We have the following equation for height:
h (t) = (1/2) * (a) * t ^ 2 + vo * t + h0
Where,
a: acceleration
vo: initial speed
h0: initial height.
The value of the acceleration is:
a = -g = -9.8 m / s ^ 2
For t = 0 we have:
h (0) = (1/2) * (a) * 0 ^ 2 + vo * 0 + h0
h (0) = h0
h0 = 0 (reference system equal to zero when the ball is hit).
For t = 5.8 we have:
h (5.8) = (1/2) * (- 9.8) * (5.8) ^ 2 + vo * (5.8) + 0
(1/2) * (- 9.8) * (5.8) ^ 2 + vo * (5.8) + 0 = 0
vo = (1/2) * (9.8) * (5.8)
vo = 28.42
Substituting values we have:
h (t) = (1/2) * (a) * t ^ 2 + vo * t + h0
h (t) = (1/2) * (- 9.8) * t ^ 2 + 28.42 * t + 0
Rewriting:
h (t) = -4.9 * t ^ 2 + 28.42 * t
The maximum height occurs when:
h '(t) = -9.8 * t + 28.42
-9.8 * t + 28.42 = 0
t = 28.42 / 9.8
t = 2.9 seconds.
Answer:
The ball was at maximum elevation when:
t = 2.9 seconds.
Answer:
0.0065 m³
Explanation:
Apparent weight = weight − buoyancy
32 N = 96 N − (1000 kg/m³) (9.8 m/s²) V
V = 0.0065 m³
Answer: 20.2 m/s
Explanation:
From the question above, we have the following data;
M1 = 800kg
M2 = 1200kg
V1 = 13m/s
V2 = 25m/s
U (common velocity) =?
M1V1 + M2V2 = (M1 + M2). U
(800*13) + (1200*25) = (800+1200) * U
10400 + 30000 = 2000u
40400 = 2000u
U = 40400 / 2000
U = 20.2 m/s
Answer:
Explanation:
Given:
- mass of the object on a horizontal surface,
- coefficient of static friction,
- coefficient of kinetic friction,
- horizontal force on the object,
<u>Now the value of limiting frictional force offered by the contact surface tending to have a relative motion under the effect of force:</u>
where:
normal force of reaction acting on the body= weight of the body
As we know that the frictional force acting on the body is always in the opposite direction:
So, the frictional force will not be at its maximum and will be equal in magnitude to the applied external force and hence the body will not move.
so, the frictional force will be:
Answer:
C
Explanation:
gravity is a pulling force according to Newton
