Answer:
The final velocity of the runner at the end of the given time is 2.7 m/s.
Explanation:
Given;
initial velocity of the runner, u = 1.1 m/s
constant acceleration, a = 0.8 m/s²
time of motion, t = 2.0 s
The velocity of the runner at the end of the given time is calculate as;

where;
v is the final velocity of the runner at the end of the given time;
v = 1.1 + (0.8)(2)
v = 2.7 m/s
Therefore, the final velocity of the runner at the end of the given time is 2.7 m/s.
Hello!
Using Hooke's law, F spring=k delta x, find the distance a spring with an elastic constant of 4 N/cm will stretch if a 2 newton force is applied to it.
Data:
Hooke represented mathematically his theory with the equation:
F = K * Δx
On what:
F (elastic force) = 2 N
K (elastic constant) = 4 N/cm
Δx (deformation or elongation of the elastic medium or distance from a spring) = ?
Solving:




simplify by 2


Answer:
B.) 1/2 cm
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I Hope this helps, greetings ... Dexteright02! =)
Answer:
14.0 cm
Explanation:
Draw a free body diagram of the block. There are three forces: weight force mg pulling down, elastic force k∆L pulling down, and buoyancy ρVg pushing up.
Sum of forces in the y direction:
∑F = ma
ρVg − mg − k∆L = 0
(1000 kg/m³) (4.63 kg / 648 kg/m³) (9.8 m/s²) − (4.63 kg) (9.8 m/s²) − (176 N/m) ∆L = 0
∆L = 0.140 m
∆L = 14.0 cm
If the car is going at constant speed than the velocity isnt changing, only the direction of velocity.
<span>The correct answer is: Oxygen
Explanation:
In order to function properly (movement etc.) during exercise, muscles require oxygen. During exercise, the depth as well as the rate of breathing increase, which in turn increases the amount of oxygen inhaled. In order to expand and contract lungs and for other bodily movements, muscles require oxygen, and for that, more oxygen is carried in the blood to muscles. Hence, the correct answer is Oxygen.</span>