The pet store would be the reference point because it is where he started and it will not move. Hope this helped.
Answer: displacement of airplane is 172 km in direction 34.2 degrees East of North
Explanation:
In constructing the two displacements it is noticed that the angle between the 75 km vector and the 155 km vector is a right angle (90 degrees).
Hence if the plane starts out at A, it travels to B, 75 km away, then turns 90 degrees to the right (clockwise) and travels to C, 155 km away from B. Angle ABC is 90 degrees, hence we can use Pythagoras theorem to solve for AC
AC2 = AB2 + BC2 ; AC^2 = 752 + 1552 ; from this we get AC = 172 km (3 significant figures)
Angle BAC = Tan-1(155/75) ; giving angle BAC = 64.2 degrees
Hence AC is in a direction (64.2 - 30) = 34.2 degrees East of North
Therefore the displacement of the airplane is 172 km in a direction 34.2 degrees East of North
Answer:
Given that the block have two applied masses 250 g at East and 100 g at South. In order to make a situation in which block moves towards point A, we have to apply minimum number of masses to the blocks. In order to prevent block moving toward East, we have to apply a mass at West, equal to the magnitude of mass at East but opposite in direction. Therefore, mass of 250 g at West is the required additional mass that has to be added. There is already 100 g of mass acting at South, that will attract block towards South or point A. No need to add further mass in North-South direction.
Answer:
A
Explanation:
Solution:-
- According to the law of relativity the relative speed between two moving objects is inversely proportional to the the time taken.
- Ignoring Doppler Effect.
- So if the relative speeds of two objects in motion i.e ( swing and spaceship) are positive then the time frame of reference for both object relative to other other decreases. So in other words if spaceship approaches the swing i.e relative velocity is positive then the time period of oscillation observed would be less than actual i.e less than 4 seconds.
- Similarly, if spaceship moves away from the swing i.e relative velocity is negative then the time period of oscillation observed would be more than actual i.e more than 4 seconds.
Answer:
The time constant and its uncertainty is t ± Δt = 0.526 ± 0.057 s
Explanation:
If we make a comparison we have to:
y = A*(1-e^-(C*x)) + B
If the time remains constant we have to:
t = R*C = 1/C
In this way we calculate the time constant and its uncertainty. this will be equal to:
t ± Δt = (1/1.901) ± (0.2051/1.901)*(1/1.901) = 0.526 ± 0.057 s
