claim cold air weighs more than hot air
reasoning is expt
Answer:
The water particles move perpendicular to the source of the sound wave.
Explanation:
Water particles are transverse waves
Answer:
2.06 x 10⁴ J
Explanation:
The process takes place in three steps. First, the ice is heated from -20 °C to 0 °C. Then the ice undergoes a phase change to water. Finally, the water is heated from 0 °C to 50 °C.
The heat energy required for the first step is as follows:
Q = mcΔT = (36.0 g)(2.00 Jg⁻¹°C⁻¹)(0 °C - (-20 °C)) = 1440 J
The heat energy required for the phase change (where L is the heat of fusion) is then calculated. Grams are converted to moles using the molar weight of water (18.02 g/mol)
Q = ML = (36.0 g)(mol/18.02g)(6000 J/mol) = 11987 J
Finally, the heat energy required to raise the temperature of the water to 50°C is calculated:
Q = mcΔT = (36.0 g)(4.00 Jg⁻¹°C⁻¹)(50 °C - 0 °C) = 7200 J
Adding all of the heat energy values together gives:
(1440 + 11987 + 7200) J = 20627 J
The final answer is 2.06 x 10⁴ J
Answer:
The child represented by a star on the outside path.
Explanation:
Answer:
I = 1.06886 N s
Explanation:
The expression for momentum is
I = F t = Δp
therefore the momentum is a vector quantity, for which we define a reference system parallel to the floor
Let's find the components of the initial velocity
sin 28.2 = v_y / v
cos 28.2= vₓ / v
v_y = v sin 282
vₓ = v cos 28.2
v_y = 42.8 sin 28.2 = 20.225 m / s
vₓ = 42.8 cos 28.2 = 37.72 m / s
since the ball is heading to the ground, the vertical velocity is negative and the horizontal velocity is positive, it can also be calculated by making
θ = -28.2
v_y = -20.55 m / s
v_x = 37.72 m / s
X axis
Iₓ = Δpₓ =
since the ball moves in the x-axis without changing the velocity, the change in moment must be zero
Δpₓ = m - m v₀ₓ = 0
v_{fx} = v₀ₓ
therefore
Iₓ = 0
Y axis
I_y = Δp_y = p_{fy} -p_{oy}
when the ball reaches the floor its vertical speed is downwards and when it leaves the floor its speed has the same modulus but the direction is upwards
v_{fy} = - v_{oy}
Δp_y = 2 m v_{oy}
Δp_y = 2 0.0260 (20.55)
= 1.0686 N s
the total impulse is
I = Iₓ i ^ + I_y j ^
I = 1.06886 j^ N s