Answer:
a) a = 3.72 m / s², b) a = -18.75 m / s²
Explanation:
a) Let's use kinematics to find the acceleration before the collision
v = v₀ + at
as part of rest the v₀ = 0
a = v / t
Let's reduce the magnitudes to the SI system
v = 115 km / h (1000 m / 1km) (1h / 3600s)
v = 31.94 m / s
v₂ = 60 km / h = 16.66 m / s
l
et's calculate
a = 31.94 / 8.58
a = 3.72 m / s²
b) For the operational average during the collision let's use the relationship between momentum and momentum
I = Δp
F Δt = m v_f - m v₀
F = 
F = m [16.66 - 31.94] / 0.815
F = m (-18.75)
Having the force let's use Newton's second law
F = m a
-18.75 m = m a
a = -18.75 m / s²
A gas heated to millions of degrees would emit mostly x-rays.
Answer:
Wave speed, frequency and wavelength in refraction
Explanation:
The diagram shows that as a wave travels into a denser medium, such as water, it slows down and the wavelength decreases. Although the wave slows down, its frequency remains the same, due to the fact that its wavelength is shorter. Hope this helps :>
Answer:
a
The height is 
b
The horizontal distance is 
Explanation:
From the question we are told that
The speed is 
The angle is 
The height of the cannon from the ground is h = 2 m
The distance of the net from the ground is k = 1 m
Generally the maximum height she reaches is mathematically represented as
=> ![H = \frac{(15)^2 [sin (40)]^2 }{2 * 9.8} + 2](https://tex.z-dn.net/?f=H%20%20%3D%20%20%5Cfrac%7B%2815%29%5E2%20%5Bsin%20%2840%29%5D%5E2%20%7D%7B2%20%2A%209.8%7D%20%20%2B%20%202)
=>
Generally from kinematic equation
Here s is the displacement which is mathematically represented as
s = [-(h-k)]
=> s = -(2-1)
=> s = -1 m
There reason why s = -1 m is because upward motion canceled the downward motion remaining only the distance of the net from the ground which was covered during the first half but not covered during the second half
a = -g = -9.8
So
=> 
using quadratic formula to solve the equation we have
Generally distance covered along the horizontal is
=> 
=>
ΔVl = L di/dt
i = i₀e -t/T
di/dt = i₀ × (-1/T) e -t/T
ΔVl = L× (-I/T i₀e -t/T
ΔVl = -L/T i₀e -t/T
b. 15mm, i₀ = 36mA, T = 1.1m
t= Os
ΔVl = 0,491V
C. t = 1ms
ΔVl = 0.198V
t = 2ms
ΔVl = 0.08V
E. t = ms
ΔVl = 0.032V
