Answer:
7 meters per second or 7 m/s
Answer:
h> 2R
Explanation:
For this exercise let's use the conservation of energy relations
starting point. Before releasing the ball
Em₀ = U = m g h
Final point. In the highest part of the loop
Em_f = K + U = ½ m v² + ½ I w² + m g (2R)
where R is the radius of the curl, we are considering the ball as a point body.
I = m R²
v = w R
we substitute
Em_f = ½ m v² + ½ m R² (v/R) ² + 2 m g R
em_f = m v² + 2 m g R
Energy is conserved
Emo = Em_f
mgh = m v² + 2m g R
h = v² / g + 2R
The lowest velocity that the ball can have at the top of the loop is v> 0
h> 2R
He produced the first orderly arrangement of known elements, he used patterns to predict undiscovered elements
Answer:
<u><em>a. True</em></u>
Explanation:
<em>Vectors are an important part of the language of science, mathematics, and engineering.</em>
Answer:
The statement is not correct.
Explanation:
To know if the statement is correct, we shall determine the velocity of the car after 3 s. This is illustrated below.
Data obtained from the question include:
Initial velocity (u) = 0 m/s
Acceleration due to gravity (g) = 9.8 m/s²
Time (t) = 3 s
Final velocity (v) =?
v = u + gt
v = 0 + (9.8 × 3)
v = 0 + 29.4
v = 29.4 m/s
Thus, the velocity of the car after 3 s is 29.4 m/s.
Hence, the statement made by the friend is not correct as the car has a falling velocity of 29.4 m/s after 3 s.
