If the mass of an object is doubled, its gravitational force will:
2 answers:
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Answer and Explanation:
with reference to Einstein's theory of special relativity, the speed of an electromagnetic radiation, here, laser will not change in any inertial frame or remains same irrespective of any change in inertial frame.
Therefore, the speed of light measured in both the cases, i.e., in astronaut's reference frame and spaceship's reference frame will be equal to the speed of light in vacuum, i.e., .
The laser gun's speed in astronaut's reference frame is the same as the speed of the spaceship as it mounted on it, i.e., the speed of the laser gun is 200 million m/s.
The laser gun's speed measured in spaceship's reference frame will be zero, as it is mounted on the spaceship and is stationary in the spaceship's reference frame.
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°
