Answer:
Vf = 23 m/s
Explanation:
First we need to find the distance covered by the motorcycle 2 when it passes motorcycle 1. Using the uniform speed equation for motorcycle 1:
s₁ = v₁t₁
where,
s₁ = distance covered by motorcycle 1 = ?
v₁ = speed of motorcycle 1 = 6.5 m/s
t₁ = time = 10 s
Therefore,
s₁ = (6.5 m/s)(10 s)
s₁ = 65 m
Now, for the distance covered by motorcycle 2 at the meeting point. Since, the motorcycle started 50 m ahead of motorcycle 2. Therefore,
s₂ = s₁ + 50 m
s₂ = 65 m + 50 m
s₂ = 115 m
Now, using second equation of motion for motorcycle 2:
s₂ = Vi t + (1/2)at²
where,
Vi = initial velocity of motorcycle 2 = 0 m/s
Therefore,
115 m = (0 m/s)(10 s) + (1/2)(a)(10 s)²
a = 230 m/100 s²
a = 2.3 m/s²
Now, using 1st equation of motion:
Vf = Vi + at
Vf = 0 m/s + (2.3 m/s²)(10 s)
Vf = 23 m/s
Answer:
Explanation:
Capacitance of capacitor
= ε₀ A / d , ε₀ is permittivity of space , A is area of plate , d is distance between plates.
= 8.85 x 10⁻¹² x 2 x 10⁻³ / (6 x 10⁻³)
= 2.95 x 10⁻¹² F
Electric field E = V / d , V is potential difference
V = E x d
= 100 x 10³ x 6 x 10⁻³
= 600 V
Charge on the capacitor
= capacitance x potential difference
= 2.95 x 10⁻¹² x 600
= 17.7 x 10¹⁰ C
Answer:
The focal length of eye piece is 6.52 cm.
Explanation:
Given that,
Angular Magnification of the microscope M = -46
the distance between the lens in microscope L= 16 cm
The focal length of objective f₀ = 1.5 cm
Normal near point N = 25 cm
Have to find focal length of eye piece f ₙ =?
The angular magnification is given by
M ≈ - (L-fₙ)N/f₀fₙ
Rearranging for fₙ
fₙ =L(1 - Mf₀/N)⁺¹
=18/2.76
fₙ = 6.52 cm
The focal length of eye piece is 6.52 cm.
Answer:
Gamma rays have the highest energies, the shortest wavelengths, and the highest frequencies. Radio waves, on the other hand, have the lowest energies, longest wavelengths, and lowest frequencies of any type of EM radiation.
Explanation:
Hope this helps:)
