False because they let it in to the atmosphere and then its condenced
Answer:
ΔD = 2.29 10⁻⁵ m
Explanation:
This is a problem of thermal expansion, if the temperature changes are not very large we can use the relation
ΔA = 2α A ΔT
the area is
A = π r² = π D² / 4
we substitute
ΔA = 2α π D² ΔT/4
as they do not indicate the initial temperature, we assume that ΔT = 75ºC
α = 1.7 10⁻⁵ ºC⁻¹
we calculate
ΔA = 2 1.7 10⁻⁵ pi (1.8 10⁻²) ² 75/4
ΔA = 6.49 10⁻⁷ m²
by definition
ΔA = A_f- A₀
A_f = ΔA + A₀
A_f = 6.49 10⁻⁷ + π (1.8 10⁻²)² / 4
A_f = 6.49 10⁻⁷ + 2.544 10⁻⁴
A_f = 2,551 10⁻⁴ m²
the area is
A_f = π D_f² / 4
A_f =
D_f =
D_f = 1.80229 10⁻² m
the change in diameter is
ΔD = D_f - D₀
ΔD = (1.80229 - 1.8) 10⁻² m
ΔD = 0.00229 10⁻² m
ΔD = 2.29 10⁻⁵ m
To calculate the change in kinetic energy, you must know the force as a function of position. The work done by the force causes the kinetic energy change
Explanation:
The work-energy theorem states that the change in kinetic enegy of an object is equal to the work done on the object:

where the work done is the integral of the force over the position of the object:

As we see from the formula, the magnitude of the force F(x) can be dependent from the position of the object, therefore in order to solve correctly the integral and find the work done on the object, it is required to know the behaviour of the force as a function of the position, x.
Answer:
F=1.65 x 10²⁶ N
Explanation:
Given that
Distance ,R= 3.34 x 10¹² m
Mass m₁= 2.78 x 10³⁰ kg
Mass ,m₂= 9.94 x 10³⁰ kg
we know that gravitational force F given as

G=Constant
G=6.67 x 10⁻¹¹ Nm²/kg²
Now by putting the values

F=1.65 x 10²⁶ N
Therefore the force between these two mass will be 1.65 x 10²⁶ N.
(a) 
The energy of a photon is given by:

where
is the Planck constant
is the speed of light
is the wavelength
For the microwave photon,

So the energy is

And converting into electronvolts,

(b) 
For the energy of the photon, we can use the same formula:

For the visible light photon,

So the energy is

And converting into electronvolts,

(c) 
For the energy of the photon, we can use the same formula:

For the x-ray photon,

So the energy is

And converting into electronvolts,
