There was no net force on the stuffed toy, because the kids might have the same strength, The same force is on both sides of it. T<span>hey cancel each other out. They exert a force on the stuffed toy equal in strength but opposite in direction. The forces are balanced and the stuffed toy does not move. </span>Its like a game of tug-o-war, but you and I have the same strength. the rope would be still and not moving.
We have that The ratio U1/U2 of their potential energies due to their interactions with Q is
From the question we are told that
Question 1
Charge q1 is distance r from a positive point charge Q.
Question 2
Charge q2=q1/3 is distance 2r from Q.
Charge q1 is distance s from the negative plate of a parallel-plate capacitor.
Charge q2=q1/3 is distance 2s from the negative plate.
Generally the equation for the potential energy is mathematically given as

Therefore
The Equations of U1 and U2 is
For U1

For U2

Since
U is a function of q and q2=q1/3
Therefore

For Question 2
For U1

Therefore

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Answer:
T₂ =602 °C
Explanation:
Given that
T₁ = 227°C =227+273 K
T₁ =500 k
Gauge pressure at condition 1 given = 100 KPa
The absolute pressure at condition 1 will be
P₁ = 100 + 100 KPa
P₁ =200 KPa
Gauge pressure at condition 2 given = 250 KPa
The absolute pressure at condition 2 will be
P₂ = 250 + 100 KPa
P₂ =350 KPa
The temperature at condition 2 = T₂
We know that

T₂ = 875 K
T₂ =875- 273 °C
T₂ =602 °C
Answer:
v=8m/s
Explanation:
To solve this problem we have to take into account, that the work done by the friction force, after the collision must equal the kinetic energy of both two cars just after the collision. Hence we have
![W_{f}=E_{k}\\W_{f}=\mu N=\mu(m_1+m_1)g\\E_{k}=\frac{1}{2}[m_1+m_2]v^2](https://tex.z-dn.net/?f=W_%7Bf%7D%3DE_%7Bk%7D%5C%5CW_%7Bf%7D%3D%5Cmu%20N%3D%5Cmu%28m_1%2Bm_1%29g%5C%5CE_%7Bk%7D%3D%5Cfrac%7B1%7D%7B2%7D%5Bm_1%2Bm_2%5Dv%5E2)
where
mu: coefficient of kinetic friction
g: gravitational acceleration
We can calculate the speed of the cars after the collision by using

Now , we can compute the speed of the second car by taking into account the conservation of the momentum

the car did not exceed the speed limit
Hope this helps!!
As we know by the formula of bulk modulus

now we can rearrange it as


now the final volume after pressure is applied is given as

now we know that



now plug in all data


so volume is above after pressure is applied over it