Answer:
From -15⁰ to 0⁰
H=mc¶
where H= heat absorbed or evolved
m=mass involved
c=specific heat capacity
¶=change in temperature
H=mc¶
Answer:
h> 2R
Explanation:
For this exercise let's use the conservation of energy relations
starting point. Before releasing the ball
Em₀ = U = m g h
Final point. In the highest part of the loop
Em_f = K + U = ½ m v² + ½ I w² + m g (2R)
where R is the radius of the curl, we are considering the ball as a point body.
I = m R²
v = w R
we substitute
Em_f = ½ m v² + ½ m R² (v/R) ² + 2 m g R
em_f = m v² + 2 m g R
Energy is conserved
Emo = Em_f
mgh = m v² + 2m g R
h = v² / g + 2R
The lowest velocity that the ball can have at the top of the loop is v> 0
h> 2R
Answer:
Explanation:
Force between charge is given by the following expression
F = k Q₁ Q₂ / R² , k = 9 x 10⁹ , Q₁ and Q₂ are charges , R is distance between charges .
Putting the given values ,
.85 = 9 x 10⁹ x 7.1 x 10⁻⁶ x Q₂ / 5.4²
Q₂ = .85 x 5.4² / (9 x 10⁹ x 7.1 x 10⁻⁶ )
= .38788 x 10⁻³ C .
= 387.88 x 10⁻⁶ C .
Answer:I would guess a plane
Assuming they all Thad the same velocity....
50km / 2.5 = 20 km per hour