The final velocity of the train after 8.3 s on the incline will be 12.022 m/s.
Answer:
Explanation:
So in this problem, the initial speed of the train is at 25.8 m/s before it comes to incline with constant slope. So the acceleration or the rate of change in velocity while moving on the incline is given as 1.66 m/s². So the final velocity need to be found after a time period of 8.3 s. According to the first equation of motion, v = u +at.
So we know the values for parameters u,a and t. Since, the train slows down on the slope, so the acceleration value will have negative sign with the magnitude of acceleration. Then
v = 25.8 + (-1.66×8.3)
v =12.022 m/s.
So the final velocity of the train after 8.3 s on the incline will be 12.022 m/s.
Answer:
2.2 m/s^2
Explanation:
Acceleration = Force / Mass
= 7.92 / 3.6 = 2.2m/s^2
Hope this help you :3
Answer: 1.55 x 10⁴ Nm²c^-1
Explanation: The electric flux, electric field intensity and area are related by the formulae below.
Φ= EAcosθ,
Where Φ= electric flux (Nm²c^-1)
E =electric field intensity (N/m²)
A = Area (m²)
θ= this is angle between the planar area and the magnetic flux
For our question E=3.80KN/c= 3800 N/c
A= 0.700 x 0.350= 0.245m²
θ= 0° ( this is because the electric field was applied along the x axis, thus the electric flux will be parallel to the area).
Hence Φ= 3800 x 0.245 x cos(0)
= 3800 x 0.245 x 1 (value of cos 0° =1)
= 1.55 x 10⁴ Nm²c^-1
Thus the electric field is 1.55 x 10⁴ Nm²c^-1
I think D. It starts at (0.0) and goes to the correct points so it makes sense
Answer:
1.4 billion light years away
Explanation:
v = Recessional velocity = 30000 km/s[/tex]
= Hubble constant = 
D = Distance to the galaxy
According to Hubble's law
The galaxy is 1.4 billion light years away
