Q= mcΔT
Where Q is heat or energy
M is mass, c is heat capacitance and t is temperature
You have to convert Celsius into kelvin in order to use this formula I believe
Celsius + 273 = Kelvin
21 + 273 = 294K
363 + 273 = 636K
Now...
Q= (0.003)(0.129)(636-294)
Q= 0.132 J if you are using kilograms, in terms of grams which seems more appropriate the answer would be 132J of energy.
Answer:
α = 395 rad/s²
Explanation:
Main features of uniformly accelerated circular motion
A body performs a uniformly accelerated circular motion when its trajectory is a circle and its angular acceleration is constant (α = cte). In it the velocity vector is tangent at each point to the trajectory and, in addition, its magnitude varies uniformly.
There is tangential acceleration (at) and is constant.
at = α*R Formula (1)
where
α is the angular acceleration
R is the radius of the circular path
There is normal or centripetal acceleration that determines the change in direction of the velocity vector.
Data
R = 0.0600 m :blade radius
at = 23.7 m/s² : tangential acceleration of the blades
Angular acceleration of the blades (α)
We replace data in the formula (1)
at = α*R
23.7 = α*(0.06)
α = (23.7) / (0.06)
α = 395 rad/s²
This process shows alpha decay so X represents an alpha particle. The characteristic of alpha decay is that the mass number stays constant while the proton number increases.
The weight of the object on mars is about 80n
Answer:
a) R₁ = 14.1 Ω, b) R₂ = 19.9 Ω
Explanation:
For this exercise we must use ohm's law remembering that in a series circuit the equivalent resistance is the sum of the resistances
all resistors connected
V = i (R₁ + R₂)
with R₁ connected
V = (i + 0.5) R₁
with R₂ connected
V = (i + 0.25) R₂
We have a system of three equations with three unknowns for which we can solve it
We substitute the last two equations in the first
V = i ( 
)
1 = i ( 
)
1 = i ( 
) = 
i² + 0.75 i + 0.125 = 2i² + 0.75 i
i² - 0.125 = 0
i = √0.125
i = 0.35355 A
with the second equation we look for R1
R₁ = 
R₁ = 12 /( 0.35355 +0.5)
R₁ = 14.1 Ω
with the third equation we look for R2
R₂ = 
R₂ =
R₂ = 19.9 Ω
