The kinetic energy as measured in the Earth reference frame is 6.704*10^22 Joules.
To find the answer, we have to know about the Lorentz transformation.
<h3>What is its kinetic energy as measured in the Earth reference frame?</h3>
It is given that, an alien spaceship traveling at 0.600 c toward the Earth, in the same direction the landing craft travels with a speed of 0.800 c relative to the mother ship. We have to find the kinetic energy as measured in the Earth reference frame, if the landing craft has a mass of 4.00 × 10⁵ kg.

- Let us consider the earth as S frame and space craft as S' frame, then the expression for KE will be,

- So, to
find the KE, we have to find the value of speed of the approaching landing craft with respect to the earth frame. - We have an expression from Lorents transformation for relativistic law of addition of velocities as,

- Substituting values, we get,


Thus, we can conclude that, the kinetic energy as measured in the Earth reference frame is 6.704*10^22 Joules.
Learn more about frame of reference here:
brainly.com/question/20897534
SPJ4
The difference between temperature and thermal energy is that temperature measures the average kinetic speed of molecule and thermal energy is the total kinetic energy of all particles in a given substance.
Answer: <u><em>A</em></u>
Explanation:
A pneumatic control system uses compressed air as a method of control for HVAC systems. ... Each senor responds to changes in temperature, humidity, and static pressure as examples, to provide feedback in a control loop to open or close the actuator to meet the control set point.
Answer: Less than 4 ohms
Explanation:
We have three resistors with the following resistance:



Now, when the resistors are connected in parallel, the total resistance
is calculated as follows:

Isolating
:

Rewriting with th known values:

Finally:

Hence, the correct option is less than 4 ohms.
Answer:
Explanation:
Using the magnification formula.
Magnification = Image distance(v)/object distance(u) = Image Height(H1)/Object Height(H2)
M = v/u = H1/H2
v/u = H1/H2...1
3) Given the radius of curvature of the concave lens R = 20cm
Focal length F = R/2
f = 20/2
f = 10cm
Object distance u = 5cm
Object height H2= 5cm
To get the image distance v, we will use the mirror formula
1/f = 1/u+1/v
1/v = 1/10-1/5
1/v = (1-2)/10
1/v =-1/10
v = -10cm
Using the magnification formula
(10)/5 = H1/5
10 = H1
H1 = 10cm
Image height of the peg is 10cm
4) If u = 15cm
1/v = 1/f-1/u
1/v = 1/10-1/15
1/v = 3-2/30
1/v = 1/30
v = 30cm
30/15 = H1/5
15H1 = 150
H1/= 10cm
5) if u = 20cm
1/v = 1/f-1/u
1/v = 1/10-1/20
1/v = 2-1/20
1/v = 1/20
v = 20cm
20/20 = H1/5
20H1 = 100
H1 = 5cm
6) If u = 30cm
1/v = 1/f-1/u
1/v = 1/10-1/30
1/v = 3-1/30
1/v = 2/30
v = 30/2 cm
v =>15cm
15/30 = Hi/5
30H1 = 75
H1 = 75/30
H1 = 2.5cm