The banking angle of the curved part of the speedway is determined as 32⁰.
<h3>
Banking angle of the curved road</h3>
The banking angle of the curved part of the speedway is calculated as follows;
V(max) = √(rg tanθ)
where;
- r is radius of the path
- g is acceleration due to gravity
V² = rg tanθ
tanθ = V²/rg
tanθ = (34²)/(190 x 9.8)
tanθ = 0.62
θ = arc tan(0.62)
θ = 31.8
θ ≈ 32⁰
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(6) first choice: the frequency appears higher and wavelength is shorter.
The car approaches a stationary observer and so the sound will appear to have shorter wavelength. This creates an effect of its siren to sound with higher frequency than it would do if both were stationary.
(7) The Doppler formula for frequency in the case of a stationary observer and source approaching it is as follows:

The wavelength is then

The third choice "0.80m; 431Hz" is correct
Answer:
Approximately
to the right (assuming that both astronauts were originally stationary.)
Explanation:
If an object of mass
is moving at a velocity of
, the momentum
of that object would be
.
Since momentum of this system (of the astronauts) conserved:
.
Assuming that both astronauts were originally stationary. The total initial momentum of the two astronauts would be
since the velocity of both astronauts was
.
Therefore:
.
The final momentum of the first astronaut (
,
to the left) would be
to the left.
Let
denote the momentum of the astronaut in question. The total final momentum of the two astronauts, combined, would be
.
.
Hence,
. In other words, the final momentum of the astronaut in question is the opposite of that of the first astronaut. Since momentum is a vector quantity, the momentum of the two astronauts magnitude (
) but opposite in direction (to the right versus to the left.)
Rearrange the equation
to obtain an expression for velocity in terms of momentum and mass:
.
.
Hence, the velocity of the astronaut in question (
) would be
to the right.
M = 40 Kg , g=9.8 m/s² , h = 2 m
PE = m g h
PE = (40) (9.8) (2)
PE = 784 J
KE = PE
½m v² = m g h
½ v² = g h
½ v² = (9.8) (2)
½ v² = 19.6
v² = 19.6×2
v² = 39.2
V = √39.2
V = 6.26 m/s
KE = ½mv²
KE = ½(40) (6.26)²
KE =783.8 J
Answer:
The correct answer is: B area = area
Explanation: You can solve this problem through the use of laws of Kepler's planetary motion. There are Kepler the laws ( 3 of them) In this exercise, we will use the second law. According to this law, a line segment joining a planet and the sun sweeps out areas is equal to that during the equal intervals of time