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2 years ago

Determine the potential difference between the ends of the wire of resistance 5 Ω if 720 C passes through it per minute.

Physics

1 answer:

Strike441 [17]2 years ago

8 0

Answer:

The potential difference between the ends of a wire is 60 volts.

Explanation:

It is given that,

Resistance, R = 5 ohms

Charge, q = 720 C

Time, t = 1 min = 60 s

We know that the charge flowing per unit charge is called current in the circuit. It is given by :

I = 12 A

Let V is the potential difference between the ends of a wire. It can be calculated using Ohm's law as :

V = IR

V = 60 Volts

So, the potential difference between the ends of a wire is 60 volts. Hence, this is the required solution.

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Korvikt [17]

Answer: iDc im in 3grade

Explanation:sorry

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2 years ago

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The coefficient of kinetic friction between an object and the surface upon which it is sliding is 0.10. The mass of the object i

denis-greek [22]

Answer:

Explanation:

1) Force Friction = Normal Force * Coefficient of Friction

Force Friction = Mass * Gravity * Coefficient of Friction

2) F = ma

Force = mass * acceleration

Force Friction (from #1) = mass * acceleration

acceleration = Force Friction / Mass

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2 years ago

The equation of a progressive

Juliette [100K]

Answer:

Amplitude = 0.02m

Frequency = 640 Hz

Wavelength, λ = 0.5m

v = 320 m/s

Explanation:

Given the wave equation :

y=0.02 sin2π/0.5 (320t - x) where x and y are in

meters and t is in second

Comparing the above relation with the general wave equation :

y(x, t) = Asin2π/λ(wt - kx)

The amplitude, A = 0.02

From the equation :

2π/0.5 = 2π/λ

λ = 0.5 m

320t = vt

Hence, v = 320 m/s

Recall :

v = fλ

320 = f * 0.5

f = 320 / 0.5

f = 640 Hz

6 0

3 years ago

A transverse standing wave is set up on a string that is held fixed at both ends. The amplitude of the standing wave at an antin

ZanzabumX [31]

Answer:

a) the maximum transverse speed of a point on the string at an antinode is 5.9899 m/s

b) the maximum transverse speed of a point on the string at x = 0.075 m is 4.2338 m/s

Explanation:

Given the data in the question;

as the equation of standing wave on a string is fixed at both ends

y = 2AsinKx cosωt

but k = 2π/λ and ω = 2πf

λ = 4 × 0.150 = 0.6 m

and f = v/λ = 260 / 0.6 = 433.33 Hz

ω = 2πf = 2π × 433.33 = 2722.69

given that A = 2.20 mm = 2.2×10⁻³

so $V_{max1}$ = A × ω

$V_{max1}$ = 2.2×10⁻³ × 2722.69 m/s

$V_{max1}$ = 5.9899 m/s

therefore, the maximum transverse speed of a point on the string at an antinode is 5.9899 m/s

A' = 2AsinKx

= 2.20sin( 2π/0.6 ( 0.075) rad )

= 2.20 sin( 0.7853 rad ) mm

= 2.20 × 0.706825 mm

A' = 1.555 mm = 1.555×10⁻³

$V_{max2}$ = A' × ω

$V_{max2}$ = 1.555×10⁻³ × 2722.69

$V_{max2}$ = 4.2338 m/s

Therefore, the maximum transverse speed of a point on the string at x = 0.075 m is 4.2338 m/s

8 0

2 years ago

5.<br>why does the pointer rotate?

morpeh [17]

Answer: find the answer in the explanation.

Explanation:

From the experiment set up in the diagram, the pointer is resting on the drinking straw while the rod is resting on the drinking straw.

When the rod is being heated through the bursen burner, there will be linear expansion in the rod. As the rod increases its length, this causes the drinking straw to roll and thereby causing the pointer to rotate.

The pointer therefore rotates because of the thermer expansion that happen in the rod due to the heat from the bursen burner.

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3 years ago