The first answer should be correct if not then the second one
The launch velocity of the marble launcher is 34.65 m/s
Given that the launch velocity of marble launcher, launches a 25g marble to a distance of 73 cm (0.73 m) and the marble roll up to 6.2 meters before stopping. The launch height is 20 cm (0.2 m).
The time for landing can be calculated by the second equation of motion formula:
h = ut + 
g
Let u = 0
0.2 = 0×t + × 9.8 ×
= 
= 0.04
t = 0.2s
Now, the launch velocity of the marble launcher can be calculated by:
Speed = Distance / Time
Speed = 
Speed = 
Speed = 34.65 m/s
Therefore, the launch velocity of the marble launcher is 34.65 m/s
Know more about Launch velocity: -brainly.com/question/18883779
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We want to find how much momentum the dumbbell has at the moment it strikes the floor. Let's use this kinematics equation:
Vf² = Vi² + 2ad
Vf is the final velocity of the dumbbell, Vi is its initial velocity, a is its acceleration, and d is the height of its fall.
Given values:
Vi = 0m/s (dumbbell starts falling from rest)
a = 10m/s² (we'll treat downward motion as positive, this doesn't affect the result as long as we keep this in mind)
d = 80×10⁻²m
Plug in the values and solve for Vf:
Vf² = 2(10)(80×10⁻²)
Vf = ±4m/s
Reject the negative root.
Vf = 4m/s
The momentum of the dumbbell is given by:
p = mv
p is its momentum, m is its mass, and v is its velocity.
Given values:
m = 10kg
v = 4m/s (from previous calculation)
Plug in the values and solve for p:
p = 10(4)
p = 40kg×m/s
<span>We can use an equation to find the gravitational force exerted on the HST.
F = GMm / r^2
G is the gravitational constant
M is the mass of the Earth
m is the mass of the HST
r is the distance to the center of the Earth
This force F provides the centripetal force for the HST to move in a circle. The equation we use for circular motion is:
F = mv^2 / r
m is the mass of the HST
v is the tangential speed
r is the distance to the center of the Earth
Now we can equate these two equations to find v.
mv^2 / r = GMm / r^2
v^2 = GM / r
v = sqrt{GM / r }
v = sqrt{(6.67 x 10^{-11})(5.97 x 10^{24}) / 6,949,000 m}
v = 7570 m/s which is equal to 7.570 km/s
HST's tangential speed is 7570 m/s or 7.570 km/s</span>
Once the car is on top of the hill it contains potential energy witch means it is storing enough energy to slide the hill until acted on. Once the car moves and slides down the hill it creates kinetic energy witch means it's in motion. The car then turns the kinetic energy into mechanical energy witch means it's working. I'm not sure if that helped but good luck!
