The answer would be:
D.
X: Low potential energy
Y: High Potential energy
Z: Flow of electrons
Z is clearly the flow of electrons, as shown by the arrow demonstrating the direction of the flow. So you can easily cross out choices B and C. Now, you can see that Y has more energy stored and X has a lot less, so you can conclude that Y has high potential energy while X has low potential energy.
D. Inclined plane is the correct answer, since the lines on the screw spiral up like a staircase.
Answer:
By applying a force of one Newton, one can hold a body of mass of 102 gram.
Explanation:
- Force is the pull or push of an object. It can be mathematically measured as, F= m* g.
where, F= force in newton
m= mass in kg
g= acceleration due to gravity (
)
F= m* 9.8
or, m= 
= 0.102 kg
or, m= 102 gram.
- Hence, 102 gram mass can be hold by one Newton force.
Answer:
Explanation:
Incomplete question but for understanding.
We want to find the electrical force between two charges, then you can use the coulombs law which states that the force of attraction or repulsion between two charges is directly proportional to the product of the two charges and inversely proportional to the square of their distance apart,
So,
F = kq1•q2 / r²
Where k is a constant and it is given as
K = 8.99 × 10^9 Nm²/C²
q1 and q2 are the charges and in this question it is not given, so the question is incomplete. Let assume that,
q1 = - 1.609 × 10^-19 C electron
q2 = 1.609 × 10^-19 C proton
Since unlike charges attract, then it is force of attraction
Also, r is the distance apart and it is not given, let assume the distance between the two charges is 2 × 10^-5m
Then,
F = kq1•q2 / r²
F = 8.99 × 10^9 × 1.609 × 10^-19 × 1.609 × 10^-19 / (2 × 10^-5)²
F = 5.82 × 10^-19 N
Answer:
v = 31.84 cm/s or 0.318 m/s
the speed of the water leaving the end of the hose is 31.84 cm/s or 0.318 m/s
Explanation:
Given;
Diameter of hose d = 2.76 cm
Volume filled V = 20.0 L = 20,000 cm^3
Time t = 1.45 min = 105 seconds
The volumetric flow rate of water is;
F = V/t = 20,000cm^3 ÷ 105 seconds
F = 190.48 cm^3/s
The volumetric flow rate is equal the cross sectional area of pipe multiply by the speed of flow.
F = Av
v = F/A
Area A = πd^2/4
Speed v = F/(πd^2/4)
v = 4F/πd^2 ......1
Substituting the given values;
v = (4×190.48)/(π×2.76^2)
v = 31.83767439628 cm/s
v = 31.84 cm/s or 0.318 m/s
the speed of the water leaving the end of the hose is 31.84 cm/s or 0.318 m/s
