Answer:
It can only display one record at a time
Explanation:
Form ;
1. This is a document with spaces (also called placeholders or fields ) in which a series of documents with similar content can be written or selected.
2.This is the most popular method of data entry
3.It may contain images in the background.
4.This can be sorted data regardless of its source of information.
Only option C is wrong.
Therefore the answer C is correct.
actually the answer is B because Chlorine, sulfur, and silicon. Chlorine is a halogen and gas. Sulfur forms an ion with a -2 charge in ionic bonds. Silicon is a well-known metalloid.
Answer:
it depends on a person's own weight
Answer:
a) 5 N b) 225 N c) 5 N
Explanation:
a) Per Coulomb's Law the repulsive force between 2 equal sign charges, is directly proportional to the product of the charges, and inversely proportional to the square of the distance between them, acting along the line that joins the charges, as follows:
F₁₂ = K Q₁ Q₂ / r₁₂²
So, if we make Q1 = Q1/5, the net effect will be to reduce the force in the same factor, i.e. F₁₂ = 25 N / 5 = 5 N
b) If we reduce the distance, from r, to r/3, as the factor is squared, the net effect will be to increase the force in a factor equal to 3² = 9.
So, we will have F₁₂ = 9. 25 N = 225 N
c) If we make Q2 = 5Q2, the force would be increased 5 times, but if at the same , we increase the distance 5 times, as the factor is squared, the net factor will be 5/25 = 1/5, so we will have:
F₁₂ = 25 N .1/5 = 5 N
Answer:
(A). The work done by friction in crossing the patch is -637.98 J.
(B). The speed of skier is 10.57 m/s.
Explanation:
Given that,
Mass of skier = 62 kg
Speed = 6.5 m/s
Length = 3.50 m
Coefficient kinetic friction = 0.30
Height = 2.5 m
(A) we need to calculate the work done by friction in crossing the patch
Using formula of work done
Put the value into the formula
The work done by friction in crossing the patch is -637.98 J.
(B) we need to calculate the speed of skier
Using conservation of energy
Final potential energy is zero
So,
Put the value into the formula
The speed of skier is 10.57 m/s.
Hence, (A).The work done by friction in crossing the patch is -637.98 J.
(B).The speed of skier is 10.57 m/s.