Answer:
Logically yes, because Newton's Third law state "When one body exerts a force on a second body, the second body simultaneously exerts a force equal in magnitude and opposite in direction on the first body."
If force wasn't pushing up then neither gravity is pulling down.
Answer:
D. 43° 52`
Explanation:
A bearing is an angle, measured clockwise from the north direction. When solving a bearing problem, it is good to represent the bearings in the given question with diagram.
The diagrammatically representation of the bearing of lines A and B, 16° 10` and 332° 18` respectively given in the question is shown in the figure attached.
At Point A, we will calculate angle ∠BAO.
Calculating the angle ∠BAO
∠BAO = 90° - 16° 10`
= 73° 50`
At Point B, we will calculate angle ∠ABO.
Calculating the angle ∠ABO
∠ABO = 332° 18` - 270° 0`
= 62° 18`
At Point O, we will calculate the include angle ∠BOA.
Calculating the angle ∠BOA
∠BAO + ∠ABO + ∠BOA = 180° (sum of angles in a triangle)
73° 50` + 62° 18` + ∠BOA = 180°
136° 8` + ∠BOA = 180°
∠BOA = 180° - 136° 8`
∠BOA = 43° 52`
The value of the included angle BOA is 43° 52
Answer:
The outer core is a liquid
Explanation:
The outer core is a fluid composed of iron and nickel
Answer:
Explanation:
Given
1 mole of perfect, monoatomic gas
initial Temperature


Work done in iso-thermal process
=initial pressure
=Final Pressure

Since it is a iso-thermal process therefore q=w
Therefore q=39.64 J
(b)if the gas expands by the same amount again isotherm-ally and irreversibly
work done is




