Answer:
Because the wavelengths of macroscopic objects are too short for them to be detectable.
Explanation:
Wavelength of an object is given by de Broglie wavelength as:
![\lambda=\frac{h}{mv}](https://tex.z-dn.net/?f=%5Clambda%3D%5Cfrac%7Bh%7D%7Bmv%7D)
Where, 'h' is Planck's constant, 'm' is mass of object and 'v' is its velocity.
So, for macroscopic objects, the mass is very large compared to microscopic objects. As we can observe from the above formula, there is an inverse relationship between the mass and wavelength of the object.
So, for vary larger masses, the wavelength would be too short and one will find it undetectable. Therefore, we don't observe wave properties in macroscopic objects.
Answer:
68.6 J
Explanation:
Applying,
P.E = mgh............... Equation 1
Where P.E = Potential Energy of the basket, m = mass of the basket, g = acceleration due to gravity of the basket, h = height of the basket
From the question,
Given: m = 3.5 kg, h = 2.00 m
Constant: g = 9.8 m/s²
Substitute these values into equation 1
P.E = 3.5×2×9.8
P.E = 68.6 J
Hence the potential energy of the basket is 68.6 J
Assume the ball was thrown at ground level, so its initial position vector is
. Let
denote the ball's initial speed and
the angle at which the ball is thrown.
Over time, its position changes according to
, where the components
are determined by
![x=v_0\cos\theta\,t](https://tex.z-dn.net/?f=x%3Dv_0%5Ccos%5Ctheta%5C%2Ct)
![y=v_0\sin\theta\,t-\dfrac g2t^2](https://tex.z-dn.net/?f=y%3Dv_0%5Csin%5Ctheta%5C%2Ct-%5Cdfrac%20g2t%5E2)
where
is the acceleration due to gravity. We know the ball is in the air for 4.5 seconds and that it travels a horizontal distance of 45 meters, which means
![\begin{cases}45\,\mathrm m=v_0\cos\theta(4.5\,\mathrm s)\\0=v_0\sin\theta(4.5\,\mathrm s)-\frac g2(4.5\,\mathrm s)^2\end{cases}\implies\begin{cases}v_0\cos\theta=10\,\frac{\mathrm m}{\mathrm s}\\v_0\sin\theta=22.05\,\frac{\mathrm m}{\mathrm s}\end{cases}](https://tex.z-dn.net/?f=%5Cbegin%7Bcases%7D45%5C%2C%5Cmathrm%20m%3Dv_0%5Ccos%5Ctheta%284.5%5C%2C%5Cmathrm%20s%29%5C%5C0%3Dv_0%5Csin%5Ctheta%284.5%5C%2C%5Cmathrm%20s%29-%5Cfrac%20g2%284.5%5C%2C%5Cmathrm%20s%29%5E2%5Cend%7Bcases%7D%5Cimplies%5Cbegin%7Bcases%7Dv_0%5Ccos%5Ctheta%3D10%5C%2C%5Cfrac%7B%5Cmathrm%20m%7D%7B%5Cmathrm%20s%7D%5C%5Cv_0%5Csin%5Ctheta%3D22.05%5C%2C%5Cfrac%7B%5Cmathrm%20m%7D%7B%5Cmathrm%20s%7D%5Cend%7Bcases%7D)
Divide the second equation by the first to eliminate
, and we can solve for
, expecting some solution within
:
![\dfrac{v_0\sin\theta}{v_0\cos\theta}=\tan\theta=2.205\implies\theta\approx66^\circ](https://tex.z-dn.net/?f=%5Cdfrac%7Bv_0%5Csin%5Ctheta%7D%7Bv_0%5Ccos%5Ctheta%7D%3D%5Ctan%5Ctheta%3D2.205%5Cimplies%5Ctheta%5Capprox66%5E%5Ccirc)
Then plugging this into either equation above will tell us the ball's initial speed:
![v_0\cos66^\circ=10\,\dfrac{\mathrm m}{\mathrm s}\implies v_0\approx25\,\dfrac{\mathrm m}{\mathrm s}](https://tex.z-dn.net/?f=v_0%5Ccos66%5E%5Ccirc%3D10%5C%2C%5Cdfrac%7B%5Cmathrm%20m%7D%7B%5Cmathrm%20s%7D%5Cimplies%20v_0%5Capprox25%5C%2C%5Cdfrac%7B%5Cmathrm%20m%7D%7B%5Cmathrm%20s%7D)
Answer:
student 1 is right and student 2 is wrong
Explanation:
In the graph that the two students are observing is a graph of energy and frequency vs. wavelength, this graph is constructed using the relationship between a wave and its wavelength and frequency.
c = λ f
therefore for all waves we have the speed salami.
Therefore student 1 is right and student 2 is wrong