Answer:
865.08 m
Explanation:
From the question given above, the following data were obtained:
Initial velocity (u) = 243 m/s
Height (h) of the cliff = 62 m
Horizontal distance (s) =?
Next, we shall determine the time taken for the cannon to get to the ground. This can be obtained as follow:
Height (h) of the cliff = 62 m
Acceleration due to gravity (g) = 9.8 m/s²
Time (t) =?
h = ½gt²
62 = ½ × 9.8 × t²
62 = 4.9 × t²
Divide both side by 4.9
t² = 62/4.9
Take the square root of both side.
t = √(62/4.9)
t = 3.56 s
Finally, we shall determine the horizontal distance travelled by the cannon ball as shown below:
Initial velocity (u) = 243 m/s
Time (t) = 3.56 s
Horizontal distance (s) =?
s = ut
s = 243 × 3.56 s
s = 865.08 m
Thus, the cannon ball will impact the ground 865.08 m from the base of the cliff.
A positive charge and a negative charge held a certain distance apart are released. as they move, the force on each particle increases
The most common charge carriers are the positively charged proton and the negatively charged electron. The movement of any of these charged particles constitutes an electric current
<h3>What is a Charge ?</h3>
When there are more or fewer protons in an atom than electrons, the substance has an electric charge. Protons have a positive charge, while electrons have a negative charge. If a substance has more protons than electrons, it is positively charged; if it has more electrons, it is negatively charged.
- The SI units for charge are ampere-second or coulomb. When one ampere of electric current goes through the conductor for one second, one coulomb of charge passes through it. Charge is denoted by the formula Q = I t.
Learn more about Charge here:
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Angry sound level = 70 db
Soothing sound level = 50 db
Frequency, f = 500 Hz
Assuming speed of sound = 345 m/s
Density (assumed) = 1.21 kg/m^3
Reference sound intensity, Io = 1*10^-12 w/m^2
Part (a): Initial sound intensity (angry sound)
10log (I/Io) = Sound level
Therefore,
For Ia = 70 db
Ia/(1*10^-12) = 10^(70/10)
Ia = 10^(70/10)*10^-12 = 1*10^-5 W/m^2
Part (b): Final sound intensity (soothing sound)
Is = 50 db
Therefore,
Is = 10^(50/10)*10^-12 = 18*10^-7 W/m^2
Part (c): Initial sound wave amplitude
Now,
I (W/m^2) = 0.5*A^2*density*velocity*4*π^2*frequency^2
Making A the subject;
A = Sqrt [I/(0.5*density*velocity*4π^2*frequency^2)]
Substituting;
A_initial = Sqrt [(1*10^-5)/(0.5*1.21*345*4π^2*500^2)] = 6.97*10^-8 m = 69.7 nm
Part (d): Final sound wave amplitude
A_final = Sqrt [(1*10^-7)/(0.5*1.21*345*4π^2*500^2)] = 6.97*10^-9 m = 6.97 nm
To solve this problem it is necessary to apply the concepts related to Newton's second law, the definition of density and sum of forces in bodies.
From Newton's second law we understand that
Gravity at this case)
Where,
m = mass
a= acceleration
Also we know that
Part A) The buoyant force acting on the balloon is given as
As mass is equal to the density and Volume and acceleration equal to Gravity constant
PART B) The forces acting on the balloon would be given by the upper thrust force given by the fluid and its weight, then
PART C) The additional mass that can the balloon support in equilibrium is given as
P1v1/t1 = p2v2/t2
p1=475, v1=4, t1=290
v2=6.5, t2=277
solve for p2 in kpa
