The correct answer is C.) a leaf
Answer:
x = 6.94 m
Explanation:
For this exercise we can find the speed at the bottom of the ramp using energy conservation
Starting point. Higher
Em₀ = K + U = ½ m v₀² + m g h
Final point. Lower
= K = ½ m v²
Em₀ = Em_{f}
½ m v₀² + m g h = ½ m v²
v² = v₀² + 2 g h
Let's calculate
v = √(1.23² + 2 9.8 1.69)
v = 5.89 m / s
In the horizontal part we can use the relationship between work and the variation of kinetic energy
W = ΔK
-fr x = 0- ½ m v²
Newton's second law
N- W = 0
The equation for the friction is
fr = μ N
fr = μ m g
We replace
μ m g x = ½ m v²
x = v² / 2μ g
Let's calculate
x = 5.89² / (2 0.255 9.8)
x = 6.94 m
Answers:
a)The balloon is 68 m away of the radar station
b) The direction of the balloon is towards the radar station
Explanation:
We can solve this problem with the Doppler shift equation:
(1)
Where:
is the actual frequency of the sound wave
is the "observed" frequency
is the velocity of sound
is the velocity of the observer, which is stationary
is the velocity of the source, which is the balloon
Isolating
:
(2)
(3)
(4) This is the velocity of the balloon, note the negative sign indicates the direction of motion of the balloon: It is moving towards the radar station.
Now that we have the velocity of the balloon (hence its speed, the positive value) and the time (
) given as data, we can find the distance:
(5)
(6)
Finally:
(8) This is the distance of the balloon from the radar station
Answer:
the banking angle of the road is 24.2⁰
Explanation:
Given;
speed of the vehicles considered, v = 75 mi/h
Speed in m/s ⇒ 1 mi/h --------> 0.44704 m/s
75 mi/h --------> ?
= 75 x 0.44704 m/s = 33.528 m/s
radius of the curve, r = 255 m
The banking angle of the road is calculated as;

Therefore, the banking angle of the road is 24.2⁰