Answer:
U = - G m M / r
Explanation:
The gravitational potential energy is given by the expression
U = - G m₁ m₂ / r
dodne G is the gravitational cosntnate (G = 6.67 10⁻¹¹¹), m and m are the mass of the bodies involved
subtype the given values
U = - G m M / r
Answer:
True
Explanation:
This is a universal rule for all standard motor vehicles.
Answer:
1. Urea, (d)Fertilizer
2. Combines, (c) Harvesting and threshing
3. Silos, (b) Storage of grains
4. Seed drills, (a) Sowing seeds
5. Irrigation, (f) Sprinklers
6. Tilling, (e) Preparation of soil
Explanation:
1) Urea is a widely used and important fertilizer in the agricultural industry
2) The combine harvester combines three categories of harvesting grain drops such as threshing reaping and winnowing
3) Silos are used to store grains
4) A seed drill is a seed planting mechanism for burying seeds to a particular depth during seed planting
5) Irrigation is the application of required volume of water to plants
6) Tilling is the digging, turning, and staring operations meant to prepare the soil.
Answer: E/4 ( one - fourth of it electric field)
Explanation:
The electric field of a point charge is given below as
E =kq/r²
E = electric field,
K = electric constant
q = magnitude of electric charge
r = distance between point charge and electric field.
It can be seen that only E and r are the only variable here and also, E is inversely proportional to r²
Which implies that
E = k/r² , k = E × r²
E1 ×(r1)² = E2 × (r2)²
Let E1 = E, r1 =1, r2 = 2 and E2 =?
Let us substitute the parameters
E × 1 = E2 × 2²
E × 1 = E2 × 4
E = E2 × 4
E2 = E/4
Which implies that the electric field at the second distance (r =4) is one fourth of the initial electric field.
Answer:
a. E = 122.4 N/C
b. E = 58.2 N/C
c. E = 0
Explanation:
The electric field at an arbitrary point away from the axis of the cylinder can found by applying Gauss’ Law, which states that an electric flux through a closed surface is equal to the total charge enclosed by this surface divided by electric permittivity.
In order to apply this law, we have to draw an imaginary cylindrical surface of arbitrary height ‘h’ and radius ‘r’, which is equal to the point where the E-field is asked.
A. For the outside of the cylinder, we will draw our imaginary surface with r = 1.97.
B. This time our imaginary surface should be inside the cylinder, therefore the enclosed charge will be less than that of part A.
C. In this case our imaginary surface will be inside the cylinder, where there is no charge at all. Therefore, the enclosed charge will be zero and the electric field will be zero.
