The answer is c. +2.0 µC
To calculate this, we will use Coulomb's Law:
F = k*Q1*Q2/r²
where F is force, k is constant, Q is a charge, r is a distance between charges.
k = 9.0 × 10⁹ N*m/C²
It is given:
F = 7.2 N
d = 0.1 m = 10⁻¹ m
Q1 = -4.0 µC = 4 * 1.0 × 10⁻⁶ = 4.0 × 10⁻⁶
Q2 = ?
Thus, let's replace this in the formula for the force:
7.2 = 9.0 × 10⁹ * 4.0 × 10⁻⁶ * Q2/(10⁻¹)²
7.2 = 9 * 4 * 10⁹⁻⁶ * Q2/10⁻¹°²
7.2 = 36 × 10³ * Q2 / 10⁻²
Multiply both sides of the equation by 10⁻²:
7.2 × 10⁻² = 36 × 10³ * Q2
⇒ Q2 = 7.2 × 10⁻² / 36 × 10³ = 7.2/36 × 10⁻²⁻³ = 0.2 × 10⁻⁵ = 2 × 10⁻⁶
Since µC = 1.0 × 10^-6:
Q2 = 2 * 1.0 × 10^-6 = 2 µC
Hot body will lose heat from it, and that heat will goes out from it through radiation, so it's temperature will decrease after some time.
In same manner, cold body will take the heat, and it's temperature will increase
Hope this helps!
Answer:
Part a)

Part b)

Part c)

Part d)

Part e)

Part f)

Explanation:
As we know that catapult is projected with speed 19.9 m/s
so here we have


similarly we have


Part a)
Horizontal displacement in 1.03 s



Part b)
Vertical direction we have
![y = v_y t - \frac{1]{2}gt^2](https://tex.z-dn.net/?f=y%20%3D%20v_y%20t%20-%20%5Cfrac%7B1%5D%7B2%7Dgt%5E2)


Part c)
Horizontal displacement in 1.71 s



Part d)
Vertical direction we have
![y = v_y t - \frac{1]{2}gt^2](https://tex.z-dn.net/?f=y%20%3D%20v_y%20t%20-%20%5Cfrac%7B1%5D%7B2%7Dgt%5E2)


Part e)
Horizontal displacement in 5.44 s



Part f)
Vertical direction we have
![y = v_y t - \frac{1]{2}gt^2](https://tex.z-dn.net/?f=y%20%3D%20v_y%20t%20-%20%5Cfrac%7B1%5D%7B2%7Dgt%5E2)

