Answer:
v = 1.28 m/s
Explanation:
Given that,
Maximum compression of the spring, 
Spring constant, k = 800 N/m
Mass of the block, m = 0.2 kg
To find,
The velocity of the block when it first reaches a height of 0.1 m above the ground on the ramp.
Solution,
When the block is bounced back up the ramp, the total energy of the system remains conserved. Let v is the velocity of the block such that,
Initial energy = Final energy
Substituting all the values in above equation,
v = 1.28 m/s
Therefore the velocity of block when it first reaches a height of 0.1 m above the ground on the ramp is 1.28 m/s.
The maximum speed of the donkey is 10.72m/s
The question is based on the principle of motion in one dimension and hence formulas of motion in one dimension can be applied.
It is given that donkey attains an acceleration of 1.6 m/s^2
The time taken to accelerate to given speed is 6.7 seconds
We use the formula v=u + at to find the fastest speed
v is the final or maximum speed
u is the initial speed which in this case is 0 as the donkey is at rest
a is the acceleration of the donkey
t is the time taken in seconds
v = u + at
v= 0 + 1.6 x 6.7
= 10.72 m/s
Hence the donkey obtains the speed of 10.72 m/s
For further reference:
brainly.com/question/24478168?referrer=searchResults
#SPJ9
The answer is <span>C. 49 m/s
The kinetic equation is:
v2 = v1 + a * t
v1 - initial velocity
v2 - final velocity
a - gravitational acceleration
t - time
We know:
v2 = ?
v1 = 0 (in free fall
a = 9.8 m/s
t = 5
</span>v2 = v1 + a * t
v2 = 0 + 9.8 * 5
v2 = 0 + 49
v2 = 49 m/s
Answer:
The new height the ball will reach = (1/4) of the initial height it reached.
Explanation:
The energy stored in any spring material is given as (1/2)kx²
This energy is converted to potential energy, mgH, of the ball at its maximum height.
If the initial height reached is H
And the initial compression of the spring = x
So, mgH = (1/2)kx²
H = kx²/2mg
The new compression, x₁ = x/2
New energy of loaded spring = (1/2)kx₁²
And the new potential energy = mgH₁
mgH₁ = (1/2)kx₁²
But x₁ = x/2
mgH₁ = (1/2)k(x/2)² = kx²/8
H₁ = kx²/8mg = H/4 (provided all the other parameters stay constant)
