The acceleration of the runner in the given time is 2.06m/s².
Given the data in the question;
Since the runner begins from rest,
- Initial velocity;
- Final velocity;
- Time elapsed;
Acceleration of the runner; 
<h3>Velocity and Acceleration</h3>
Velocity is the speed at which an object moves in a particular direction.
Acceleration is simply the rate of change of the velocity of a particle or object with respect to time. Now, we can see the relationship from the First Equation of Motion
Where v is final velocity, u is initial velocity, a is acceleration and t is time elapsed.
To determine the acceleration of the runner, we substitute our given values into the equation above.
Therefore, the acceleration of the runner in the given time is 2.06m/s².
Learn more about Equations of Motion: brainly.com/question/18486505
The force of the air resistance is 4 N.
The given parameters;
- mass of the flower pot, m = 2 kg
- weight of the flower pot, W = 20 N
Let the air resistance = F
Apply Newton's second law of motion to determine the force of the air resistance acting upward to oppose the motion of the pot falling downwards.
Thus, the force of the air resistance is 4 N.
Learn more here: brainly.com/question/19887955
Explanation:
(a) Given:
Δx = 150 m
v₀ = 27 m/s
v = 54 m/s
Find: a
v² = v₀² + 2aΔx
(54 m/s)² = (27 m/s)² + 2a (150 m)
a = 7.29 m/s²
(b) Given:
Δx = 150 m
v₀ = 0 m/s
a = 7.29 m/s²
Find: t
Δx = v₀ t + ½ at²
150 m = (0 m/s) t + ½ (7.29 m/s²) t²
t = 6.42 s
(c) Given:
v₀ = 0 m/s
v = 27 m/s
a = 7.29 m/s²
Find: t
v = at + v₀
27 m/s = (7.29 m/s²) t + 0 m/s
t = 3.70 s
(d) Given:
v₀ = 0 m/s
v = 27 m/s
a = 7.29 m/s²
Find: Δx
v² = v₀² + 2aΔx
(27 m/s)² = (0 m/s)² + 2 (7.29 m/s²) Δx
Δx = 50 m
Answer:
new temperature of the tire will be 278.76 K
Explanation:
when the temperature increases, the particles will have greater kinetic energy and also the pressure will be increase for the gas particles.
so when the temperature increases, pressure will also increase and vice versa
T is directly proportional to P
T1 = initial temperature= 303 k
P1= Initial pressure = 325000 pa
T2= Final temperature= ?
P2= Final pressure = 299000 pa
mathematically
P1/T1= P2/T2
T2= P2 x T1/P1
T2 = 299000 x 303/ 325000= 278.76 k
