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2 years ago

How do the current and coltage readings compare to before the wires were switched. Explain how to interpret them physically?

Physics

1 answer:

Blababa [14]2 years ago

7 0

The potential difference between two locations in an electric circuit is measured using a voltmeter.
If the electricity passes through the voltmeter it shows deflection.

<h3>What is the purpose of a voltmeter?</h3>

A voltage meter, usually referred to as a voltmeter, is a device that measures the voltage, or potential difference, between two points in an electrical or electronic circuit.
volts is the unit of voltmeter(volts, millivolts, kilovolts)

<h3>What is the explanation for the link between current and voltage?</h3>

Ohm's law states that the voltage across a conductor is directly proportional to the current flowing through it, provided all physical conditions and temperatures remain constant.

<h3>What is ohm's law in circuit?</h3>

V = IR, where V is voltage, I is current, and R is resistance, is known as Ohm's Law.
If you know the voltage of the battery in the circuit and how much resistance is in the circuit, you may use Ohm's Law to identify properties of a circuit, such as how much current is flowing through it.

To learn more about current and voltage visit:

brainly.com/question/10254698

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A 4 kg bowling ball rolls at a speed of 5 m/s on the roof of a building that is 30 meters tall. What is its kinetic energy?

klasskru [66]

ans: B

Kinetic energy = $\frac{1}{2} mv^{2}$

= $\frac{1}{2} (4)(5^{2} )$

= 50J

ps. the height will only affect its potential energy, not kinetic.

3 0

3 years ago

What is in the boron family with 3 valence electrons

olganol [36]

Boron, Aluminum, Gallium, Indium, Thallium

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3 years ago

A small ball with a mass of 0.6 kg and a velocity of 12 m/s hits another ball with the same mass. The firs forward and hits a th

Ket [755]

I think is 30m/s I am not sure

8 0

2 years ago

if you drop a stone from height of 2.5m. what is the speed of the stone right before it hits the ground?

KonstantinChe [14]

Since the stone was dropped from height, its initial velocity = 0 m/s

Using v² = u² + 2gs.

Where g ≈ 10 m/s², u = initial velocity = 0 m/s, s = height from drop = 2.5 m

v² = u² + 2gs

v² = 0² + 2*10*2.5

v² = 0 + 50

v² = 50

v = √50

v ≈ 7.07 m/s

Hence velocity just before hitting the ground is ≈ 7.07 m/s

6 0

3 years ago

A cannon is fired from the edge of a cliff, which is 60m above the sea. The cannonball's initial velocity is 88.3m/s and it is f

wel

Answer:

a. 11.29 s b. 94.72 m/s at -39.8° c. 821.57 m

Explanation:

a. Using y - y₀ = ut - 1/2gt² where u = vertical component of velocity = v₀sinθ where v₀ = 88.3 m/s and θ = 34.5°, y₀ = + 60 m and y = water surface = 0 m, g = 9.8 m/s² and t = time it takes the cannon to reach the water surface.

So y - y₀ = ut - 1/2gt²

y - y₀ = (v₀sinθ)t - 1/2gt²

substituting the values of the variables into the equation, we have

0 - 60 = (88.3 m/s × sin34.5°)t - 1/2 × 9.8 m/s²× t²

- 60 = 50t - 4.9t²

So, 4.9t² - 50t - 60 = 0

Using the quadratic formula to find t,

$t = \frac{-(-50) +/- \sqrt{(-50)^{2} - 4 X 4.9 X -60} }{2 X 4.9} \\t = \frac{50 +/- \sqrt{2500 + 1176} }{9.8} \\t = \frac{50 +/- \sqrt{3676} }{9.8} \\t = \frac{50 +/- 60.63 }{9.8} \\t = \frac{50 + 60.63 }{9.8} or t = \frac{50 - 60.63 }{9.8} \\t = \frac{110.63 }{9.8} or t = \frac{-10.63 }{9.8} \\t = 11.29 sor -1.085$

Since t cannot be negative, t = 11.29 s

b. We first need to find the impact vertical velocity component. Using

v = u - gt where u = initial vertical velocity component = v₀sinθ and t = 11.29 s and g = 9.8 m/s². So,

v = v₀sinθ - gt

= 88.3 m/s × sin34.5° - 9.8 m/s² × 11.29 s

= 50.01 m/s - 110.64 m/s

= -60.63 m/s

Since the horizontal velocity is constant u' = v₀cosθ = 88.3 m/s × cos34.5° = 72.77 m/s.

The impact velocity is thus the resultant of the horizontal velocity and final impact velocity. So, V = √(v² + u'²)

= √((-60.63 m/s)² + (72.77 m/s)²)

= √((3676 m²/s² + 5295.48 m²/s²)

= √(8971.48 m²/s²

= 94.72 m/s

The angle θ = tan⁻¹(v/u') = tan⁻¹(-60.63 m/s ÷ 72.77 m/s) = tan⁻¹(-0.8332) = -39.8°

So the impact velocity is 94.72 m/s at -39.8°

c. The horizontal distance out from the base of the cliff that the ball strikes the water is the range, R = u't = 72.77 m/s × 11.29 s = 821.57 m

5 0

3 years ago