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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."

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A metal sphere of radius 2.0 cm carries an excess charge of 3.0 μC. What is the electric field 6.0 cm from the center of the sph

Nonamiya [84]

Answer:

The electric field is $E = 7.5 *10^{6} \ N/C$

Explanation:

From the question we are told that

The radius of the metal sphere is $R = 2.0 \ cm = 0.02 \ m$

The excess charge which the metal sphere carries is $q = 3.0 \mu C = 3.0*10^{-6} \ C$

The distance of the position being to the center is $D = 6.0 \ cm = 0.06 \ m$

The coulomb constant is $k =9*10^{9} \ N \cdot m^2 /C^2$

Generally the electric field is mathematically represented as

$E = \frac{k * q}{D^2}$

substituting values

$E = \frac{9*10^{9} * 30.*10^{-6}}{(0.06)^2}$

$E = 7.5 *10^{6} \ N/C$

5 0

2 years ago

A car traveling at a velocity v can stop in a minimum distance d. What would be the car's minimum stopping distance if it were t

alexira [117]

Answer:

a. 4d.

If the car travels at a velocity of 2v, the minimum stopping distance will be 4d.

Explanation:

Hi there!

The equations of distance and velocity of the car are the following:

x = x0 + v0 · t + 1/2 · a · t²

v = v0 + a · t

Where:

x = position of the car at time t.

x0 = initial position.

v0 = initial velocity.

t = time.

a = acceleration.

v = velocity of the car at time t.

Let´s find the time it takes the car to stop using the equation of velocity. When the car stops, its velocity is zero. Then:

velocity = v0 + a · t v0 = v

0 = v + a · t

Solving for t:

-v/a = t

Since the acceleration is negative because the car is stopping:

v/a = t

Now replacing t = v/a in the equation of position:

x = x0 + v0 · t + 1/2 · a · t² (let´s consider x0 = 0)

x = v · (v/a) + 1/2 · (-a) (v/a)²

x = v²/a - 1/2 · v²/a

x = 1/2 v²/a

At a velocity of v, the stopping distance is 1/2 v²/a = d

Now, let´s do the same calculations with an initial velocity v0 = 2v:

Using the equation of velocity:

velocity = v0 + a · t

0 = 2v - a · t

-2v/-a = t

t = 2v/a

Replacing in the equation of position:

x1 = x0 + v0 · t + 1/2 · a · t²

x1 = 2v · (2v/a) + 1/2 · (-a) · (2v/a)²

x1 = 4v²/a - 2v²/a

x1 = 2v²/a

x1 = 4(1/2 v²/a)

x1 = 4x

x1 = 4d

If the car travels at a velocity 2v, the minimum stopping distance will be 4d.

5 0

2 years ago

The inventor of the photographic process in which a photograph produced without a negative by exposing objects to light on light

Nikolay [14]

A. Man Ray --------------------

7 0

1 year ago

Read 2 more answers

A toy rocket launcher can project a toy rocket at a speed as high as 35.0 m/s.

Anestetic [448]

Answer:

(a) 62.5 m

(b) 7.14 s

Explanation:

initial speed, u = 35 m/s

g = 9.8 m/s^2

(a) Let the rocket raises upto height h and at maximum height the speed is zero.

Use third equation of motion

$v^{2}=u^{2}+2as$

$0^{2}=35^{2}- 2 \times 9.8 \times h$

h = 62.5 m

Thus, the rocket goes upto a height of 62.5 m.

(b) Let the rocket takes time t to reach to maximum height.

By use of first equation of motion

v = u + at

0 = 35 - 9.8 t

t = 3.57 s

The total time spent by the rocket in air = 2 t = 2 x 3.57 = 7.14 second.

8 0

2 years ago

An actor who memorizes his lines word-for-word has demonstrated _____ learning.

Nana76 [90]

Mechanical association learning used by an actor to memorize his lines

8 0

2 years ago

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