Answer:
1,700feet
Explanation:
If an object in free fall travels a distance s that is directly proportional to the square of the time t, this can be represented mathematically as;
S = kt²where;
k is the proportionality constant
K = s/t²
s1/t1²= s2/t2²= Sn/tn²= k for values of the distance and time. Using the formula
s1/t1² = s2/t2² where;
s1 is the falling distance in time t1 s2 is the falling distance in time t2
Given s1 = 1088feet, t1 = 8secs, s2 = ? t2 = 10secs
Substituting this value in the formula to get s2, we have;
1088/8²= s2/10²
64s2= 108800
s2 = 108800/64
s2 = 1,700feet
This means the object will fall a distance of 1,700feet in 10seconds
120 km / 60 min
2 km / 1 min
1 km / 30 sec
0.5 km / 15 sec
0.03333 km / sec
0.03333 km = 33.33 m
F = G m1*m2 / r^2 => [G] = [F]*[r]^2 /([m1]*[m2]) = N * m^2 / kg^2
That is one answer.
Also, you can use the fact that N = kg*m/s^2
[G] = kg * m / s^2 * m^2 / kg^2 = m^3 /(s^2 * kg)
Answer:
The change in the charge on the positve plate when the Teflon is inserted is +2.5 nC.
Explanation:
- It can be showed that the capacitance of a parallel-plate capacitor, can be expresssed as follows:

- Where ε, is the dielectric constant of the material that fills the space between plates.
- When this space is filled with air, ε= ε₀ = 8,85*10⁻¹² F/m.
- At the same time, the capacitance of a capacitor, by definition, is as follows:

- If we insert a Teflon slab, in such a way that fills completely the gap between the plates, all other parameters being equal, if ε = 2*ε₀, this means that C₂ = 2* C₁. = 50 pF
- As V₂=V₁ (due to the capacitor remains connected to the same battery) the charge must be the double, so Q₂ = 2* Q₁ = 5 nC.
- So, the change in the charge of the positive plate is +2.5 nC.