Explanation:
(a) Hooke's law:
F = kx
7.50 N = k (0.0300 m)
k = 250 N/m
(b) Angular frequency:
ω = √(k/m)
ω = √((250 N/m) / (0.500 kg))
ω = 22.4 rad/s
Frequency:
f = ω / (2π)
f = 3.56 cycles/s
Period:
T = 1/f
T = 0.281 s
(c) EE = ½ kx²
EE = ½ (250 N/m) (0.0500 m)²
EE = 0.313 J
(d) A = 0.0500 m
(e) vmax = Aω
vmax = (0.0500 m) (22.4 rad/s)
vmax = 1.12 m/s
amax = Aω²
amax = (0.0500 m) (22.4 rad/s)²
amax = 25.0 m/s²
(f) x = A cos(ωt)
x = (0.0500 m) cos(22.4 rad/s × 0.500 s)
x = 0.00919 m
(g) v = dx/dt = -Aω sin(ωt)
v = -(0.0500 m) (22.4 rad/s) sin(22.4 rad/s × 0.500 s)
v = -1.10 m/s
a = dv/dt = -Aω² cos(ωt)
a = -(0.0500 m) (22.4 rad/s)² cos(22.4 rad/s × 0.500 s)
a = -4.59 m/s²
Answer:![F_{net}=\frac{kq^2}{(L)^2}\left [ \frac{1}{2}+\sqrt{2}\right ]](https://tex.z-dn.net/?f=F_%7Bnet%7D%3D%5Cfrac%7Bkq%5E2%7D%7B%28L%29%5E2%7D%5Cleft%20%5B%20%5Cfrac%7B1%7D%7B2%7D%2B%5Csqrt%7B2%7D%5Cright%20%5D)
Explanation:
Given
Three charges of magnitude q is placed at three corners and fourth charge is placed at last corner with -q charge
Force due to the charge placed at diagonally opposite end on -q charge

where
Distance between the two charges

negative sign indicates that it is an attraction force
Now remaining two charges will apply the same amount of force as they are equally spaced from -q charge

The magnitude of force by both the charge is same but at an angle of 
thus combination of two forces at 2 and 3 will be

Now it will add with force due to 1 charge
Thus net force will be
![F_{net}=\frac{kq^2}{(L)^2}\left [ \frac{1}{2}+\sqrt{2}\right ]](https://tex.z-dn.net/?f=F_%7Bnet%7D%3D%5Cfrac%7Bkq%5E2%7D%7B%28L%29%5E2%7D%5Cleft%20%5B%20%5Cfrac%7B1%7D%7B2%7D%2B%5Csqrt%7B2%7D%5Cright%20%5D)
True would be the Correct answer
Answer:
Lorsque l'on détend l'air son volume augmente et sa pression diminue. L'air qui est un mélange de gaz est compressible et expansible. – Lorsque l'on comprime l'air, son volume diminue et sa pression augmente. – Lorsque l'on détend l'air, son volume augmente et sa pression diminue.
Answer:
699.67ft
Explanation:
We are given with,
- α = 1.2×10⁻⁵ / °C
- L₀ = 700 ft
- ΔT = -10°C − 30°C = -40°C
Now, We have to find ΔL:
- ΔL = (1.2×10⁻⁵ / °C) (700 ft) (-40°C)
- ΔL = −0.336
Rounded to two significant figures, the change in length is −0.33ft.
<u>Therefore, the final length is approximately 700 ft − 0.33 ft = 699.67ft</u>.