Answer:
The cannon ball was not able to hit the target because the target is located at a height of 50 m whereas the cannon ball was only above to get to a height of 20 m.
Explanation:
From the question given above, the following data were obtained:
Height to which the target is located = 50 m
Initial velocity (u) = 20 m/s
To know whether or not the cannon ball is able to hit the target, we shall determine the maximum height to which the cannon ball attained. This can be obtained as follow:
Initial velocity (u) = 20 m/s
Final velocity (v) = 0 (at maximum height)
Acceleration due to gravity (g) = 10 m/s²
Maximum height (h) =?
v² = u² – 2gh (since the ball is going against gravity)
0² = 20² – (2 × 10 × h)
0 = 400 – 20h
Collect like terms
0 – 400 = – 20h
– 400 = – 20h
Divide both side by – 20
h = – 400 / – 20
h = 20 m
Thus, the the maximum height to which the cannon ball attained is 20 m.
From the calculations made above, we can conclude that the cannon ball was not able to hit the target because the target is located at a height of 50 m whereas the cannon ball was only above to get to a height of 20 m.
The relationship between the charge flowing through a conductor, the current flowing through the conductor and the time is:
Q = It
Where Q is the charge, I is the current and t is the time of application of the current. Substituting the values:
48.96 = 1.39 x t
t = 35.2 seconds
To answer this question, we must use the equation for the volumetric expansion of gases at constant pressure. This equation is given by:
We know:
is the initial volume 
ΔT is the temperature change = 45 ° -20 ° = 25 °
is the coefficient of gas expansion and is equal to 1/273
Then the final volume of the gas is:
Answer:
Explanation:
Electric field at the surface of the the lead 208 = KQ/ R²
where K = 8.99 × 10⁹ Nm² /C²
Q ( total charge inside the nucleus) and e is the charge of a proton = Ne = 82 × 1.6 × 10⁻¹⁹ C = 1.312 × 10⁻¹⁷ C
V of the lead = 208 v of a proton assuming they both are sphere
4/3 πR³ =208( 4/3 πr³) where R is the radius of the sphere and r is the radius of the proton
R³ = 208 r³
R = ∛( 208 r³) = 5.92r
replace r with 1.20 x 10-15 m
R = 5.92 ×1.20 x 10-15 m = 7.11 × 10⁻¹⁵ m
E = ( 8.99 × 10⁹ Nm² /C² × 1.312 × 10⁻¹⁷ C ) / (7.11 × 10⁻¹⁵ m)² = 0.233 × 10²² N/C = 2.33 × 10²¹ N/C
1.If a compound is made from a metal and a non-metal, its bonding will be ionic.
2.If a compound is made from two non-metals, its bonding will be covalent.
