Answer:
a) 1 m tall, 3 m wide
b) 1 m tall, 1.31 m wide
Explanation:
According to the captain of the spaceship, the dimensions of the picture is the same i.e 1.0 m tall along the y' axis and 3.0 m wide along the x' axis.
b) The dimensions of the picture as seen by an observer on the Earth along the y axis will remain the same, 1.0 m tall, for the direction of the y axis is perpendicular to the spaceship movement.
The dimensions of the picture as seen by an observer on the Earth along the x axis will reduce if we are to go by the Lorentz contraction:
L(x) = L(x)' * √[1 - (v²/c²)]
where
L(x)' = the dimensions of the picture along the x axis on the spaceship,
v² = the speed of the spaceship and c² = the speed of light in the vacuum.
On substituting, we have
L(x) = 3 * √[1 - (0.81c²/c²)]
L(x) = 1.31 m
If an object experiences no net force, its velocity will remain constant. The object is either at rest and the velocity is zero, or it moves in a straight line with a constant speed.
Answer:
20.62 N
4.123 m/s^2
Explanation:
A = 20 N west
B = 5 N North
m = 5 kg
Both the forces acting at right angle
Use the formula of resultant of two vectors.
Let r be the magnitude of resultant of two vectors.
R = 20.62 N
Let a be the acceleeration.
a = Net force / mass = R / m = 20.62 / 5
a = 4.123 m/s^2
I f im not mistaken, I think the answer is B.