You've managed somehow to post the mirror image of the circuit diagram, including the numbers and values of the resistors. I'm curious to know how you did that.
The three resistors at the left end of the diagram are 3Ω , 2Ω , and 1Ω all in series. They behave like a single resistor of (3+2+1) = 6Ω .
That 6Ω resistor is in parallel with the 2Ω drawn vertically in the middle of the diagram. That combination acts like a single resistor of 1.5Ω in that position.
Finally, we have that 1.5Ω resistor in series with 1Ω and 4Ω . That series combination behaves like a single resistor of <em>6.5Ω</em> across the battery V.
Answer:
reviewing the final statements, the correct one is the quarter
The nail exerts an equal force on the hammer in the opposite direction.
Explanation:
This is an action-reaction problem or Newton's third law, which states that forces in naturals occur in pairs.
This is the foregoing, the hammer exerts a force on the nail of magnitude F and it will direct downwards, if we call this action and the nail exerts a force on the hammer of equal magnitude but opposite direction bone directed upwards, each force is applied in one of the bodies.
The difference in result that each force is that the force between the nail exerts a very high pressure (relation between the force between the nail area), instead the area of the hammer is much greater, therefore the pressure is small.
When reviewing the final statements, the correct one is the quarter
The nail exerts an equal force on the hammer in the opposite direction.
The molecular weight of carbon dioxide (CO2) is 44.00 a.m.u., and the molecular weight of propane gas (C3H8) is 44.10 a.m.u. Thus CO2 diffuses_______ C3H8.
Slower Than
Faster Than
The Same As
Slower than
I think the correct answer would be the child quietly sits until the timer goes off. It is ineffective because the child would be used to this process and would come to a time where he will not get something from it.Also, it would make a child feel isolated and alone which would affect hi self-esteem in a long run.
Answer:
Explanation:
a ) It is given that bomb was at rest initially , so , its momentum before the explosion was zero.
b ) We shall apply law of conservation of momentum along x and y direction separately because no external force acts on the bomb.
If v be the velocity of the third part along a direction making angle θ
with x axis ,
x component of v = vcosθ
So momentum along x axis after explosion of third part = mv cosθ
= 10 v cosθ
Momentum along x of first part = - 5 x 42 m/s
momentum of second part along x direction =0
total momentum along x direction before explosion = total momentum along x direction after explosion
0 = - 5 x 42 + 10 v cosθ
v cosθ = 21
Similarly
total momentum along y direction before explosion = total momentum along y direction after explosion
0 = - 5 x 38 + 10 v sinθ
v sinθ= 21
squaring and and then adding the above equation
v² cos²θ +v² sin²θ = 21² +19²
v² = 441 + 361
v = 28.31 m/s
Tanθ = 21 / 19
θ = 48°
