Which is true concerning the acceleration due to gravity? A. It decreases with increasing altitude. B. It is different for diffe
1 answer:
Answer:
The correct answer is option 'a': It decreases with increase in altitude
Explanation:
Acceleration due to gravity is the acceleration that a body is subjected to when it is freely dropped from a height from surface of any planet, ignoring the resistance that the object may face in it's motion such as drag due to any fluid.. The acceleration due to gravity is same for all the objects and is independent of their masses, it only depends on the mass of the planet and the radius of the planet on which the object is dropped. it's values varies with:
1) Depth from surface of planet.
2)Height from surface of planet.
3) Latitude of the object.
Hence it neither is a fundamental quantity nor an universal constant.
The variation of acceleration due to gravity with height can be mathematically written as:
where,
R is the radius of the planet
is value of acceleration due to gravity at surface.
hence we can see that upon increase in altitude the value of 'g' goes on decreasing.
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Answer:
A)Magnitude of the net electric field at the origin due q₁ and q₂
E₀= 131.6 N/C
B) Direction of the net electric field at the origin due q₁ and q₂
β=3.91°
Explanation:
Conceptual analysis
The electric field at a point P due to a point charge is calculated as follows:
E = k*q/d²
E: Electric field in N/C
q: charge in Newtons (N)
k: electric constant in N*m²/C²
d: distance from charge q to point P in meters (m)
The electric field at a point P due to several point charges is the vector sum of the electric field due to individual charges.
Equivalences
1nC= 10⁻⁹ C
Data
q₁ = -4 nC = -4*10⁻⁹ C
q₂ = +6nC =+ 6*10⁻⁹ C
k = 9*10⁹ N*m²/C²
d₂ = 0.6 m
Graphic attached
The attached graph shows the field in x=0, y=0, due to the charges q₁ and q₂:
E₁: Total field at point x=0 , y=0 due to charge q₁. As the charge q₁ is negative (q₁-), the field enters the charge.
E₂: Total field at point x=0 , y=0 due to charge q₂. As the charge q2 is positive (q₂+) ,the field leaves the charge.
Calculation of the electric field at the origin of x-y coordinates due to the charge q₁
E₁=K*q₁/d₁²=9*10⁹*4*10⁻⁹/1²=36N/C
Components (x-y) of the field due to q1:
E₁x=E₁*cosα = 36*(0.6/1) = 36*(0.6) = 21.6 N/C
E₁y=E₁*sinα = 36*(0.8/1) = 36*(0.8) =28.8 N/C
Calculation of the electric field at the origin of x-y coordinates due to the charge q₂
E₂=K*q₂/d₂² = -9*10⁹*6*10⁻⁹/0.6² = -150 N/C
Calculation of the electric field components at the origin of x-y coordinates
E₀ is the electric field at the origin of x-y coordinates (x=0,y=0)
E₀x= E₁x+E₂= 21.6-150 = -128.4 N/C
E₀y= E₁y = 28.8 N/C
A )Magnitude of the net electric field at the origin due q₁ and q₂
E₀= 131.6 N/C
B) Direction of the net electric field at the origin due q₁ and q₂
β=3.91°
