The answer is C because when the chemicals on the tip of a match are heated up from friction it ignites.
The frequency of the light observed from the Earth is 
Explanation:
First of all, we start by noticing that the galaxy is receding from Earth (moving away): this means that according to the Doppler effect, the frequency of the light as seen from the Earth must be shorter than the real frequency of the light emitted by the galaxy.
Furthermore, we can quantify the change in frequency of the light using the following equation:
where
is the change in frequency
f is the real frequency
v is the velocity of recession of the galaxy (negative if the galaxy is moving away)
c is the speed of light
In this problem, we have:
Substituting and solving for , we find
And therefore, the frequency of the light observed from the Earth is
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Answer:
Thrust developed = 212.3373 kN
Explanation:
Assuming the ship is stationary
<u>Determine the Thrust developed</u>
power supplied to the propeller ( Punit ) = 1900 KW
Duct distance ( diameter ; D ) = 2.6 m
first step : <em>calculate the area of the duct </em>
A = π/4 * D^2
= π/4 * ( 2.6)^2 = 5.3092 m^2
<em>next : calculate the velocity of propeller</em>
Punit = (A*v*β ) / 2 * V^2 ( assuming β = 999 kg/m^3 ) also given V1 = 0
∴V^3 = Punit * 2 / A*β
= ( 1900 * 10^3 * 2 ) / ( 5.3092 * 999 )
hence V2 = 8.9480 m/s
<em>Finally determine the thrust developed </em>
F = Punit / V2
= (1900 * 10^3) / ( 8.9480)
= 212.3373 kN
Answer:
a)
reaction time = 0.70 s
distance travelled in reaction time = v*t
= 20 m/s * 0.70 s
= 14 m
So, when brake is applied, distance remaining= 110 m - 14 m = 96 m
Answer: 96 m
b)
vf = 0 m/s
d = 96 m
vi = 20 m/s
use:
vf^2 = vi^2 + 2*a*d
0 = 20^2 + 2*a*96
-400 = 2*a*96
a = -2.08 m/s^2
Answer: -2.08 m/s^2
c)
use:
vf = vi + a*t
0 = 20 - 2.08*t
t = 9.6 s
Answer: 9.6 s
Explanation:
Answer:
As sound waves move (or more accurately, when they travel by transferring their energy) they interact with physical objects. Soft surfaces will absorb sound while hard surfaces will reflect it. .Hard surfaces reflect sound and soft surfaces absorb sound.
The convex mirror has a reflecting surface that curves outward, resembling a portion of the exterior of a sphere. Light rays parallel to the optical axis are reflected from the surface in a direction that diverges from the focal point, which is behind the mirror
