V = (1 / p) × p0×V0 = (1 / 219kPa) × 117kPa × 3L = 1.06L
Answer:
Part a)
![a= 0.32 m/s^2](https://tex.z-dn.net/?f=a%3D%200.32%20m%2Fs%5E2)
Part b)
![F_c = 3.6 N](https://tex.z-dn.net/?f=F_c%20%3D%203.6%20N)
Part c)
![F_c = 5.5 N](https://tex.z-dn.net/?f=F_c%20%3D%205.5%20N)
Explanation:
Part a)
As we know that the friction force on two boxes is given as
![F_f = \mu m_a g + \mu m_b g](https://tex.z-dn.net/?f=F_f%20%3D%20%5Cmu%20m_a%20g%20%2B%20%5Cmu%20m_b%20g)
![F_f = 0.02(10.6 + 7)9.81](https://tex.z-dn.net/?f=F_f%20%3D%200.02%2810.6%20%2B%207%299.81)
![F_f = 3.45 N](https://tex.z-dn.net/?f=F_f%20%3D%203.45%20N)
Now we know by Newton's II law
![F_{net} = ma](https://tex.z-dn.net/?f=F_%7Bnet%7D%20%3D%20ma)
so we have
![F_p - F_f = (m_a + m_b) a](https://tex.z-dn.net/?f=F_p%20-%20F_f%20%3D%20%28m_a%20%2B%20m_b%29%20a)
![9.1 - 3.45 = (10.6 + 7) a](https://tex.z-dn.net/?f=9.1%20-%203.45%20%3D%20%2810.6%20%2B%207%29%20a)
![a = \frac{5.65}{17.6}](https://tex.z-dn.net/?f=a%20%3D%20%5Cfrac%7B5.65%7D%7B17.6%7D)
![a= 0.32 m/s^2](https://tex.z-dn.net/?f=a%3D%200.32%20m%2Fs%5E2)
Part b)
For block B we know that net force on it will push it forward with same acceleration so we have
![F_c - F_f = m_b a](https://tex.z-dn.net/?f=F_c%20-%20F_f%20%3D%20m_b%20a)
![F_c = \mu m_b g + m_b a](https://tex.z-dn.net/?f=F_c%20%3D%20%5Cmu%20m_b%20g%20%2B%20m_b%20a)
![F_c = 0.02(7)(9.8) + 7(0.32)](https://tex.z-dn.net/?f=F_c%20%3D%200.02%287%29%289.8%29%20%2B%207%280.32%29)
![F_c = 3.6 N](https://tex.z-dn.net/?f=F_c%20%3D%203.6%20N)
Part c)
If Alex push from other side then also the acceleration will be same
So for box B we can say that Net force is given as
![F_p - F_f - F_c = m_b a](https://tex.z-dn.net/?f=F_p%20-%20F_f%20-%20F_c%20%3D%20m_b%20a)
![9.1 - 0.02(7)(9.8) - F_c = 7(0.32)](https://tex.z-dn.net/?f=9.1%20-%200.02%287%29%289.8%29%20-%20F_c%20%3D%207%280.32%29)
![F_c = 9.1 - 0.02 (7)(9.8) - 7(0.32)](https://tex.z-dn.net/?f=F_c%20%3D%209.1%20-%200.02%20%287%29%289.8%29%20-%207%280.32%29)
![F_c = 5.5 N](https://tex.z-dn.net/?f=F_c%20%3D%205.5%20N)
Answer:
Δm Δt> h ’/ 2c²
Explanation:
Heisenberg uncertainty principle, stable uncertainty of energy and time, with the expressions
ΔE Δt> h ’/ 2
h’= h / 2π
to relate this to the masses let's use Einstein's relationship
E = m c²
let's replace
Δ (mc²) Δt> h '/ 2
the speed of light is a constant that we can condense exact, so
Δm Δt> h ’/ 2c²
Answer:
Increased by 16 times
Explanation:
F = Gravitational force between two bodies
G = Gravitational constant = 6.67408 × 10⁻¹¹ m³/kg s²
m₁ = Mass of one body
m₂ = Mass of other body
d = distance between the two bodies
![F=\frac{Gm_1m_2}{d^2}\\ F=\frac{1}{d^2}\quad \text {(as G and masses are constant)}](https://tex.z-dn.net/?f=F%3D%5Cfrac%7BGm_1m_2%7D%7Bd%5E2%7D%5C%5C%20F%3D%5Cfrac%7B1%7D%7Bd%5E2%7D%5Cquad%20%5Ctext%20%7B%28as%20G%20and%20masses%20are%20constant%29%7D)
![F_{new}=\frac{1}{\left (\frac{d}{4}\right )^2}\\\Rightarrow F_{new}=\frac{1}{\frac{d^2}{16}}\\\Rightarrow F_{new}={16}\times \frac{1}{d^2}\\\Rightarrow F_{new}=16\times F](https://tex.z-dn.net/?f=F_%7Bnew%7D%3D%5Cfrac%7B1%7D%7B%5Cleft%20%28%5Cfrac%7Bd%7D%7B4%7D%5Cright%20%29%5E2%7D%5C%5C%5CRightarrow%20F_%7Bnew%7D%3D%5Cfrac%7B1%7D%7B%5Cfrac%7Bd%5E2%7D%7B16%7D%7D%5C%5C%5CRightarrow%20F_%7Bnew%7D%3D%7B16%7D%5Ctimes%20%5Cfrac%7B1%7D%7Bd%5E2%7D%5C%5C%5CRightarrow%20F_%7Bnew%7D%3D16%5Ctimes%20F)
∴Force will increase 16 times
<span>Acceleration is a change in motion ... slowing down, speeding up,
or changing direction even if speed doesn't change</span>.