Answer:
New gravitational force, F' = 216 units
Explanation:
Given that,
Gravitational force between object 1 and 2, F = 72 N
The gravitational force between two objects is given by :

............(1)
If the mass of object 1, 

If the distance between objects is halved, 
Let new gravitational force is F'. It is changed to :





F' = 216 units
So, the new gravitational force will be 216 units.
Answer:
Qsinθ/4πε₀R²θ
Explanation:
Let us have a small charge element dq which produces an electric field E. There is also a symmetric field at P due to a symmetric charge dq at P. Their vertical electric field components cancel out leaving the horizontal component dE' = dEcosθ = dqcosθ/4πε₀R² where r is the radius of the arc.
Now, let λ be the charge per unit length on the arc. then, the small charge element dq = λds where ds is the small arc length. Also ds = Rθ.
So dq = λRdθ.
Substituting dq into dE', we have
dE' = dqcosθ/4πε₀R²
= λRdθcosθ/4πε₀R²
= λdθcosθ/4πε₀R
E' = ∫dE' = ∫λRdθcosθ/4πε₀R² = (λ/4πε₀R)∫cosθdθ from -θ to θ
E' = (λ/4πε₀R)[sinθ] from -θ to θ
E' = (λ/4πε₀R)[sinθ]
= (λ/4πε₀R)[sinθ - sin(-θ)]
= (λ/4πε₀R)[sinθ + sinθ]
= 2(λ/4πε₀R)sinθ
= (λ/2πε₀R)sinθ
Now, the total charge Q = ∫dq = ∫λRdθ from -θ to +θ
Q = λR∫dθ = λR[θ - (-θ)] = λR[θ + θ] = 2λRθ
Q = 2λRθ
λ = Q/2Rθ
Substituting λ into E', we have
E' = (Q/2Rθ/2πε₀R)sinθ
E' = (Q/θ4πε₀R²)sinθ
E' = Qsinθ/4πε₀R²θ where θ is in radians
Hand grip test an example of an isometric muscle contraction.
Isometric(increased pressure but no shortening)
Explanation:
- Isometric contraction, is one in which the muscle is activated, but instead of being allowed to lengthen or shorten, it is held at a constant length.
- An isometric exercise is a form of exercise involving the static contraction of a muscle without any visible movement in the angle of the joint.
- Isometric exercises are contractions of a particular muscle or group of muscles. During isometric exercises, the muscle doesn't noticeably change length and the affected joint doesn't move. Isometric exercises help maintain strength. They can also build strength, but not effectively.
- Isometric exercise is also known as static strength training. (Examples include the plank and side bridge as well as the wall sit and many yoga poses such as chair and tree poses).
Home is like a shelter where it keeps us safe.