Wouldn't everything fall?
Answer:
Av = 25 [m/s]
Explanation:
To solve this problem we must use the definition of speed, which is defined as the relationship between distance over time. for this case we have.

where:
Av = speed [km/h] or [m/s]
distance = 180 [km]
time = 2 [hr]
Therefore the speed is equal to:
![Av = \frac{180}{2} \\Av = 90 [km/h]](https://tex.z-dn.net/?f=Av%20%3D%20%5Cfrac%7B180%7D%7B2%7D%20%5C%5CAv%20%3D%2090%20%5Bkm%2Fh%5D)
Now we must convert from kilometers per hour to meters per second
![90[\frac{km}{h}]*1000[\frac{m}{1km}]*1[\frac{h}{3600s} ]= 25 [m/s]](https://tex.z-dn.net/?f=90%5B%5Cfrac%7Bkm%7D%7Bh%7D%5D%2A1000%5B%5Cfrac%7Bm%7D%7B1km%7D%5D%2A1%5B%5Cfrac%7Bh%7D%7B3600s%7D%20%5D%3D%2025%20%5Bm%2Fs%5D)
I am not sure but i think the answer is C
0.02020 ohm is the resistance of a carbon rod at 25.8 ∘C if its resistance is 0.0200 Ω at 0.0 ∘C.
<h3 /><h3>What is a resistor?</h3>
A resistor is an electrical component that controls or restricts how much electrical current can pass across a circuit in an electronic device. A specified voltage can be supplied via resistors to an active device like a transistor.
The temperature of the resistor varies based on the variation in the temperature. The equation that describes the relationship between the two of them is:
R = R0[1+ alpha(T-T0)] where:
R is the new resistance we are looking for
alpha is the temperature coefficient of resistance. For carbon rod, alpha = ₋ 4.8 x
(1/°c)
T0 is the standard temperature =25.8°C
R0 is the resistance at T0 = 0.0200 ohms
T is the temperature at which we want to get R = 0
Substitute in the equation to get R as follows:
R = 0.0200 [1+( ₋ 4.8 x
) (0-25.8)] = 0.02020 ohm
To know more about resistance refer to: brainly.com/question/11431009
#SPJ1
The statement about pointwise convergence follows because C is a complete metric space. If fn → f uniformly on S, then |fn(z) − fm(z)| ≤ |fn(z) − f(z)| + |f(z) − fm(z)|, hence {fn} is uniformly Cauchy. Conversely, if {fn} is uniformly Cauchy, it is pointwise Cauchy and therefore converges pointwise to a limit function f. If |fn(z)−fm(z)| ≤ ε for all n,m ≥ N and all z ∈ S, let m → ∞ to show that |fn(z)−f(z)|≤εforn≥N andallz∈S. Thusfn →f uniformlyonS.
2. This is immediate from (2.2.7).
3. We have f′(x) = (2/x3)e−1/x2 for x ̸= 0, and f′(0) = limh→0(1/h)e−1/h2 = 0. Since f(n)(x) is of the form pn(1/x)e−1/x2 for x ̸= 0, where pn is a polynomial, an induction argument shows that f(n)(0) = 0 for all n. If g is analytic on D(0,r) and g = f on (−r,r), then by (2.2.16), g(z) =