Answer:
t = √2y/g
Explanation:
This is a projectile launch exercise
a) The vertical velocity in the initial instants (
= 0) zero, so let's use the equation
y =
t -1/2 g t²
y= - ½ g t²
t = √2y/g
b) Let's use this time and the horizontal displacement equation, because the constant horizontal velocity
x = vox t
x = v₀ₓ √2y/g
c) Speeds before touching the ground
vₓ = vox = constant
=
- gt
= 0 - g √2y/g
= - √2gy
tan θ = Vy / vx
θ = tan⁻¹ (vy / vx)
θ = tan⁻¹ (√2gy / vox)
d) The projectile is higher than the cliff because it is a horizontal launch
The correct answer is: Option (D) length, speed
Explanation:
According to Faraday's Law of Induction:
ξ = Blv
Where,
ξ = Emf Induced
B = Magnetic Induction
l = Length of the conductor
v = Speed of the conductor.
As you can see that ξ (Emf/voltage induction) is directly proportional to the length and the speed of the conductor. Therefore, the correct answer will be Option (D) Length, Speed
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
True : <span>There are numerous third-class </span>levers<span> in the human </span>body<span>; one example can be illustrated in the elbow joint</span>