Answer: A projectile which is fired horizontally is being constantly acted upon by acceleration due to gravity, acting vertically downwards. Hence, it does not follow a straight line path. Also Why a projectile fixed along the horizontal not follow a straight line path? Because the projectile fired horizontally is constantly acts upon by acceleration due to gravity acting vertically downwards.
Explanation:
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Answer:
The correct option is;
Still constant
Explanation:
The relative refractive index ₁n₂ between the two medium can be as follows;
Therefore, given that the speed of light in medium 1 is constant and the speed of light on medium 2 is also constant, the relative refractive index ₁n₂ = c₁/c₂ is always constant.
Answer:
Explanation:
The rod will act as pendulum for small oscillation .
Time period of oscillation
angular frequency ω = 2π / T
= 
b )
ω = 20( given )
velocity = ω r = ω l
Let the maximum angular displacement in terms of degree be θ .
1/2 m v ² = mgl ( 1 - cosθ ) ,
[ l-lcosθ is loss of height . we have applied law of conservation of mechanical energy .]
.5 ( ω l )² = gl( 1 - cos θ )
.5 ω² l = g ( 1 - cosθ )
1 - cosθ = .5 ω² l /g
cosθ = 1 - .5 ω² l /g
θ can be calculated , if value of l is given .
Answer:
The acceleration is 1 cm/s^2.
Explanation:
The acceleration is defined as the rate of change of velocity.
Here, initial velocity, u = 3/1 = 3 cm/s
final velocity, v = 4/1 = 4 cm/s
time, t = 1 s
Let the acceleration is a.
Use first equation of motion
v = u + at
4 = 3 + 1 x a
a = 1 cm/s^2
<u>Explanation</u>
- The relationship between the strength of a bond (single vs double vs triple) and its wave-number on an IR spectrum as the bond strength increases the wave number increases.
STRENGTH OF BONDS TRIPLE>DOUBLE>SINGLE
WAVE NUMBER SINGLE>DOUBLE>TRIPLE
- wave number for single bond is greatest because it has greatest bond frequency among the three( more the frequency greater is the wave number).
