Answer:
The galaxy is moving away from the observer
Explanation: when a galaxy is moving away from us, the light we percieve from it is "streched". Since the wavelength has an inverse raltionship whith frequency, the longer the wavelength is, the lower the frequency. And lower frequencies correspond to red and infrarred light.
So when we see the light has shifted to the infrarred part of the spectrum, it means the source is traveling away from us, making the light waves we percieve streched and move from visible light to infrarred.
Answer:
88.8 m/s= Speed of wave propagation in the required mode.(3 loops)
Explanation:
When there are 3 loops.
the total length = L = 3 λ /2
⇒ λ = 2 L / 3 = 2 ( 1.11 ) / 3 = 0.74 m
Velocity = v = f λ = (120)(0.74) = 88.8 m/s
Answer:
Tt = 70 + 135e^-0.031t
13 minutes
Explanation:
Given that :
Initial temperature, Ti = 205°
Temperature after 2.5 minutes = 195°
Temperature of room, Ts= 70
Using the relation :
Tt = Ts + Ce^-kt
Temperature after time, t
When freshly poured, t = 0
205 = 70 + Ce^-0k
205 = 70 + C
C = 205 - 70 = 135°
T after 2.5 minutes to find proportionality constant, k
Tt = Ts + Ce^-kt
195 = 70 + 135e^-2.5k
125 = 135e^-2.5k
125 / 135 = e^-2.5k
0.9259 = e^-2.5k
Take In of both sides :
−0.076989 = - 2.5k
k = −0.076989 / - 2.5
k = 0.031
Equation becomes :
Tt = 70 + 135e^-0.031t
t when Tt = 160
160 = 70 + 135e^-0.031k
90 = 135e^-0.031t
90/135 = e^-0.031t
0.6667 = e^-0.031t
In(0.6667) = - 0.031t
−0.405465 = - 0.031t
t = 0.405465/ 0.031
t = 13.071
t = 13 minutes
The frog's launch speed and the time spends in the air are 22.5m/s and 2.73s respectively.
To find the answer, we need to know about the time of flight and range of projectile motion.
<h3>What's the expression of range of a projectile motion?</h3>
- Range = U²× sin(2θ)/g
- U= initial velocity, θ= angle of projectile and g= acceleration due to gravity
- U=√{Range×g/sin(2θ)}
- Here, range= 2.20m, = 36.5°
- U= √{2.20×9.8/sin(73)}
U= √{2.20×9.8/sin(73)} = 22.5m/s
<h3>What's the expression of time of flight in projectile motion?</h3>
- Time of flight= (2×U×sinθ)/g
- So, T= (2×22.5×sin36.5°)/9.8
= 2.73 s
Thus, we can conclude that the frog's launch speed and the time spends in the air are 22.5m/s and 2.73s respectively.
Learn more about the range and time period of projectile motion here:
brainly.com/question/24136952
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