Answer:
10 °C
Explanation:
Applying
q = cm(t₂-t₁)............... Equation 2
Where q = heat energy, c = specific heat of ethanol, m = mass of ethanol, t₁ = initial temperature, t₂ = Final temperature.
Given: c = 2.44 J/g.°C, m = 300 g, q = 14640 J, t₂ = 30°C
Substitute into equation 2 and solve for t₁
14640 = 2.44×300(30-t₁)
14640 = 732(30-t₁)
732(30-t₁) = 14640
(30-t₁) = 14640/732
(30-t₁) = 20
t₁ = 30-20
t₁ = 10 °C
Answer:
Explanation:
This problem indicates that the speed of light in a material medium is 0.5 10⁸ m / s, they ask to find the critical angle between the material and the vacuum
Let's find the refractive index of the material
n = c / v
n = 3 10⁸ / 0.5 10⁸
n = 6
When the material passes from one medium to another, it must comply with the law of refraction
n₁ sin θ = n₂ sin θ₂
for the angle criticize the angles tea2 = 90
tea = sin⁻¹n₂/ n₁
The vacuum replacement index is n₂ = 1
tea = sin⁻¹ (1/6)
tea = 9.59º
When atmospheric pressure is higher than the absolute pressure of a gas in a container.
Answer:
7.11x10^-3
Explanation:
We are to get the volume rate of flows
1/2pv1² + pl = 1/2pv2²
Such that A1V1 = A2V2
V1 = A2V2/A1
From the attachment I uploaded, we have a formula named equation 1 from which I have plugged in these values
P2 = 33000
P2 = 24000
P = 1000
r2 = 2.25
r1 = 4
When we put these values into the equation,
V2 = 4.47
A2V2 = pi(0.0225)²x4.47
= 7.7x19^-3m³/s
