Answer:
Explanation:
All the displacement will be converted into vector, considering east as x axis and north as y axis.
5.3 km north
D = 5.3 j
8.3 km at 50 degree north of east
D₁= 8.3 cos 50 i + 8.3 sin 50 j.
= 5.33 i + 6.36 j
Let D₂ be the displacement which when added to D₁ gives the required displacement D
D₁ + D₂ = D
5.33 i + 6.36 j + D₂ = 5.3 j
D₂ = 5.3 j - 5.33i - 6.36j
= - 5.33i - 1.06 j
magnitude of D₂
D₂²= 5.33² + 1.06²
D₂ = 5.43 km
Angle θ
Tanθ = 1.06 / 5.33
= 0.1988
θ =11.25 ° south of due west.
935,500 joules because when we use the KE formula KE=1/2mv^2;
KE=1/2(750)(50)^2
KE=375(2500)
KE=935,500 Joules
Hope it helps
A. Diagram A
B. Diagram C & D
C. Diagram B
D. Diagram C & D
E. Diagram B
F. Diagram C & D
These are simplified representations of an object's body and the force vectors acting on it. Some of the main forces that are involve are normal force, friction, push or pull and gravity.
In fact, entropy of an isolated system never decreases (2nd law of thermodynamics), unless some external energy is provided in order to "restore" order in the system and decrease its entropy.
(note that when external energy is added to the system, it is no longer "isolated").
*This is only true if the question is referring to a certain system within the universe. If we are considering the universe itself as the system, then this option is no longer correct, because no external energy can be provided to the universe, and since the universe is an isolated system, its entropy can never decrease. If we are considering the universe itself as the system, none of the options is true.
Answer:
1 day
Explanation:
The spinning movement of Earth around its axis is known as Earth's rotation.
The Earth rotates around the Sun in 1 day that is 24 hours but once in every 23 hours, 56 minutes, and 4 seconds with respect to other stars.
According to the Indian astronomer Aryabhata, Earth rotates about its axis daily.
