Answer:
egrfeirugherhgourehgabgwehgoehborghrewuhgelkg
Explanation:
The Doppler Effect provides the equation for the
calculation of apparent frequency:
f=fo[vo/(vo-vr)]
where:<span>
vo=source wave velocity
vr=relative speed between source and observer
f=apparent frequency
fo=source frequency </span>
<span>
The velocity of the doppler wave is
v=λf</span>
where λ is light wavelength. Hence,
v=λfo[vo/(vo-vr)]
Based on the equation, we can say that wave
velocity will always be defined by one and only one wavelength.
Therefore the answer is letter C.
<span> </span>
Let F1=Force exerted by the brother (+F1)
F1= Force exerted by the sister (-F2)
Fnet=(+F1) + (-F2)
Fnet= (+F1) + (-F2)
Fnet=F1 - F2
Fnet= (+3N)+(-5N)
Fnet= -2N
-F
towards the sister (-F) (greater force applied)
Answer:
(a) 1.414 km
(b) 1.06 m/s
Explanation:
(a) For John:
Distance = 1 km north and then 1 km east
Speed = 1.5 m/s
total distance traveled = 1 + 1 = 2 km = 2000 m
Time taken to travel = Distance / speed
t = 2000 / 1.5 = 1333.3 seconds
Displacement = 
(b) For jane :
Time is same as john = 1333.33 second
Distance = 1.414 km = 1414 m
Speed = distance / time = 1414 / 1333.33 = 1.06 m/s
Answer:
The shortest braking distance is 35.8 m
Explanation:
To solve this problem we must use Newton's second law applied to the boxes, on the vertical axis we have the norm up and the weight vertically down
On the horizontal axis we fear the force of friction (fr) that opposes the movement and acceleration of the train, write the equation for each axis
Y axis
N- W = 0
N = W = mg
X axis
-Fr = m a
-μ N = m a
-μ mg = ma
a = μ g
a = - 0.32 9.8
a = - 3.14 m/s²
We calculate the distance using the kinematics equations
Vf² = Vo² + 2 a x
x = (Vf² - Vo²) / 2 a
When the train stops the speed is zero (Vf = 0)
Vo = 54 km/h (1000m/1km) (1 h/3600s)= 15 m/s
x = ( 0 - 15²) / 2 (-3.14)
x= 35.8 m
The shortest braking distance is 35.8 m
