Answer: 7.38 km
Explanation: The attachment shows the illustration diagram for the question.
The range of the bomb's motion as obtained from the equations of motion,
H = u(y) t + 0.5g(t^2)
U(y) = initial vertical component of velocity = 0 m/s
That means t = √(2H/g)
The horizontal distance covered, R,
R = u(x) t = u(x) √(2H/g)
Where u(x) = the initial horizontal component of the bomb's velocity = 287 m/s, H = vertical height at which the bomb was thrown = 3.24 km = 3240 m, g = acceleration due to gravity = 9.8 m/s2
R = 287 √(2×3240/9.8) = 7380 m = 7.38 km
A. through a relatively short distance.
The speed is actually called the drift speed of the electron.
Answer:
2.5m/s2
Explanation:
The following were obtained from the question:
F = 600N
M = 240 kg
a =?
Recall: F = Ma
a = F/M
a = 600/240
a = 2.5m/s2
Therefore, the acceleration of the motorcycle is 2.5m/s2
Answer:
5. dispersion
6. 49.8°
Explanation:
5. Dispersion is the name given to the phenomenon of light of different wavelengths being bent differently. A rainbow is the result of light from a point source (the sun) being spread out by wavelength (color), a nice example of dispersion.
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6. n = 1.31 is the ratio of the sine of the angle of refraction to the sine of the angle of incidence (for light passing to a medium of n = 1). When the angle of refraction is 90°, the angle of incidence is the "critical angle." So, ...
sin(90°)/sin(critical) = 1.31
critical angle = arcsin(1/1.31) ≈ 49.8°
Answer: Use Roman Numerals in answering. 11. - 27310189. ... Hiwalay! Hatol. Unang puntos, Bughaw! 12. Hinto! Pula, pangalawang laglag. Panalo ...
Explanation:. Hinto! Hiwalay! Bughaw, 1 puntos. Unang paglabag! 14. Hinto! Bughaw, pangalawang paglabag
