Fg = ma

N1 = ma + Fg ...(1)

Stopping motion of the elevator occurs after the elevator has accelerates down. The sum of forces in this case will give;

N2 - Fg = -ma

N2 = -ma+Fg ...(2)

Adding equation 1 and 2 we will have;

N1+N2 = 2Fg

592N + 400N = 2Fg

992N 2Fg

Fg = 992/2

Fg = 496N

The weight of the person is 496N

<em>\b) To get the person mass, we will use the relationship Fg = mg</em>

g = 9.81m/s

496 = 9.81m

mass m = 496/9.81

mass = 50.56kg

c) To get the magnitude of acceleration of the elevator, we will subtract equation 1 from 2 to have;

N1-N2 = 2ma

592-400 = 2(50.56)a

192 = 101.12a

a = 192/101.12

a = 1.90m/s²

3 0
4 years ago
9.58 A spring of equilibrium length L1 and spring constant k1 hangs from the ceiling. Mass m1 is suspended from its lower end. T
andrey2020 [161]

Answer:

The distance of m2 from the ceiling is L1 +L2 + m1g/k1 + m2g/k1 + m2g/k2.

See attachment below for full solution

Explanation:

This is so because the the attached mass m1 on the spring causes the first spring to stretch by a distance of m1g/k1 (hookes law). This plus the equilibrium lengtb of the spring gives the position of the mass m1 from the ceiling. The second mass mass m2 causes both springs 1 and 2 to stretch by an amout proportional to its weight just like above. The respective stretchings are m2g/k1 for spring 1 and m2g/k2 for spring 2. These plus the position of m1 and the equilibrium length of spring 2 L2 gives the distance of L2 from the ceiling.

4 0
4 years ago
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