Answer:
<em>The balloon is 66.62 m high</em>
Explanation:
<u>Combined Motion
</u>
The problem has a combination of constant-speed motion and vertical launch. The hot-air balloon is rising at a constant speed of 14 m/s. When the camera is dropped, it initially has the same speed as the balloon (vo=14 m/s). The camera has an upward movement for some time until it runs out of speed. Then, it falls to the ground. The height of an object that was launched from an initial height yo and speed vo is

The values are


We must find the values of t such that the height of the camera is 0 (when it hits the ground)


Multiplying by 2

Clearing the coefficient of 

Plugging in the given values, we reach to a second-degree equation

The equation has two roots, but we only keep the positive root

Once we know the time of flight of the camera, we use it to know the height of the balloon. The balloon has a constant speed vr and it already was 15 m high, thus the new height is



Answer:
Angle of Refraction = 28.9 degrees
Explanation:
<u><em>We'll use Snell's law for this. It's mathematical form is:</em></u>
=> 
Where 
=>
and
are the refractive indexes of the air and water respectively.
<u><em>Solution:</em></u>
=> 1 * sin (40) = 1.33 * sin(
)
=> sin
= 
=> sin
= 0.4821
=>
= 28.9 degrees
Answer:
-30 °C
Explanation:
First, we have to calculate the molality (m) of the solution. If the solution is 50% C₂H₆O₂ by mass. It means that in 100 g of solution, the are 50 g of solute (C₂H₆O₂) and 50 g of solvent (water).
The molar mass of C₂H₆O₂ is 62.07 g/mol. The moles of solute are:
50 g × (1 mol / 62.07 g) = 0.81 mol
The mass of the solvent is 50 g = 0.050 kg.
The molality is:
m = 0.81 mol / 0.050 kg = 16 m
The freezing-point depression (ΔT) can be calculated using the following expression.
ΔT = Kf × m = (1.86 °C/m) × 16 m = 30 °C
where,
Kf: freezing-point constant
The normal freezing point for water is 0°C. The freezing point of the radiator fluid is:
0°C - 30°C = -30 °C