To solve the answer use the equation: a = fnet / m
a = 300 N / 25 kg
300 N / 25 kg = 12m/s
The acceleration of the object is 12m/s
Answer
Given,
Time period of star,T = 3.37 x 10⁷ s
Radius of circular orbit,R = 1.04 x 10¹¹ m
a) Angular speed of the planet
b) tangential speed
v = 1.94 x 10⁴ m/s
c) centripetal acceleration magnitude
a = 3.62 x 10⁻³ m/s²
<span>c. low frequency and low energy</span>
Oooooo, that statement is not true. Countries create quotas and tariffs to <em>LIMIT</em> the volume of trade with other countries, including their neighbors.
Answer:
λ = 623.2 nm
Explanation:
We are given;
separation distance; d = 0.195 mm = 0.195 × 10^(-3) m
interference pattern distance; D = 4.85 m
Width of two adjacent bright interference; w = 1.55 cm = 1.55 × 10^(-2) m
Formula for fringe width is given as;
w = λD/d
Where λ is wavelength
Thus;
λ = dw/D
λ = (0.195 × 10^(-3) × 1.55 × 10^(-2))/4.85
λ = 0.0000006232 m
Converting to nm gives;
λ = 623.2 nm
