Answer:
The minimum distance between two points on the object that are barely resolved is 0.26 mm
The corresponding distance between the image points = 0.0015 m
Explanation:
Given
focal length f = 50 mm and maximum aperture f>2
s = 9.0 m
aperture = 25 mm = 25 *10^-3 m
Sin a = 1.22 *wavelength /D
Substituting the given values, we get –
Sin a = 1.22 *600 *10^-9 m /25 *10^-3 m
Sin a = 2.93 * 10 ^-5 rad
Now
Y/9.0 m = 2.93 * 10 ^-5
Y = 2.64 *10^-4 m = 0.26 mm
Y’/50 *10^-3 = 2.93 * 10 ^-5
Y’ = 0.0015 m
Answer:
B
Explanation:
Speed is the magnitude of the velocity vector, so it can never be negative.
Answer:
1. 8437500 N
2. The force between the two charges is attractive.
Explanation:
1. Determination of the force between the two charges.
Charge 1 (q₁) = –2.0 C
Charge 2 (q₂) = 3.0 C
Distance apart (r) = 80 m
Electrical constant (K) = 9×10⁹ Nm²/C²
Force (F) =?
F = Kq₁q₂ / r²
F = 9×10⁹ × 2 × 3 / 80²
F = 5.4×10¹⁰ / 6400
F = 8437500 N
Thus, the force of attraction between the two charges is 8437500 N
2. From the question given, the charges are:
Charge 1 (q₁) = –2.0 C
Charge 2 (q₂) = 3.0 C
We understood that like charges repels while unlike charges attract. Since the two charges (i.e –2 C and 3 C) has opposite signs, it means they will attract each other.
Thus the force between them is attractive.
F = m*a
30 N = (ma + mb) * a
30 = 5*a
a = 6 m/s ^2
F de B em A
30 - F de B,A = ma * a
30 - F de B em A = 3 * 6
30 - 18 = F de B em A
12 = F de B em A
Resposta: 6 m/s^2 e 12N
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Answer:
Explanation:
if
and g=9.81 m/s2=32.16 ft/s2
and
W=m*g
we can just replace de mass and gravity and we have
