Answer:
Explanation:
We shall represent displacement in vector form .Consider east as x axes and north as Y axes west as - ve x axes and south as - ve Y axes . 255 km can be represented by the following vector
D₁ = - 255 cos 49 i + 255 sin49 j
= - 167.29 i + 192.45 j
Let D₂ be the further displacement which lands him 125 km east . So the resultant displacement is
D = 125 i
So
D₁ + D₂ = D
- 167.29 i + 192.45 j + D₂ = 125 i
D₂ = 125 i + 167.29 i - 192.45 j
= 292.29 i - 192.45 j
Angle of D₂ with x axes θ
tan θ = -192.45 / 292.29
= - 0.658
θ = 33.33 south of east
Magnitude of D₂
D₂² = ( 192.45)² + ( 292.29)²
D₂ = 350 km approx
Tan
4 blocks north because he is it not asking for north east
Answer:
Torque on the rocket will be 1.11475 N -m
Explanation:
We have given that muscles generate a force of 45.5 N
So force F = 45.5 N
This force acts on the is acting on the effective lever arm of 2.45 cm
So length of the lever arm d = 2.45 cm = 0.0245 m
We have to find torque
We know that torque is given by
So torque on the rocket will be 1.11475 N -m
Answer:
Explanation:
The kinetic energy will convert to heat energy (provided the car has friction brakes and not regenerative brakes as might be found on an electric or hybrid) Also<u> assuming level road</u>.
E = ½mv² = ½(1000)30² = 450,000 J