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DochEvi [55]
3 years ago
13

It is well known that bullets fired at Superman simply bounce off his chest. Suppose that a gangster sprays Superman's chest wit

h bullets of mass m = 1.7 g at a rate of R = 165 bullets/min. The speed of each bullet is v = 380 m/s. Suppose the bullets rebound with no change in speed. What is the average force exerted by the stream of bullets on Superman's chest?
Physics
1 answer:
Zanzabum3 years ago
6 0

Answer:

1.77 N

Explanation:

Mass of bullet m = 1.7 x 10⁻³ kg

velocity v = 380 m /s .

Momentum of one bullet

= 1.7 x 10⁻³ x 380

= 646 x 10⁻³ kg m/s

momentum of 165 bullets

= 165 x 646 x 10⁻³ kg m/s

= 106.59 kg m/s

Final momentum after bouncing

= 0

change in momentum

= 106.59  - 0

= 106.59 kg m/s

This change occurs in one minute

so rate of change in momentum

= 106.59 / 60 kg m/s per second

= 1.77 kg m/s per second

rate of change in momentum = force

This is force on superman's chest

Force required = 1.77 N .

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3 years ago
Which vector goes from (-1, -3) to (-4, -1)
Alisiya [41]
Its this (couldn’t write it down on here properly so i had to ss it)
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6 0
2 years ago
irius, the brightest star in the sky, is 2.6 parsecs (8.6 light-years) from Earth, giving it a parallax of 0.379 arcseconds. Ano
Norma-Jean [14]

The actual distance of Regulus from Earth is 23.81 parsecs.

Given:

Parallax of Regulus, p = 0.042 arc seconds

Calculation:

When an observer changes their position, an apparent change in the object's position takes place. This change can be calculated using the angle ( or semi-angle) made by the observer and object i.e. the angle made between the two lines of observation from the object to the observer.

Thus from the relation of parallax of a celestial body we get:

S = 1/ tan p ≈ 1 / p

where S is the actual distance between the object and the observer

            p is the parallax angle observed

Here for Regulus, we get:

S = 1 / p

  = 1 / (0.042)                                     [ 1 parsecs = 1 arcseconds ]

  = 23.81 parsecs

We know that,

1 parsecs = 3.26 light-years = 206,000 AU

Converting the actual distance into light years we get:

23.81 parsecs = 23.81 × (3.26 light yrs) = 77.658 light-years

Therefore, the actual distance of Regulus from Earth is 23.81 parsecs which is 77.658 in light years.

Learn more about astronomical units here:

<u>brainly.com/question/16471213</u>

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6 0
2 years ago
A battery charger is connected to a dead battery and delivers a current of 8.9 A for 4.7 hours, keeping the voltage across the b
adell [148]

Answer:

1807.56 kJ

Explanation:

Parameters given:

Current, I = 8.9A

Time, t = 4.7hrs = 4.7 * 3600 = 16920 secs

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Electrical energy is given as:

E = I*V*t

Where I = Current

V = Voltage/Potential differenxe

t = time in seconds.

E = 8.9 * 12 * 16920

E = 1807056 J = 1807.056 kJ

3 0
3 years ago
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