Answer:
need i help with 1-7 please
Answer:
1) d
2) 5 m/s
3) 100
Explanation:
The equation of position x for a constant acceleration a and an initial velocity v₀, initial position x₀, time t is:
(i) ![x=\frac{1}{2}at^2+v_0t+x_0](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B1%7D%7B2%7Dat%5E2%2Bv_0t%2Bx_0)
The equation for velocity v and a constant acceleration a is:
(ii) ![v=at+v_0](https://tex.z-dn.net/?f=v%3Dat%2Bv_0)
1) Solve equation (ii) for acceleration a and plug the result in equation (i)
(iii) ![a = \frac{v -v_0}{t}](https://tex.z-dn.net/?f=a%20%3D%20%5Cfrac%7Bv%20-v_0%7D%7Bt%7D)
(iv) ![x = \frac{v-v_0}{2t}t^2+v_0t + x_0](https://tex.z-dn.net/?f=x%20%3D%20%5Cfrac%7Bv-v_0%7D%7B2t%7Dt%5E2%2Bv_0t%20%2B%20x_0)
Simplify equation (iv) and use the given values v = 0, x₀ = 0:
(v) ![x=-\frac{v_0}{2}t + v_0t= \frac{v_0}{2}t](https://tex.z-dn.net/?f=x%3D-%5Cfrac%7Bv_0%7D%7B2%7Dt%20%2B%20v_0t%3D%20%5Cfrac%7Bv_0%7D%7B2%7Dt)
2) Given v₀= 3m/s, a=0.2m/s², t=10 s. Using equation (ii) to get the final velocity v:![v=at+v_0=0.2\frac{m}{s^2} * 10s+3\frac{m}{s}=2\frac{m}{s}+3\frac{m}{s}=5\frac{m}{s}](https://tex.z-dn.net/?f=v%3Dat%2Bv_0%3D0.2%5Cfrac%7Bm%7D%7Bs%5E2%7D%20%2A%2010s%2B3%5Cfrac%7Bm%7D%7Bs%7D%3D2%5Cfrac%7Bm%7D%7Bs%7D%2B3%5Cfrac%7Bm%7D%7Bs%7D%3D5%5Cfrac%7Bm%7D%7Bs%7D)
3) Given v₀=0m/s, t₁=10s, t₂=1s and x₀=0. Looking for factor f = x(t₁)/x(t₂) using equation(i) to calculate x(t₁) and x(t₂):
![f=\frac{x(t_1)}{x(t_2)}=\frac{\frac{1}{2}at_1^2 }{\frac{1}{2}at_2^2}=\frac{t_1^2}{t_2^2}=\frac{10^2}{1^2}=\frac{100}{1}](https://tex.z-dn.net/?f=f%3D%5Cfrac%7Bx%28t_1%29%7D%7Bx%28t_2%29%7D%3D%5Cfrac%7B%5Cfrac%7B1%7D%7B2%7Dat_1%5E2%20%7D%7B%5Cfrac%7B1%7D%7B2%7Dat_2%5E2%7D%3D%5Cfrac%7Bt_1%5E2%7D%7Bt_2%5E2%7D%3D%5Cfrac%7B10%5E2%7D%7B1%5E2%7D%3D%5Cfrac%7B100%7D%7B1%7D)
To develop this problem it is necessary to apply the concepts of sum of capacitors in a circuit, either in parallel or in series.
When capacitors are connected in series, they consecutively add their capacitance.
However, when they are connected in parallel, the sum that is made is that of the inverse of the capacitance, that is
where, C is the capacitance.
For the given case, it is best to connect two of these capacitors in series and one in parallel, so
We have three 100 pF capacitors, then
![C_1 = 100 pF\\C_2 = 100 pF\\C_3 = 100 pF\\](https://tex.z-dn.net/?f=C_1%20%3D%20100%20pF%5C%5CC_2%20%3D%20100%20pF%5C%5CC_3%20%3D%20100%20pF%5C%5C)
Here we can see how two capacitors 1 and 2 are in series and the third in parallel.
![\frac{1}{C_{serie}} = \frac{1}{100pF}+\frac{1}{100pF}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7BC_%7Bserie%7D%7D%20%3D%20%5Cfrac%7B1%7D%7B100pF%7D%2B%5Cfrac%7B1%7D%7B100pF%7D)
![\frac{1}{C_{serie}} = \frac{1}{50pF}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7BC_%7Bserie%7D%7D%20%3D%20%5Cfrac%7B1%7D%7B50pF%7D)
Investing equality
![C_{serie} = 50pF](https://tex.z-dn.net/?f=C_%7Bserie%7D%20%3D%2050pF)
Adding it in parallel, then
![C_{total} = C_{serie} +C_3](https://tex.z-dn.net/?f=C_%7Btotal%7D%20%3D%20C_%7Bserie%7D%20%2BC_3)
![C_{total} = 50pF+100pF](https://tex.z-dn.net/?f=C_%7Btotal%7D%20%3D%2050pF%2B100pF)
![C_{total} = 150pF](https://tex.z-dn.net/?f=C_%7Btotal%7D%20%3D%20150pF)
<span>First let's find the acceleration required in the barrel to speed the ball up from 0 to 83 m/s in a distance of 2.17 m. We know the force the cannon exerts on the cannonball is 20000 N; if we can find this acceleration then we can use F = ma to find the mass.
We can find the acceleration using one of the kinematic equations of motion. We have:
u = initial speed = 0 m/s
v = final speed = v0 = 83 m/s
d = distance = 2.17 m
a = acceleration = ?
v² = u² + 2ad. Since u = 0, this reduces to v² = 2ad and rearranges to a = v²/2d = 83²/2*2.17 = 83²/4.34 = 1587.327 m/s².
Now F = ma, so m = F/a = (20000N)/(1587.327 m/s²) = 12.6 kg.
For part 2, use the Range Equation:
If R is the horizontal distance the cannonball travels,
v = v0 = the initial velocity = 83 m/s
g = acceleration due to gravity - 9.8 m/s²
x the launch angle relative to the horizontal, then
R = (v²sin(2x))/g.
So R = (83²sin(2*37))/9.8
= (6889sin74)/9.8 = 676 m.
So the target ship is 676 m away.</span>
Answer:
Yes it is a solid
Explanation:
It just has unusual properties.
Does it float when you drop it? then it is not a gas.
Does it pour into any container as water does? Then it is not a liquid.
So it's a solid.
And it can change it's shape