The speed of the toy when it hits the ground is 2.97 m/s.
The given parameters;
- mass of the toy, m = 0.1 kg
- the maximum height reached by the, h = 0.45 m
The speed of the toy before it hits the ground will be maximum. Apply the principle of conservation of mechanical energy to determine the maximum speed of the toy.
P.E = K.E

Substitute the given values and solve the speed;

Thus, the speed of the toy when it hits the ground is 2.97 m/s.
Learn more here: brainly.com/question/7562874
Part A.
The forces are the same because the force from the smaller ball it transferring its Energy through the basketball and it's rebounding as Connecticut Energy back up to the smaller ball
If you mean S is the distance then it is true
Velocity = Distance / time
Answer:
lowest frequency = 535.93 Hz
distance between adjacent anti nodes is 4.25 cm
Explanation:
given data
length L = 32 cm = 0.32 m
to find out
frequency and distance between adjacent anti nodes
solution
we consider here speed of sound through air at room temperature 20 degree is approximately v = 343 m/s
so
lowest frequency will be =
..............1
put here value in equation 1
lowest frequency will be =
lowest frequency = 535.93 Hz
and
we have given highest frequency f = 4000Hz
so
wavelength =
..............2
put here value
wavelength =
wavelength = 0.08575 m
so distance =
..............3
distance =
distance = 0.0425 m
so distance between adjacent anti nodes is 4.25 cm
Answer:
v = -v₀ / 2
Explanation:
For this exercise let's use kinematics relations.
Let's use the initial conditions to find the acceleration of the electron
v² = v₀² - 2a y
when the initial velocity is vo it reaches just the negative plate so v = 0
a = v₀² / 2y
now they tell us that the initial velocity is half
v’² = v₀’² - 2 a y’
v₀ ’= v₀ / 2
at the point where turn v = 0
0 = v₀² /4 - 2 a y '
v₀² /4 = 2 (v₀² / 2y) y’
y = 4 y'
y ’= y / 4
We can see that when the velocity is half, advance only ¼ of the distance between the plates, now let's calculate the velocity if it leaves this position with zero velocity.
v² = v₀² -2a y’
v² = 0 - 2 (v₀² / 2y) y / 4
v² = -v₀² / 4
v = -v₀ / 2
We can see that as the system has no friction, the arrival speed is the same as the exit speed, but with the opposite direction.