Answer:
The answer to your question is: W = 390 J
Explanation:
Work is the transfer of energy when a body is moved from one place to another.
Data
Force = 65 N
mass = 45 kg
distance = 6 meters
work = ? J
Formula
W = F x d
Process
W = 65 N x 6 m
W = 390 J
<h2>
Answer: 117.626m/s</h2>
Explanation:
The escape velocity 
is given by the following equation:
(1)
Where:
is the Gravitational Constant and its value is 
is the mass of the asteroid
is the radius of the asteroid
On the other hand, we know the density of the asteroid is 
and its volume is 
.
The density of a body is given by:
(2)
Finding :
(3)
(4) This is the mass of the spherical asteroid
In addition, we know the volume of a sphere is given by the following formula:
(5)
Finding :
![R=\sqrt[3]{\frac{3V}{4\pi}}](https://tex.z-dn.net/?f=R%3D%5Csqrt%5B3%5D%7B%5Cfrac%7B3V%7D%7B4%5Cpi%7D%7D)
(6)
![R=\sqrt[3]{\frac{3(2.17(10)^{12}m^{3})}{4\pi}}](https://tex.z-dn.net/?f=R%3D%5Csqrt%5B3%5D%7B%5Cfrac%7B3%282.17%2810%29%5E%7B12%7Dm%5E%7B3%7D%29%7D%7B4%5Cpi%7D%7D)
(7)
(8) This is the radius of the asteroid
Now we have all the necessary elements to calculate the escape velocity from (1):
(9)
Finally:
This is the minimum initial speed the rocks need to be thrown in order for them never return back to the asteroid.
Answer:
6.78 X 10³ N/C
Explanation:
Electric field near a charged infinite plate
= surface charge density / 2ε₀
Field will be perpendicular to the surface of the plate for both the charge density and direction of field will be same so they will add up.
Field due to charge density of +95.0 nC/m2
E₁ = 95 x 10⁻⁹ / 2 ε₀
Field due to charge density of -25.0 nC/m2
E₂ = 25 x 10⁻⁹ / 2ε₀
Total field
E = E₁ + E₂
= 95 x 10⁻⁹ / 2 ε₀ + 25 x 10⁻⁹ / 2ε₀
= 6.78 X 10³ N/C
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"
