Answer:
The required angle is (90-25)° = 65°
Explanation:
The given motion is an example of projectile motion.
Let 'v' be the initial velocity and '∅' be the angle of projection.
Let 't' be the time taken for complete motion.
Let 'g' be the acceleration due to gravity
Taking components of velocity in horizontal(x) and vertical(y) direction.
= v cos(∅)
= v sin(∅)
We know that for a projectile motion,
t =
Since there is no force acting on the golf ball in horizonal direction.
Total distance(d) covered in horizontal direction is -
d = ×t = vcos(∅)× = 
.
If the golf ball has to travel the same distance 'd' for same initital velocity v = 23m/s , then the above equation should have 2 solutions of initial angle 'α' and 'β' such that -
α +β = 90° as-
d = 
= ![\frac{v^{2}sin(2[90-β]) }{g}](https://tex.z-dn.net/?f=%5Cfrac%7Bv%5E%7B2%7Dsin%282%5B90-%CE%B2%5D%29%20%7D%7Bg%7D)
=
= 
.
∴ For the initial angles 'α' or 'β' , total horizontal distance 'd' travelled remains the same.
∴ If α = 25° , then
β = 90-25 = 65°
∴ The required angle is 65°.
Answer:
Explanation:
Let the extension in the spring be x .
restoring force = weight of block
kx = mg
x = 
= 23.84 cm
b )
When the elevator is going upwards
Restoring force = mg + ma
k x₁ = 10.9 ( 9.8 + 1.89 )
x₁ = 28.44 cm
( y coordinate will be - ( 28.44 - 23.84 ) = - 4.6 cm )
c ) When the cable snaps , both elevator and block undergo free fall . In this case apparent g = 0
Since the spring is stretched by 28.44 cm , a restoring force continues to act on the block which is equal to
.2844 x 448
= 127.41 N
So a net acceleration a will act on the block
a = 127.41 / 10.9
= 11.68 m / s²
The block will undergo SHM with amplitude equal to 28.44 cm .
Answer:
The angular momentum of the particle is 58.14 kg m²/s along positive z- axis and is independent of time .
Explanation:
Given that,
Mass = 1.70 kg
Position vector 
We need to calculate the angular velocity
The velocity is the rate of change of the position of the particle.
We need to calculate the angular momentum of the particle
Using formula of angular momentum
Where, p = mv
Put the value of p into the formula
Substitute the value into the formula
Hence, The angular momentum of the particle is 58.14 kg m²/s along positive z- axis and is independent of time .
They are unproven but accepted as fact.
Many experiments support them but they can be disproven by the results of a single experiment. Until then, they stand.
The third statement is correct.
Answer b like the last time should be it
