Answer:
241.24m
Explanation:
The height at which the shell explodes will be at the maximum height. In projectile motion, maximum height formula is expressed as:
H = u²sin²θ/2g
u is the initial speed = 70m/s
θ the angle of launch = 75°
g is the acceleration due to gravity = 9.81m/s²
Substitute the values into the formula and get H
H = 70²(sin75°)/2(9.81)
H = 4900sin75°/19.62
H = 4900*0.9659/19.62
H = 4733.037/19.62
H = 241.24m
Hence the height at which the shell explodes is 241.24m
Answer:
25
Explanation:
By Using Pythagoras Theorem,
h^2=p^2+b^2
or, x^2=(24)^2+7^2
or, x^2=576+49
or, x^2=625
or, x=√625
•°• x=25
Answer:
L= 1 m, ΔL = 0.0074 m
Explanation:
A clock is a simple pendulum with angular velocity
w = √ g / L
Angular velocity is related to frequency and period.
w = 2π f = 2π / T
We replace
2π / T = √ g / L
T = 2π √L / g
We will use the value of g = 9.8 m / s², the initial length of the pendulum, in general it is 1 m (L = 1m)
With this length the average time period is
T = 2π √1 / 9.8
T = 2.0 s
They indicate that the error accumulated in a day is 15 s, let's use a rule of proportions to find the error is a swing
t = 1 day (24h / 1day) (3600s / 1h) = 86400 s
e= Δt = 15 (2/86400) = 3.5 104 s
The time the clock measures is
T ’= To - e
T’= 2.0 -0.00035
T’= 1.99965 s
Let's look for the length of the pendulum to challenge time (t ’)
L’= T’² g / 4π²
L’= 1.99965 2 9.8 / 4π²
L ’= 0.9926 m
Therefore the amount that should adjust the length is
ΔL = L - L’
ΔL = 1.00 - 0.9926
ΔL = 0.0074 m
The answer to this is Longitudinal/compression waves, such as sound waves.
