Answer:
230.4 N
Explanation:
From the question given above, the following data were obtained:
Charge (q) of each protons = 1.6×10¯¹⁹ C
Distance apart (r) = 1×10¯¹⁵ m
Force (F) =?
NOTE: Electric constant (K) = 9×10⁹ Nm²/C²
The force exerted can be obtained as follow:
F = Kq₁q₂ / r²
F = 9×10⁹ × (1.6×10¯¹⁹)² / (1×10¯¹⁵)²
F = 9×10⁹ × 2.56×10¯³⁸ / 1×10¯³⁰
F = 2.304×10¯²⁸ / 1×10¯³⁰
F = 230.4 N
Therefore, the force exerted is 230.4 N
Answer:
λ = 1.4 × 10^(-7) m
Explanation:
We are given;
distance of eye piece from the source;D = 1.5 m
distance between the virtual sources;d = 7.5 × 10^(-4) m
To find the wavelength, we will use the formula for fringe width;
X = λD/d
Where X is fringe width, λ is wavelength, while d and D remain as before.
Now, fringe width = eye-piece distance moved transversely/number of fringes
Eye piece distance moved transversely = 1.88 cm = 1.88 × 10^(-2) m
Thus,
Fringe width = (1.88 × 10^(-2))/10 = 1.88 × 10^(-3) m
Thus;
1.88 × 10^(-3) = λ(1.5)/(7.5 × 10^(-4))
λ = [1.88 × 10^(-3) × (7.5 × 10^(-4))]/1.5
λ = 1.4 × 10^(-7) m
Answer:
-2.5m/s²
Explanation:
The acceleration of a body is giving by the rate of change of the body's velocity. It is given by
a = Δv / t ----------------(i)
Where;
a = acceleration (measured in m/s²)
Δv = change in velocity = final velocity - initial velocity (measure in m/s)
t = time taken for the change (measured in seconds(s))
From the question;
i. initial velocity = 5m/s
final velocity = 0 [since the body (ball) comes to rest]
Δv = 0 - 5 = -5m/s
ii. time taken = t = 2s
<em>Substitute these values into equation (i) as follows;</em>
a = (-5m/s) / (2s)
a = -2.5m/s²
Therefore, the acceleration of the ball is -2.5m/s²
NB: The negative sign shows that the ball was actually decelerating.
Answer:
805.48N/m
Explanation:
According to Hookes law
F = Ke
F is the force = mg
F = 2.4×9.8 = 23.52N
e is the extension = 2.92cm = 0.0292m
Force constant K = F/e
K = 23.52/0.0292
K = 805.48N/m
Hence the force constant of the spring is 805.48N/m
