Answer:
<em><u>solution</u></em>
<em>3</em><em>0</em><em>8</em><em>=</em><em>2</em><em>0</em><em> </em><em>swings</em><em> </em>
<em> </em><em>?</em><em>:</em><em>:</em><em>:</em><em> </em><em>=</em><em>1</em>
<em>(</em><em> </em><em>3</em><em>0</em><em>8</em><em>×</em><em>1</em><em>)</em><em>÷</em><em>2</em><em>0</em>
<em>3</em><em>0</em><em>8</em><em>÷</em><em>2</em><em>0</em>
<em>1</em><em>5</em><em>4</em><em>÷</em><em>1</em><em>0</em>
<em>=</em><em>1</em><em>5</em><em>.</em><em>4</em>
<em>=</em>15.4
Answer:
translation
Explanation:
If a ball has an approximate mass of 0.5 kg, with what force must be kicked to give it an acceleration of 1.5m / s2?
Answer:
V' = 0.84 m/s
Explanation:
given,
Linear speed of the ball, v = 2.85 m/s
rise of the ball, h = 0.53 m
Linear speed of the ball, v' = ?
rotation kinetic energy of the ball

I of the moment of inertia of the sphere

v = R ω
using conservation of energy


Applying conservation of energy
Initial Linear KE + Initial roational KE = Final Linear KE + Final roational KE + Potential energy



V'² = 0.7025
V' = 0.84 m/s
the linear speed of the ball at the top of ramp is equal to 0.84 m/s
Answer:
a. (a) grating A has more lines/mm; (b) the first maximum less than 1 meter away from the center
Explanation:
Let n₁ and n₂ be no of lines per unit length of grating A and B respectively.
λ₁ and λ₂ be wave lengths of green and red respectively , D be distance of screen and d₁ and d₂ be distance between two slits of grating A and B ,
Distance of first maxima for green light
= λ₁ D/ d₁
Distance of first maxima for red light
= λ₂ D/ d₂
Given that
λ₁ D/ d₁ = λ₂ D/ d₂
λ₁ / d₁ = λ₂ / d₂
λ₁ / λ₂ = d₁ / d₂
But
λ₁ < λ₂
d₁ < d₂
Therefore no of lines per unit length of grating A will be more because
no of lines per unit length ∝ 1 / d
If grating B is illuminated with green light first maxima will be at distance
λ₁ D/ d₂
As λ₁ < λ₂
λ₁ D/ d₂ < λ₂ D/ d₂
λ₁ D/ d₂ < 1 m
In this case position of first maxima will be less than 1 meter.
Option a is correct .