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2 years ago

This outlaw is executed by hanging "in the spring of '25" by

1 answer:

8 0

The outlaw that was executed by hanging "in the spring of '25" is identified as the HIGHWAYMAN.

This is one of the characters in the song, "American Remains", sang by The Highwaymen. The group consisted of Johnny Cash, Waylon Jennings, Willie Nelson and Kris Kristofferson. Other characters in the song were a sailor, a dam builder, and a pilot of a starship.

This is the first stanza of the song:

"I was a highwayman. Along the coach roads I did ride
With sword and pistol by my side 
Many a young maid lost her baubles to my trade 
Many a soldier shed his lifeblood on my blade 
The b*stards hung me in the spring of twenty-five 
But I am still alive."

You might be interested in

An automobile traveling at 25.0 km/h along a straight, level road accelerates to 65.0 km/h in 6.00 s. what is the magnitude of t

USPshnik [31]

Note that
1 km/h = (1000 m)/(3600 s) = 0.27778 m/s

Initial velocity, v₁ = 25 km/h = 6.9444 m/s
Final velocity, v₂ = 65 km/h = 18.0556 m/s

Time interval, dt = 6 s.

Calculate average acceleration.
a = (v₂ - v₁)/dt
= (18.0556 - 6.9444 m/s)/(6 s)
= 1.852 m/s²

Answer:
The average acceleration is 1.85 m/s² (nearest hundredth)

3 0

2 years ago

26. An ice-skater who weighs 200 N is gliding across the ice. If the force of friction is 4 N. what is the

Scrat [10]

Answer:

0.02

Explanation:

coefficient of kinetic friction = μ

force of friction = Ff

Normal Force = FN, but

FN = -W

Ff = -μFN

so μ = Ff/FN

= 4N/200N

= 0.02.

7 0

2 years ago

What common laboratory measuring device will we likely always use in experiments that measure enthalpy changes?

adelina 88 [10]

The most common measuring device to be used in measuring enthalpy changes is the thermometer.

5 0

1 year ago

A coaxial cable consists of a thin insulated straight wire carrying a current of 2.00 A surrounded by a cylindrical conductor ca

ella [17]

Answer:

B = 15μT

Explanation:

In order to calculate the magnitude of the magnetic field generated by the coaxial cable you use the Ampere's law, which is given by:

$B=\frac{\mu_oI}{2\pi r}$ (1)

μo: magnetic permeability of vacuum = 4π*10^-7 T/A

I: current

r: distance from the wire to the point in which B is calculated

In this case you have two currents with opposite directions, which also generates magnetic opposite magnetic fields. Then, you have (but only for r > radius of the cylindrical conductor) the following equation:

$B_T=B_1-B_2=\frac{\mu_o I_1}{2\pi r}-\frac{\mu_o I_2}{2\pi r}\\\\B_T=\frac{\mu_o}{2\pi r}(I_1-I_2)$ (2)

I1: current of the central wire = 2.00A

I2: current of the cylindrical conductor = 3.50A

r: distance = 2.00 cm = 0.02 m

You replace the values of all parameters in the equation (2), and you use the absolute value because you need the magnitude of B, not its direction.

$|B|=|\frac{4\pi*10^{-7}T/A}{2\pi (0.02m)}(2.00A-3.50A)|=1.5*10^{-5}T\\\\|B|=15*10^{-6}T=15\mu T$

The agnitude of the magnetic field outside the coaxial cable, at a distance of 2.00cm to the center of the cable is 15μT

3 0

2 years ago

A 0.025-kg block on a horizontal frictionless surface is attached to an ideal massless spring whose spring constant is The block

DerKrebs [107]

Complete Question

A 0.025-kg block on a horizontal frictionless surface is attached to an ideal massless spring whose spring constant is 150 N/m. The block is pulled from its equilibrium position at x = 0.00 m to a displacement x = +0.080 m and is released from rest. The block then executes simple harmonic motion along the horizontal x-axis. When the displacement is x = 0.024 m, what is the kinetic energy of the block?

Answer:

The kinetic energy is $KE = 0.4368\ J$