R = 2700 ohm
I = 2.4 mA = 2.4 × 10^(-3) A
I = Q/t
Q = I × t = 2.4 × 10^(-3) × 15 s = 36 C
Answer:
I assume that the force of 2N is applied along the direction of motion and was applied for the whole 1 meter, the formula of work is this; Work = force * distance * cosθ where θ is zero degrees. Plugging in the data to the formula;
Explanation:
Work = 2N * 1m * cos 0º.
Work = 2N * 1m * 1
Work = 2Nm
Work = 2 joules
Answer:
Pressure increases as you move deeper below earth's surface.
Tempurature increases as you move deeper below earth's surface.
Hope this helps!
Explanation:
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°
C look at how many oxygen, nitrogen’s, and hydrogens there are
