Compute first for the vertical motion, the formula is:
y = gt²/2
0.810 m = (9.81 m/s²)(t)²/2
t = 0.4064 s
whereas the horizontal motion is computed by:
x = (vx)t
4.65 m = (vx)(0.4064 s)
4.65 m/ 0.4064s = (vx)
(vx) = 11.44 m / s
So look for the final vertical speed.
(vy) = gt
(vy) = (9.81 m/s²)(0.4064 s)
(vy) = 3.99 m/s
speed with which it hit the ground:
v = sqrt[(vx)² + (vy)²]
v = sqrt[(11.44 m/s)² + (3.99 m/s)²]
v = 12.12 m / s
To develop this problem it will be necessary to apply the concepts related to the frequency of a spring mass system, for which it is necessary that its mathematical function is described as

Here,
k = Spring constant
m = Mass
Our values are given as,


Rearranging to find the spring constant we have that,




Therefore the spring constant is 1.38N/m
Answer:
noncompliance
Explanation:
If a scientist unknowingly breaks the law, he is guilty of <u>noncompliance</u>
The centripetal acceleration is 
Explanation:
For an object in uniform circular motion, the centripetal acceleration is given by

where
v is the speed of the object
r is the radius of the circle
The speed of the object is equal to the ratio between the length of the circumference (
) and the period of revolution (T), so it can be rewritten as

Therefore we can rewrite the acceleration as

For the particle in this problem,
r = 2.06 cm = 0.0206 m
While it makes 4 revolutions each second, so the period is

Substituting into the equation, we find the acceleration:

Learn more about centripetal acceleration:
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Answer:
Current: 1.0 Amperes
The minimum current is flowing through path D
Explanation:
We first find the equivalent resistance to the three resistors in parallel ( which is the total resistance of the circuit) via the equation:

with this info, we can estimate the current going through branch A using Ohm's Law, and the information that the power source is 6 V:

where the current comes in units of Amperes since all other the quantities are given in the SI system, and we can round this answer to 1.0 Amp following the request to round it to the tenth.
The current will be the lowest through the branch with the largest resistor due to the fact that less current will flow through the path of more resistance.
Than means that the lowest current will be registered through branch D where the 50
resistor is.