Answer:
128 m
Explanation:
From the question given above, the following data were obtained:
Horizontal velocity (u) = 40 m/s
Height (h) = 50 m
Acceleration due to gravity (g) = 9.8 m/s²
Horizontal distance (s) =?
Next, we shall determine the time taken for the package to get to the ground.
This can be obtained as follow:
Height (h) = 50 m
Acceleration due to gravity (g) = 9.8 m/s²
Time (t) =?
h = ½gt²
50 = ½ × 9.8 × t²
50 = 4.9 × t²
Divide both side by 4.9
t² = 50 / 4.9
t² = 10.2
Take the square root of both side
t = √10.2
t = 3.2 s
Finally, we shall determine where the package lands by calculating the horizontal distance travelled by the package after being dropped from the plane. This can be obtained as follow:
Horizontal velocity (u) = 40 m/s
Time (t) = 3.2 s
Horizontal distance (s) =?
s = ut
s = 40 × 3.2
s = 128 m
Therefore, the package will land at 128 m relative to the plane
Complete question:
if two point charges are separated by 1.5 cm and have charge values of +2.0 and -4.0 μC, respectively, what is the value of the mutual force between them.
Answer:
The mutual force between the two point charges is 319.64 N
Explanation:
Given;
distance between the two point charges, r = 1.5 cm = 1.5 x 10⁻² m
value of the charges, q₁ and q₂ = 2 μC and - μ4 C
Apply Coulomb's law;

where;
F is the force of attraction between the two charges
|q₁| and |q₂| are the magnitude of the two charges
r is the distance between the two charges
k is Coulomb's constant = 8.99 x 10⁹ Nm²/C²

Therefore, the mutual force between the two point charges is 319.64 N
Answer:

Explanation:
According to given:
- molecular mass of glycerin,

- molecular mass of water,

- ∵Density of water is

- ∴mass of water in 316 mL,

- mass of glycerin,

- pressure of mixture,

- temperature of mixture,

<em>Upon the formation of solution the vapour pressure will be reduced since we have one component of solution as non-volatile.</em>
<u>moles of water in the given quantity:</u>



<u>moles of glycerin in the given quantity:</u>



<u>Now the mole fraction of water:</u>



<em>Since glycerin is non-volatile in nature so the vapor pressure of the resulting solution will be due to water only.</em>



Answer:
Absolute pressure of the oil will be 102822.8 Pa
Explanation:
We have given height h = 31 cm = 0.31 m
Acceleration due to gravity 
Specific gravity of oil = 0.600
So density of oil 
We know that absolute pressure is given by
, here 
So absolute pressure will be equal to 
So absolute pressure of the oil will be 102822.8 Pa