We will measure all angles from West, the negative x-axis and divide the journey into 3 parts:
P1 = 370y
P2 = 410cos(45)x + 410sin(45)y = 290x + 290y
P3 = 370cos(270 - 28)x + 370sin(270 - 28) = -174x - 327y
Overall displacement:
x = 290 - 174 = 116 m
y = 370 + 290 - 327 = 333 m
displacement = √(116² + 333²)
= 353 m
Direction:
tan(∅) = y/x
∅ = tan⁻¹ (333 / 116)
∅ = 70.8° from West.
Answer:
The final temperature of both objects is 400 K
Explanation:
The quantity of heat transferred per unit mass is given by;
Q = cΔT
where;
c is the specific heat capacity
ΔT is the change in temperature
The heat transferred by the object A per unit mass is given by;
Q(A) = caΔT
where;
ca is the specific heat capacity of object A
The heat transferred by the object B per unit mass is given by;
Q(B) = cbΔT
where;
cb is the specific heat capacity of object B
The heat lost by object B is equal to heat gained by object A
Q(A) = -Q(B)
But heat capacity of object B is twice that of object A
The final temperature of the two objects is given by

But heat capacity of object B is twice that of object A

Therefore, the final temperature of both objects is 400 K.
Yes. Acceleration means any change in speed or direction of motion. When an object coasts in a circular orbit at constant speed around the Earth, its direction is constantly changing. The acceleration is "CENTRIPETAL", which points toward the center of the circle.
In Electrostatics the electrical force between Two charged objects is inversely Related to the distance of separation between the two objects .
a) It is absolute, so it does not change.
b) Inertial ones.
c) Inside the train the time will slow down relatively to the outside clock. So if one travel at nearly the speed if light for 2 hours on his clock, for outdoor observers it will look like 3 hours.