HELP! A 25.0-gram bullet enters a 2.35-kg watermelon and embeds itself in the melon. The melon is...?
immediately set into motion with a speed of 3.82 m/s. The bullet remains lodged inside the melon. What was the entry speed of the bullet?:
a. 4.18 m/s
b. 7.41 m/s
c. 359 m/s
d. 363 m/s
A plane is traveling at a speed of 250 miles/hr at an altitude of 600 feet. How far before the target must the plane drop an object to insure it hits the target?: *
a.6.12 miles
b.1.530 miles
c.0.425 miles
d.8.35 miles
Please respond to the question above.
14. A ray of light incident upon a mirror makes an angle of 36? with the mirror (54o to the normal). What is the angle between the incident ray and the reflected ray?: *
a. 36°
b. 72°
c. 108°
d. 144°
Two forces act on an object. A 36 N force acts at 225°. A 48 N force acts at 315°. Find their equilibrant force.: *
a. 60 N due west
b. 60 N due east
c. 60 N at 278°
d. 60 N at 99°
Direct answers:
(1) The bullet's initial velocity = 363 m/s
(2) Range from drop point = 2,242 ft
(3) Angle between rays = 108°
(4) No idea how to do this one. Sorry.
Math and reasoning below...
(1) Bullet-watermelon is an inelastic collision, putting things in SI units
m1= 0.025 kg
m2 = 2.35 kg
v2 = 3.82 m/s
v1 * { m1 / [m1 + m2 ] } = v2
v1 * { (0.025 kg) / [ (0.025 kg) + (2.35 kg) ] } = 3.82 m/s
v1 * { (0.025 kg) / [ 2.375 kg ] } = 3.82 m/s
v1 * { 0.01053 } = 3.82 m/s
v1 = 363 m/s
(2) Aircraft drop to target
Vo = 250 mi/h (367 ft/s)
H = 600 ft
g = 32.15 ft/s^2
R = Vo * SQRT { [2H] / g }
R = (367 ft/s) * SQRT { [ 2 * (600 ft) ] / (32.15 ft/s^2) }
R = (367 ft/s) * SQRT { [ 1,200 ft ] / (32.15 ft/s^2) }
R = (367 ft/s) * SQRT { 37.3 s^2 }
R = (367 ft/s) * (6.12 s)
R = 2,242 ft
(3) Ray reflection
The reflected ray will have the exact same geometry, but in the opposite direction. So the first ray is 54° versus normal, so is the second.
Total angle between rays = (54°) * 2 = 108°
(4) Sorry, I don't know how to do these ones.
STDS9S905E15 By Inferno's Light Star 35
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