Answer:
answer should be C.) Independent variable since its the one being changed
Explanation:
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 =
=
=
=
.
∴ 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:
27.44 J
Explanation:
We can find the energy at the top of the slide by using the potential energy equation:
At the top of the slide, the swimmer has 0 kinetic energy and maximum potential energy.
The swimmer's mass is given as 7.00 kg.
The acceleration due to gravity is 9.8 m/s².
The (vertical) height of the water slide is 0.40 m.
Substitute these values into the potential energy equation:
- PE = (7.00)(9.8)(0.40)
- PE = 27.44
Since there is 0 kinetic energy at the top of the slide, the total energy present is the swimmer's potential energy.
Therefore, the answer is 27.44 J of energy when the swimmer is at the top of the slide.
C., used in power plants I think.
Answer:
0.17 s
Explanation:
Given:
Δy = 0.15 m
v₀ = 0 m/s
a = 9.8 m/s²
Find: t
Δy = v₀ t + ½ at²
0.15 m = (0 m/s) t + ½ (9.8 m/s²) t²
t = 0.17 s