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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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Calculate the de broglie wavelength (in picometers) of a hydrogen atom traveling at 440 m/s.
Aleonysh [2.5K]

De broglie wavelength, \lambda = \frac{h}{mv}, where h is the Planck's constant,  m is the mass and v is the velocity.

h = 6.63*10^{-34}

Mass of hydrogen atom,  m = 1.67*10^{-27}kg

v = 440 m/s

Substituting

   Wavelength \lambda = \frac{h}{mv} = \frac{6.63*10^{-34}}{1.67*10^{-27}*440} = 0.902 *10^{-9}m = 902 *10^{-12}m

1 pm = 10^{-12}m\\ \\ So , \lambda =902 pm

So  the de broglie wavelength (in picometers) of a hydrogen atom traveling at 440 m/s is 902 pm

7 0
2 years ago
When waves enter a denser medium, they bend the normal.<br><br> A) Toward<br> B) Away From
Andreyy89
The answer is bend towards normal.
7 0
3 years ago
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a 10 kilogram steel ball is dropped from the top of a tower 100 meters high the kinetic energy of the ball just before it sttike
EleoNora [17]

Explanation:

If we assume negligible air resistance and heat loss, we can assume that all of the Gravitational potential energy of the ball will turn into Kinetic energy as it falls toward the ground.

Therefore our Kinetic energy = mgh = (10kg)(9.81N/kg)(100m) = 9,810J.

4 0
3 years ago
A 350-kg roller coaster car starts from rest at point A and slides down a frictionless loop-the-loop (Pig. P7.41). (a) How fast
Leona [35]

Answer:

a)Velocity of car =v=16 m/s

b)Force against the track at point B=1.15*10^{4}N

Explanation:

Given mass of roller coaster=m=350 kg

Position of A=Ha=25 m

Position of B=Hb=12 m

Net potential energy=mg(ha-hb)

Net potential energy=(350)(9.80)(25-12)

Net potential energy=44590 J

Using energy conservation

net kinetic energy=net potential energy

(1/2)mv^2=mg(ha-hb)

m=350

velocity=v=16 m/s

b)There two force acting,centripetal force upward and gravity downward.

Thus net force acting will be

Net force=(mv^2/r)-mg

Net force=14933.33-3430

Net force=1.15*10^{4} N

3 0
2 years ago
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What is the distance of separation between objects of masses 5.6 x 10 5 kg and 8.8 x 10 6 kg if the force of gravity between the
AlladinOne [14]

Answer:

2.87m

Explanation:

Using the law of gravitation to solve this question

F = GMm/r²

G is the gravitational constant

M and m are the masses

r is the distance between the masses

Substitute the given values

G = 6.67×10^-11 m³/kgs²

M =8.8 x 10^6 kg

m = 5.6 x 10^5 kg

F =440N

400 = 6.67×10^-11×8.8 x 10^6 ×5.6 x 10^5/r²

400r² = 328.698×10

400r² = 3286.98

r² = 3286.98/400

r² = 8.21745

r = √8.21745

r = 2.87m

Hence the distance of separation is 2.87m

7 0
3 years ago
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