Answer:
10.347 minutes.
Explanation:
According to F = ma, she exerts force on camera of the magnitude
F = 0.67Kg*12m/
= 8.04N, assuming it took her one second to accelerate camera to 12m/s, then by newtons third law, which says every action has equal and opposite reaction , the camera exerts the same amount of force on the astronaut which gives her acceleration of a = 
.
and velocity of V = 0.1130801680m/s.
at this velocity , the astronaut has to cover the distance of 70.2 meters, it will take her 620.7985075s = 10.347 min to get to the shuttle (using S = vt).
Answer:
σ₁ = 
C/m²
σ₂ = 
C/m²
Explanation:
The given data :-
i) The radius of smaller sphere ( r ) = 5 cm.
ii) The radius of larger sphere ( R ) = 12 cm.
iii) The electric field at of larger sphere ( E₁ ) = 358 kV/m. = 358 * 1000 v/m
Q₁ = 572.8 
C
Since the field inside a conductor is zero, therefore electric potential ( V ) is constant.
V = constant
∴
= 
C
Surface charge density ( σ₁ ) for large sphere.
Area ( A₁ ) = 4 * π * R² = 4 * 3.14 * 0.12 = 0.180864 m².
σ₁ = 
= 
= C/m².
Surface charge density ( σ₂ ) for smaller sphere.
Area ( A₂ ) = 4 * π * r² = 4 * 3.14 * 0.05² =0.0314 m².
σ₂ =
= 
= C/m²
Answer:
(b) 10 Wb
Explanation:
Given;
angle of inclination of magnetic field, θ = 30°
initial area of the plane, A₁ = 1 m²
initial magnetic flux through the plane, Φ₁ = 5.0 Wb
Magnetic flux is given as;
Φ = BACosθ
where;
B is the strength of magnetic field
A is the area of the plane
θ is the angle of inclination
Φ₁ = BA₁Cosθ
5 = B(1 x cos30)
B = 5/(cos30)
B = 5.7735 T
Now calculate the magnetic flux through a 2.0 m² portion of the same plane
Φ₂ = BA₂Cosθ
Φ₂ = 5.7735 x 2 x cos30
Φ₂ = 10 Wb
Therefore, the magnetic flux through a 2.0 m² portion of the same plane is is 10 Wb.
Option "b"
Answer:
it is essential that the charge on the plates are of the same magnitude, but in the opposite direction
Explanation:
The configuration of parallel plates is called a capacitor and is widely used to create constant electric fields inside.
To obtain this field it is essential that the charge on the plates are of the same magnitude, but in the opposite direction
This is so that the fields created by each plate can be added inside and subtracted from the outside of the plates
