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:
![v = r\omega](https://tex.z-dn.net/?f=%20v%20%3D%20r%5Comega)
Explanation:
If the object is rolling without slipping, every unit of rotated angle equals to a distance perimeter rotated.
Suppose the object complete 1 revolution within time t. The angular distance is 2π rad. Its angular velocity is 2π/t
The distance it covered is its circumference, which is 2πr, and so the speed is 2πr/t
So the linear speed compared to angular speed is
![\frac{v}{\omega} = \frac{2\pi r/t}{2\pi /t} = r](https://tex.z-dn.net/?f=%20%5Cfrac%7Bv%7D%7B%5Comega%7D%20%3D%20%5Cfrac%7B2%5Cpi%20r%2Ft%7D%7B2%5Cpi%20%2Ft%7D%20%3D%20r)
![v = r\omega](https://tex.z-dn.net/?f=%20v%20%3D%20r%5Comega)
I'd say a weekly news magazine.
Answer:
The temperature of the metal is ![T_m = 376.8 ^o C](https://tex.z-dn.net/?f=T_m%20%20%3D%20%20376.8%20%5Eo%20C)
Explanation:
From the question we are told that
The mass of the metal is ![M = 60 \ kg](https://tex.z-dn.net/?f=M%20%3D%20%2060%20%5C%20kg)
The specific heat of the metal is ![c_p = 0.1027 kcal/(kg \cdot ^oC)](https://tex.z-dn.net/?f=c_p%20%20%3D%20%200.1027%20kcal%2F%28kg%20%5Ccdot%20%5EoC%29)
The mass of the oil is ![M_o = 810 \ kg](https://tex.z-dn.net/?f=M_o%20%20%3D%20%20810%20%5C%20kg)
The temperature of the oil is ![T_o = 35^oC](https://tex.z-dn.net/?f=T_o%20%20%3D%20%2035%5EoC)
The specific heat of oil is ![c_o = 0.7167 kcal/(kg \cdot ^oC )](https://tex.z-dn.net/?f=c_o%20%20%3D%20%200.7167%20kcal%2F%28kg%20%5Ccdot%20%5EoC%20%29)
The equilibrium temperature is ![T_e = 39 ^oC](https://tex.z-dn.net/?f=T_e%20%20%3D%20%2039%20%5EoC)
According to the law of energy conservation
Heat lost by metal = heat gained by the oil
So
The quantity of heat lost by the metal is mathematically represented as
![Q = - Mc_p \Delta T](https://tex.z-dn.net/?f=Q%20%3D%20%20-%20Mc_p%20%5CDelta%20T)
=> ![Q = -Mc_p (T_m - T_c)](https://tex.z-dn.net/?f=Q%20%3D%20%20-Mc_p%20%28T_m%20%20-%20%20T_c%29)
Where
the temperature of metal before immersion
The negative sign show heat lost
The quantity of gained t by the metal is mathematically represented as
![Q = M_o c_o \Delta T](https://tex.z-dn.net/?f=Q%20%3D%20%20M_o%20c_o%20%5CDelta%20T)
=> ![Q = M_o c_o (T_c - T_o)](https://tex.z-dn.net/?f=Q%20%3D%20%20M_o%20c_o%20%28T_c%20-%20T_o%29)
So
![Mc_p (T_m - T_c) = M_o c_o (T_c - T_o)](https://tex.z-dn.net/?f=Mc_p%20%28T_m%20%20-%20%20T_c%29%20%20%20%3D%20%20%20M_o%20c_o%20%28T_c%20-%20T_o%29)
substituting values
![- 60 * 0.1027 (T_m - 39) = 810 * 0.7167 * (39 - 35)](https://tex.z-dn.net/?f=-%2060%20%2A%200.1027%20%28T_m%20%20-%2039%29%20%20%20%3D%20%20%20810%20%2A%200.7167%20%2A%20%20%2839%20-%2035%29)
=> ![T_m = 376.8 ^o C](https://tex.z-dn.net/?f=T_m%20%20%3D%20%20376.8%20%5Eo%20C)
<span> The boiling point of water at sea level is 100 °C. At higher altitudes, the boiling point of water will be.....
a) higher, because the altitude is greater.
b) lower, because temperatures are lower.
c) the same, because water always boils at 100 °C.
d) higher, because there are fewer water molecules in the air.
==> e) lower, because the atmospheric pressure is lower.
--------------------------
Water boils at a lower temperature on top of a mountain because there is less air pressure on the molecules.
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I hope this is helpful. </span>