Answer:
t = 4.08 s
R = 40.8 m
Explanation:
The question is asking us to solve for the time of flight and the range of the rock.
Let's start by finding the total time it takes for the rock to land on the ground. We can use this constant acceleration kinematic equation to solve for the displacement in the y-direction:
We have these known variables:
- (v_0)_y = 0 m/s
- a_y = -9.8 m/s²
- Δx_y = -20 m
And we are trying to solve for t (time). Therefore, we can plug these values into the equation and solve for t.
- -20 = 0t + 1/2(-9.8)t²
- -20 = 1/2(-9.8)t²
- -20 = -4.9t²
- t = 4.08 sec
The time it takes for the rock to reach the ground is 4.08 seconds.
Now we can use this time in order to solve for the displacement in the x-direction. We will be using the same equation, but this time it will be in terms of the x-direction.
List out known variables:
- v_0 = 10 m/s
- t = 4.08 s
- a_x = 0 m/s
We are trying to solve for:
By using the same equation, we can plug these known values into it and solve for Δx.
- Δx = 10 * 4.08 + 1/2(0)(4.08)²
- Δx = 10 * 4.08
- Δx = 40.8 m
The rock lands 40.8 m from the base of the cliff.
Ok i apologise for the messy working but I'll try and explain my attempt at logic
Also note i ignore any air resistance for this.
First i wrote the two equations I'd most likely need for this situation, the kinetic energy equation and the potential energy equation.
Because the energy right at the top of the swing motion is equal to the energy right in the "bottom" of the swing's motion (due to conservation of energy), i made the kinetic energy equal to the potential energy as indicated by Ek = Ep.
I also noted the "initial" and "final" height of the swing with hi and hf respectively.
So initially looking at this i thought, what the heck, there's no mass. Then i figured that using the conservation of energy law i could take the mass value from the Ek equation and use it in the Ep equation. So what i did was take the Ek equation and rearranged it for m as you can hopefully see. Then i substituted the rearranged Ek equation into the Ep equation.
So then the equation reads something like Ep = (rearranged Ek equation for m) × g (which is -9.81) × change in height (hf - hi).
Then i simplify the equation a little. When i multiply both sides by v^2 i can clearly see that there is one E on each side (at that stage i don't need to clarify which type of energy it is because Ek = Ep so they're just the same anyway). So i just canceled them out and square rooted both sides.
The answer i got was that the max velocity would be 4.85m/s 3sf, assuming no losses (eg energy lost to friction).
I do hope I'm right and i suppose it's better than a blank piece of paper good luck my dude xx
Answer: 0.01 m
Explanation: The formulae for capillarity rise or fall is given below as
h = (2T×cosθ)/rpg
Where θ = angle mercury made with glass = 50°
T = surface tension = 0.51 N/m
g = acceleration due gravity = 9.8 m/s²
r = radius of tube = 0.5mm = 0.0005m
p = density of mercury.
h = height of rise or fall
From the question, specific gravity of density = 13.3
Where specific gravity = density of mercury/ density of water, where density of water = 1000 kg/m³
Hence density of mercury = 13.3×1000 = 13,300 kg/m³.
By substituting parameters, we have that
h = 2×0.51×cos 50/0.0005×9.8×13,300
h = 0.6556/65.17
h = 0.01 m
The force and the air resistance depends on the mechanical enserfy.
Explanation:
Since its accelerating, the velocity vs time graph is linear
For displacement we need initial velocity (which is zero because it starts from rest) and final velocity (which is calculatee thro acceleration formula
A= (vf - vi)/t
a= vf-0/t
1.25=vf / 7
1.25*7=vf
8.75 = vf
Now for displacement plug all the values in
X = 1/2(vf-vi)/t formula
The displacement (x) is 30.625 m
For part 3, we know new displacement that is 22m , the final and initial velocities are the same so just plug in the values for same formula above
The answer is t = 5.02
Im pretty sure all the answers are correct
