Ans: D
In a projectile motion, horizontal velocity is constant, while the ball is travelling at a constant vertical acceleration due to the gravity (9.81 ms-2 on Earth).
Answer:
Position at t= 4 seconds is 144 m
Explanation:
It is given that acceleration, a = 18 t, where t is the time.
We know that Velocity, 
Substituting value of a,
Velocity, 
We know that at t = 0, v = -12 m/s
So, 
So velocity, 
We also know that displacement, 
Substituting value of v,
Displacement, 
We know that at t = 0, particle is at origin, x =0.
So, 
Displacement, 
At t = 4 seconds

Position at t= 4 seconds is 144 m
Answer:
r₂ = 0.316 m
Explanation:
The sound level is expressed in decibels, therefore let's find the intensity for the new location
β = 10 log
let's write this expression for our case
β₁ = 10 log \frac{I_1}{I_o}
β₂ = 10 log \frac{I_2}{I_o}
β₂ -β₁ = 10 (
)
β₂ - β₁ = 10
log \frac{I_2}{I_1} =
= 3
= 10³
I₂ = 10³ I₁
having the relationship between the intensities, we can use the definition of intensity which is the power per unit area
I = P / A
P = I A
the area is of a sphere
A = 4π r²
the power of the sound does not change, so we can write it for the two points
P = I₁ A₁ = I₂ A₂
I₁ r₁² = I₂ r₂²
we substitute the ratio of intensities
I₁ r₁² = (10³ I₁ ) r₂²
r₁² = 10³ r₂²
r₂ = r₁ / √10³
we calculate
r₂ =
r₂ = 0.316 m
Answer:
toward the normal
Explanation:
Light travels at different speed in different mediums.
Refractive index is equal to velocity of the light 'c' in empty space divided by the velocity 'v' in the substance.
Or ,
n = c/v.
Light travels at a slower speed in water as compared to air because there are more number of interfering molecules in the path of the light in case of water as compared to liquid.
When a light travels from lower denser medium say water to higher denser medium say water, it bends towards the perpendicular (normal) as its speed reduces in that medium.