Answer: To determine acceleration ,Micah also needs the Time of the total trip in seconds.
Explanation:
Acceleration can be defined as rate of change of velocity.
for calculating acceleration, initial and final velocity are required in meter per second and the total time of the trip in seconds. Then acceleration is measured in meter per second square.
Thus, Micah knows that a car had a change in velocity of 15 m/s.To determine acceleration ,Micah also needs the <u>Time</u> of the total trip in seconds.
Answer:
The percentage of its mechanical energy does the ball lose with each bounce is 23 %
Explanation:
Given data,
The tennis ball is released from the height, h = 4 m
After the third bounce it reaches height, h' = 183 cm
= 1.83 m
The total mechanical energy of the ball is equal to its maximum P.E
E = mgh
= 4 mg
At height h', the P.E becomes
E' = mgh'
= 1.83 mg
The percentage of change in energy the ball retains to its original energy,
ΔE % = 45 %
The ball retains only the 45% of its original energy after 3 bounces.
Therefore, the energy retains in each bounce is
∛ (0.45) = 0.77
The ball retains only the 77% of its original energy.
The energy lost to the floor is,
E = 100 - 77
= 23 %
Hence, the percentage of its mechanical energy does the ball lose with each bounce is 23 %
Answer:
(A) 7.9 m/s^{2}
(B) 19 m/s
(C) 91 m
Explanation:
initial velocity (U) = 0 mph = 0 m/s
final velocity (V) = 85 mph = 85 x 0.447 = 38 m/s
initial time (ti) = 0 s
final time (t) = 4.8 s
(A) acceleration = 
= 
= 7.9 m/s^{2}
(B) average velocity = 
=
= 19 m/s
(C) distance travelled (S) = ut + 
= (0 x 4.8) + 
= 91 m
Answer:
a) 69.3 m/s
b) 18.84 s
Explanation:
Let the initial velocity = u
The vertical and horizontal components of the velocity is given by uᵧ and uₓ respectively
uᵧ = u sin 40° = 0.6428 u
uₓ = u cos 40° = 0.766 u
We're given that the horizontal distance travelled by the projectile rock (Range) = 1 km = 1000 m
The range of a projectile motion is given as
R = uₓt
where t = total time of flight
1000 = 0.766 ut
ut = 1305.5
The vertical distance travelled by the projectile rocks,
y = uᵧ t - (1/2)gt²
y = - 900 m (900 m below the crater's level)
-900 = 0.6428 ut - 4.9t²
Recall, ut = 1305.5
-900 = 0.6428(1305.5) - 4.9 t²
4.9t² = 839.1754 + 900
4.9t² = 1739.1754
t = 18.84 s
Recall again, ut = 1305.5
u = 1305.5/18.84 = 69.3 m/s
Answer:
θ = 28°
Explanation:
For this exercise We will use the second law and Newton, let's set a System of horizontal and vertical.
X axis
Fₓ = m a
Nₓ = m a
Where the acceleration is centripetal
a = v² / r
The only force that we must decompose is normal, let's use trigonometry
sin θ = Nₓ / N
cos θ = 
/ N
Nₓ = N sin θ
= N cos θ
Let's replace
N sin θ = m v² / r
Y Axis
- W = 0
N cos θ = mg
Let's divide the two equations of Newton's second law
Sin θ / cos θ = v² / g r
tan θ = v² / g r
θ = tan⁻¹ (v² / g r)
We reduce the speed to the SI system
v = 61 km / h (1000 m / 1 km) (1h / 3600 s) = 16.94 m / s
Let's calculate
θ = tan⁻¹ (16.94 2 / (9.8 55.1)
θ = tan⁻¹ (0.5317)
θ = 28°
