Respuesta :
Answer:
A) 46.43 m/s
B) Acceleration = 2395.27 m/s²
C) Average force in Newton's = 5988.18 N
Ratio of this force to weight = 244.14
Explanation:
We are given that;
Mass; m = 2.5 kg
Height reached by firework; y = 110m
Length of tube;L = 0.45m
A) According to kinematic equation;
v² = u² + 2gs
In this case, v = 0 and because gravity is acting against motion. Thus,
0 = u² - 2gy
u² = 2gy
u = √2gy
Plugging in the relevant values ;
u = √(2 x 9.8 x 110)
u = √2156
u = 46.43 m/s
B) we want to find the acceleration with u = 0 m/s and v = 46.43 m/s in the tube. The tube length is 0.45m
Thus,
v² = u² + 2aL
46.43² = 0 + (2•a•0.45)
2155.7449 = 0.9a
acceleration, a = 2155.7449/0.9 = 2395.27 m/s²
C) From Newton's law of motion,
Average Force; F = ma
Thus, F = 2.5 x 2395.27 = 5988.18 N
We are told to express it as a ratio to the weight of the shell.
Thus, ratio = F/W
Weight = mg and force = ma
Thus, F/W = ma/mg = a/g = 2395.27/9.8 = 244.42
Answer:
a) the initial velocity ot the firework shell is 46.5m/s
b) the average acceleration of the shell in the tube is 2.4 × 10³m/s²
c) the average force on the shell in the tube is 2.4 × 10³m/s²
the ratio of the average force acting on the shell to the weight of the shell is 244.65 : 1
Explanation:
Given that;
Mass, m = 2.5 kg
Height, h = 110m
Length , L = 0.45m
The expression of the relationship between velocity and distance kinematic law of motion
[tex]v^2 = u^2 - 2gh[/tex]
[tex]u = \sqrt{v^2+2gh}[/tex]
At maximum height the velocity of the firework shell is zero
That is why the final velocity of the firework shell is zero
[tex]u = \sqrt{2gh}[/tex]
[tex]u = \sqrt{2\times 9.81 \times 110} \\\\u = 46.5m/s[/tex]
Hence, the initial velocity ot the firework shell is 46.5m/s
b)
[tex]v^2 = u^2 + 2as[/tex]
[tex]a = \frac{v^2 - u^2}{2s}[/tex]
The initial velocity of the shell in the tube is zero
[tex]a = \frac{v^2}{2s} \\[/tex]
substitute 46.5m/s for v and 0.450 for s
[tex]a = \frac{46.5^2}{2\times 0.450} \\\\a = 2.4 \times 10 ^3m/s^2[/tex]
Hence , the average acceleration of the shell in the tube is 2.4 × 10³m/s²
c)
From Newton's law of motion,
Average Force, F = ma
substitute 2.50kg for m and 2.4 × 10³m/s² for a
F = ( 2.50kg )(2.4 × 10³m/s²)
F = 6 × 10³N
Hence, the average force on the shell in the tube is 2.4 × 10³m/s²
Calculate the weight of the shell
W = mg
substitute 2.50kg for m and 9.81m/s² for g
W = (2.50kg)( 9.81m/s²)
W = 24.525N
Calculate the ratio of the average force acting on the shell in the tube to the weight of the shell
[tex]Ratio = \frac{F}{W}[/tex]
substitute 24.525N for W and 6 × 10³N for F
[tex]Ratio = \frac{6 \times 10^3N}{24.525N} \\\\= 244.65[/tex]
Hence , the ratio of the average force acting on the shell to the weight of the shell is 244.65 : 1
