To solve this problem it is necessary to apply the concepts related to the Stefan-Boltzman law that is responsible for calculating radioactive energy.
Mathematically this expression can be given as
Where
A = Surface area of the Object
Stefan-Boltzmann constant
e = Emissivity
T = Temperature (Kelvin)
Our values are given as
Replacing at our equation and solving to find the temperature 1 we have,
Therefore the the temperature of the coldest room in which this person could stand and not experience a drop in body temperature is 12°C
Answer:A
Explanation:
In R-L circuit current is given by
![i=i_0\left [ 1-e^{\frac{-t}{L/R}}\right ]](https://tex.z-dn.net/?f=i%3Di_0%5Cleft%20%5B%201-e%5E%7B%5Cfrac%7B-t%7D%7BL%2FR%7D%7D%5Cright%20%5D)
where i=current at any time t
R=resistance
L=Inductance
at t=0 
approaches to 1
therefore ![i=i_0\left [ 1-1\right ]](https://tex.z-dn.net/?f=i%3Di_0%5Cleft%20%5B%201-1%5Cright%20%5D)
i=0
when t approaches to 
, approaches to zero
thus 
thus we can say that initially circuit act as broken wire with zero current
and it increases exponentially with time and act as ordinary connecting wire
Answer:
a) k = 120 N / m
, b) f = 0.851 Hz
, c) v = 1,069 m / s
, d) x = 0
, e) a = 5.71 m / s²
, f) x = 0.200 m
, g) Em = 2.4 J
, h) v = -1.01 m / s
Explanation:
a) Hooke's law is
F = k x
k = F / x
k = 24.0 / 0.200
k = 120 N / m
b) the angular velocity of the simple harmonic movement is
w = √ k / m
w = √ (120 / 4.2)
w = 5,345 rad / s
Angular velocity and frequency are related.
w = 2π f
f = w / 2π
f = 5.345 / 2π
f = 0.851 Hz
c) the equation that describes the movement is
x = A cos (wt + Ф)
As the body is released without initial velocity, Ф = 0
x = 0.2 cos wt
Speed is
v = dx / dt
v = -A w sin wt
The speed is maximum for sin wt = ±1
v = A w
v = 0.200 5.345
v = 1,069 m / s
d) when the function sin wt = -1 the function cos wt = 0, whereby the position for maximum speed is
x = A cos wt = 0
x = 0
e) the acceleration is
a = d²x / dt² = dv / dt
a = - Aw² cos wt
The acceleration is maximum when cos wt = ± 1
a = A w²
a = 0.2 5.345
a = 5.71 m / s²
f) the position for this acceleration is
x = A cos wt
x = A
x = 0.200 m
g) Mechanical energy is
Em = ½ k A²
Em = ½ 120 0.2²
Em = 2.4 J
h) the position is
x = 1/3 A
Let's calculate the time to reach this point
x = A cos wt
1/3 A = A cos 5.345t
t = 1 / w cos⁻¹(1/3)
The angles are in radians
t = 1.23 / 5,345
t = 0.2301 s
Speed is
v = -A w sin wt
v = -0.2 5.345 sin (5.345 0.2301)
v = -1.01 m / s
i) acceleration
a = -A w² sin wt
a = - 0.2 5.345² cos (5.345 0.2301)
a = -1.91 m / s²
Answer:
Neptune is approximately 41 times as far from the sun as Venus
Explanation:
Estimate = distance of Neptune from the sun ÷ distance of Venus from the sun = 4.5×10^9 ÷ 1.18×10^8 = 40.9 (approximately 41)
51 inches.
This is because a stem-plot is formatted as so:
If it’s 5 on the left side of the line, anything on the right is the ones place making possible numbers 51, 53, 56, etc.
Hope this helps!
