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Serga [27]
3 years ago
15

The electric potential ( relative to infinity ) due to a single point charge Q is 400 V at a point that is 0.6 m to the right of

Q. The electric potential (relative to infinity) at a point that is 0.90 m to the left of 0 is:_____.
A. + 400 V.
B. -400 V.
C. + 200 V.
Physics
1 answer:
UkoKoshka [18]3 years ago
5 0

Answer:

The potential at a distance of 0.9 m is 266.67 V.

Explanation:

Charge = Q

Potential is 400 V at a distance 0.6 m .

Let the potential is V at a distance 0.9 m.

Use the formula of potential.

V = \frac{Kq}{r}\\\\\frac{V}{400}=\frac{0.6}{0.9}\\\\V = 266.67 V

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With the same block-spring system from above, imagine doubling the displacement of the block to start the motion. By what factor
Fofino [41]

Answer:

A)     K / K₀ = 4   b)     v / v₀ = 4

Explanation:

A) For this exercise we can use the conservation of mechanical energy

in the problem it indicates that the displacement was doubled (x = 2xo)

starting point. At the position of maximum displacement

      Em₀ = Ke = ½ k (2x₀)²

final point. In the equilibrium position

      Em_{f} = K = ½ m v²

        Em₀ = Em_{f}

        ½ k 4 x₀² = K

        (½ K x₀²) = K₀

         K = 4 K₀

          K / K₀ = 4

B) the speed value

          ½ k 4 x₀² = ½ m v²

          v = 4 (k / m) x₀

if we call

           v₀ = k / m x₀

          v = 4 v₀

         v / v₀ = 4

3 0
3 years ago
Which of the following is represented by the letter A in the diagram below?
Kruka [31]

We have that the letter A in the diagram below given as

Amplitude

Option A

<h3>Amplitude</h3>

Question Parameters:

Amplitude

Crest

Trough

Wavelength

Generally, the amplitude of a wave is the maximum  displacement of the wave in the medium from its initial position.

Amplitude is denoted with the letter A

Therefore,Amplitude

Option A

For more information on displacement visit

brainly.com/question/989117

6 0
2 years ago
The 480 g bar is rotating as shown what is the angular momentum of the bar about the axle?
Greeley [361]
On a similar problem wherein instead of 480 g, a 650 gram of bar is used:

Angular momentum L = Iω, where 
<span>I = the moment of inertia about the axis of rotation, which for a long thin uniform rod rotating about its center as depicted in the diagram would be 1/12mℓ², where m is the mass of the rod and ℓ is its length. The mass of this particular rod is not given but the length of 2 meters is. The moment of inertia is therefore </span>
<span>I = 1/12m*2² = 1/3m kg*m² </span>

<span>The angular momentum ω = 2πf, where f is the frequency of rotation. If the angular momentum is to be in SI units, this frequency must be in revolutions per second. 120 rpm is 2 rev/s, so </span>
<span>ω = 2π * 2 rev/s = 4π s^(-1) </span>

<span>The angular momentum would therefore be </span>
<span>L = Iω </span>
<span>= 1/3m * 4π </span>
<span>= 4/3πm kg*m²/s, where m is the rod's mass in kg. </span>

<span>The direction of the angular momentum vector - pseudovector, actually - would be straight out of the diagram toward the viewer. </span>

<span>Edit: 650 g = 0.650 kg, so </span>
<span>L = 4/3π(0.650) kg*m²/s </span>
<span>≈ 2.72 kg*m²/s</span>
4 0
3 years ago
A rotating fan completes 1200 revolutions every minute. Consider the tip of a blade, at a radius of 0.15 m. (a) Through what dis
olga55 [171]

Answer:

(a) 0.942 m

(b) 18.84 m/s

(c) 2366.3 m/s²

(d) 0.05 s

Explanation:

(a) In one revolution, it travels through one circumference, 2πr = 2 × 3.14 × 0.15 m = 0.942 m.

(b) Its frequency, f, is 1200 rev/min = \dfrac{1200}{60}rev/s = 20 rev/s.

Its angular frequency, ω = 2πf = 2π × 20 = 40π

The speed is given by

v = ωr = 40π × 0.15 = 6π = 18.84 m/s

(c) Its acceleration is given by, a = ω²r = (40π)² × 0.15 = 2366.3 m/s²

(d) The period is the inverse of the frequency because it is the time taken to complete one revolution.

T = \dfrac{1}{f}

T = 1/20 = 0.05 s

6 0
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