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JulijaS [17]
3 years ago
9

A 1.5m wire carries a 2 A current when a potential difference of 52 V is applied. What is the resistance of the wire?

Physics
1 answer:
pantera1 [17]3 years ago
4 0

Resistance = (voltage) / (current)

Resistance = (52 V) / (2 A)

<em>Resistance = 26 ohms</em>

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If you have a gold brick that is 2.0 cm by 3.0 cm by 4.0 cm and has a density of 19.3 g/ml, what is its mass
34kurt

Answer:

It is

Explanation:

1 Answer. The volume is 37.0 cm3Au .

8 0
3 years ago
Scenario
Anvisha [2.4K]

Answer:

1) t = 23.26 s,  x = 8527 m, 2)   t = 97.145 s,  v₀ = 6.4 m / s

Explanation:

1) First Scenario.

After reading your extensive problem, we are going to solve it, for this exercise we must use the parabolic motion relationships. Let's carry out an analysis of the situation, for deliveries the planes fly horizontally and we assume that the wind speed is zero or very small.

Before starting, let's reduce the magnitudes to the SI system

         v₀ = 250 miles/h (5280 ft / 1 mile) (1h / 3600s) = 366.67 ft/s

         y = 2650 m

Let's start by looking for the time it takes for the load to reach the ground.

         y = y₀ + v_{oy} t - ½ g t²

in this case when it reaches the ground its height is zero and as the plane flies horizontally the vertical speed is zero

         0 = y₀ + 0 - ½ g t2

          t = \sqrt{ \frac{2y_o}{g} }

          t = √(2 2650/9.8)

          t = 23.26 s

this is the horizontal scrolling time

          x = v₀ t

          x = 366.67  23.26

          x = 8527 m

the speed at the point of arrival is

         v_y = v_{oy} - g t = 0 - gt

         v_y = - 9.8 23.26

         v_y = -227.95 m / s

Module and angle form

        v = \sqrt{v_x^2 + v_y^2}

         v = √(366.67² + 227.95²)

        v = 431.75 m / s

         θ = tan⁻¹ (v_y / vₓ)

         θ = tan⁻¹ (227.95 / 366.67)

         θ = - 31.97º

measured clockwise from x axis

We see that there must be a mechanism to reduce this speed and the merchandise is not damaged.

2) second scenario. A catapult located at the position x₀ = -400m y₀ = -50m with a launch angle of θ = 50º

we look for the components of speed

           cos θ = v₀ₓ / v₀

           sin θ = v_{oy} / v₀

            v₀ₓ = v₀ cos θ

            v_{oy} = v₀ sin θ

we look for the time for the arrival point that has coordinates x = 0, y = 0

            y = y₀ + v_{oy} t - ½ g t²

            0 = y₀ + vo sin θ t - ½ g t²

            0 = -50 + vo sin 50 t - ½ 9.8 t²

            x = x₀ + v₀ₓ t

            0 = x₀ + vo cos θ t

            0 = -400 + vo cos 50 t

podemos ver que tenemos un sistema de dos ecuación con dos incógnitas

          50 = 0,766 vo t – 4,9 t²

          400 =   0,643 vo t

resolved

          50 = 0,766 ( \frac{400}{0.643 \ t}) t – 4,9 t²

          50 = 476,52 t – 4,9 t²

          t² – 97,25 t + 10,2 = 0

we solve the quadratic equation

         t = [97.25 ± \sqrt{97.25^2 - 4 \ 10.2}] / 2

         t = 97.25 ±97.04] 2

         t₁ = 97.145 s

         t₂ = 0.1 s≈0

the correct time is t1 the other time is the time to the launch point,

         t = 97.145 s

let's find the initial velocity

         x = x₀ + v₀ cos 50 t

         0 = -400 + v₀ cos 50 97.145

         v₀ = 400 / 62.44

         v₀ = 6.4 m / s

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3 years ago
A 6.0 m section of wire carries a current of 5.2 A from east to west in the earth's magnetic field of 1.0 × 105 T at a location
NNADVOKAT [17]
Well, since you only want direction, ignore the numbers. Use the right hand rule.
Current (pointer finger) points west (left). 
Magnetic field (middle finger) points south (towards you). 
Force (thumb) then points up (away from the earth)
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3 years ago
Why is acceleration measured in meters
Setler [38]
Acceleration is measured in meters per second square.
4 0
3 years ago
What is pitch related to in terms of sound? Does it affect the speed? If so, how?
Juli2301 [7.4K]

Pitch is the impression the listener gets of the <em>frequency</em> of the sound.

The speed of the sound is <em>not</em> related to its pitch/frequency.  

If the speed and frequency were related, that would be a real problem.  Bands, orchestras, and choirs could not exist !  All the instruments in the orchestra could play a note together, at the same time.  But then the higher instruments ... the flute, trumpet, violins, high guitar strings and high piano keys ... would travel to you fast, and the lower instruments ... the trombone, tuba, double bass, bass drum, low guitar strings and the low piano keys ... would travel to you slow.  They all played the note at the same time, but by the time you heard it, it would be all smeared out ... every instrument arriving at your ear at a different time !

5 0
3 years ago
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