A tennis ball with a velocity of +10.0 m/s to the right is thrown perpendicularly at a wall. After striking the wall, the ball r
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Answer:
a) F = (m / t₀) 95.33
b) θ = 70.5º
Explanation:
This is a projectile launch, as they indicate the horizontal distance this is the range of the body, let's use the expression for the range of the projectiles
R = v₀² sin 2θ / g
v₀² sin 2θ = R g
Where the range is 550.46 m
They also indicate the time that the air must remain, so this time is twice the time until reaching the maximum height.
v_{y} = - g t
At the maximum height v_{y} = 0 and the initial speed on the axis and we can find it with trigonometry
sin θ = v_{oy} / v_{o}
v_{oy} = v_{o} sin θ
v_{o} sin θ = g t
Let's write the two equations
v_{oy}² sin 2θ = g R
v_{o} sin θ = g t
We solve our accusation system
(G t / sin θ) 2 sin 2θ = g R
g t² sin 2θ = R sin θ
Let's use the trigonometric relationship
sin 2θ = 2 sin θ cos θ
We substitute
g t² (2 sin θ cos θ) = R sin θ
Cos tea = R / 2 g t²
θ = cos⁻¹ (R / 2g t²)
Let's calculate
θ = cos⁻¹ (550.46 / (2 9.8 9.17² ))
θ = 70.5º
a) Force can be Newton's second law
On the x axis the speed is constant so the force on the axis is zero
In the y axis the acceleration we have is the acceleration of gravity, so the force that acts throughout the journey is the weight of the body.
To place the body in the air from the rest we can use the equation of the impulse
F t = Δp = m v - m v₀
As kick from rest v₀ = 0
Let's find the speed of the body
= g t
v_{o} = g t / sin 70.5
v_{o} = 9.8 9.17 / sin 70.5
v_{o} = 95.33 m / s
To encocorate the force we must suppose a firing time, which in general is very short, suppose that this time is to
F = m v_{o} / t₀
F = (m / t₀) 95.33
This is the outside that should be applied, as an example suppose a body of mass 1 kg⁵ ( m = 1 kg) and a trip time to = 0.1 s
F = (1 / 0.1) 95.33
F = 953.3 N
Answer:
Part a)
Part b)
So density of atmosphere at Martian Surface is very less than the density at Earth.
Explanation:
Part a)
As per ideal gas equation we know that
here we know that Martian atmosphere is equivalent to that of carbon
so we will have
now we will have
Part b)
Now for the earth surface the density of air is given for
so we will have
So density of atmosphere at Martian Surface is very less than the density at Earth.
Answer:
a) What are the characteristics of the radiation emitted by a blackbody?
The total emitted energy per unit of time and per unit of area depends in its temperature (Stefan-Boltzmann law).
The peak of emission for the spectrum will be displaced to shorter wavelengths as the temperature increase (Wien’s displacement law).
The spectral density energy is related with the temperature and the wavelength (Planck’s law).
b) According to Wien's Law, how many times hotter is an object whose blackbody emission spectrum peaks in the blue, at a wave length of 450 nm, than a object whose spectrum peaks in the red, at 700 nm?
The object with the blackbody emission spectrum peak in the blue is 1.55 times hotter than the object with the blackbody emission spectrum peak in the red.
Explanation:
A blackbody is an ideal body that absorbs all the thermal radiation that hits its surface, thus becoming an excellent emitter, as these bodies express themselves without light radiation, and therefore they look black.
The radiation of a blackbody depends only on its temperature, thus being independent of its shape, material and internal constitution.
If it is study the behavior of the total energy emitted from a blackbody at different temperatures, it can be seen how as the temperature increases the energy will also increase, this energy emitted by the blackbody is known as spectral radiance and the result of the behavior described previously is Stefan's law:
(1)
Where is the Stefan-Boltzmann constant and T is the temperature.
The Wien’s displacement law establish how the peak of emission of the spectrum will be displace to shorter wavelengths as the temperature increase (inversely proportional):
(2)
Planck’s law relate the temperature with the spectral energy density (shape) of the spectrum:
(3)
b) According to Wien's Law, how many times hotter is an object whose blackbody emission spectrum peaks in the blue, at a wavelength of 450 nm, than a object whose spectrum peaks in the red, at 700 nm?
It is need it to known the temperature of both objects before doing the comparison. That can be done by means of the Wien’s displacement law.
Equation (2) can be rewrite in terms of T:
(4)
Case for the object with the blackbody emission spectrum peak in the blue:
Before replacing all the values in equation (4), (450 nm) will be express in meters:
⇒
Case for the object with the blackbody emission spectrum peak in the red:
Following the same approach above:
⇒
Comparison:
The object with the blackbody emission spectrum peak in the blue is 1.55 times hotter than the object with the blackbody emission spectrum peak in the red.