Answer:
The runner's acceleration was ![1\ m/s^2](https://tex.z-dn.net/?f=1%5C%20m%2Fs%5E2)
Explanation:
<u>Constant Acceleration Motion</u>
It's a type of motion in which the velocity of an object changes by an equal amount in every equal period of time.
Being a the constant acceleration, vo the initial speed, vf the final speed, and t the time, the following relation applies:
![v_f=v_o+at](https://tex.z-dn.net/?f=v_f%3Dv_o%2Bat)
Solving for a:
![\displaystyle a=\frac{v_f-v_o}{t}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20a%3D%5Cfrac%7Bv_f-v_o%7D%7Bt%7D)
The runner speeds up from vo=5 m/s to vf=9 m/s in t=4 seconds, thus:
![\displaystyle a=\frac{9-5}{4}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20a%3D%5Cfrac%7B9-5%7D%7B4%7D)
![\displaystyle a=\frac{4}{4}=1](https://tex.z-dn.net/?f=%5Cdisplaystyle%20a%3D%5Cfrac%7B4%7D%7B4%7D%3D1)
The runner's acceleration was ![1\ m/s^2](https://tex.z-dn.net/?f=1%5C%20m%2Fs%5E2)
Answer:
The correct option is C
Explanation:
From the question we are told that
The initial speed of the rocket is ![v_i = 100 m/s](https://tex.z-dn.net/?f=v_i%20%3D%20100%20m%2Fs)
The speed of the rocket engine sound is ![V](https://tex.z-dn.net/?f=V)
The final speed of the rocket is ![v_f = 200 \ m/s](https://tex.z-dn.net/?f=v_f%20%3D%20200%20%5C%20m%2Fs)
The speed of the sound at
would still remain V this because the speed of sound wave is constant and is not dependent on the speed of the observer(The mountain ) or the speed of the source (The rocket ).
A clear example when lightning strikes you will first see (that is because it travels at the speed of light which is greater than the speed of sound) but it would take some time before you hear the sound of the
lightning
Here we see that the speed of the lightning(speed of sound) does not affect the speed of the sound it generates
Answer:
option D
Explanation:
this is because it occurs in many different dimensions, including biological, cognitive and socioemotional. this is also the answer on apex.
1.549×10-19lJ is the energy of a photon emitted when an electron in a hydrogen atom undergoes a transition from =7 to =1.
The equation E= hcE =hc, where h is Planck's constant and c is the speed of light, describes the inverse relationship between a photon's energy (E) and the wavelength of light ().
The Rydberg formula is used to determine the energy change.
Rydberg's original formula used wavelengths, but we may rewrite it using units of energy instead. The result is the following.
aaΔE=R(1n2f−1n2i) aa
were
2.17810-18lJ is the Rydberg constant.
The initial and ultimate energy levels are ni and nf.
As a change of pace from
n=5 to n=3 gives us
ΔE
=2.178×10-18lJ (132−152)
=2.178×10-18lJ (19−125)
=2.178×10-18lJ×25 - 9/25×9
=2.178×10-18lJ×16/225
=1.549×10-19lJ
Learn more about Rydberg formula here-
brainly.com/question/13185515
#SPJ4
Answer:
a) Em₀ = 42.96 104 J
, b)
= -2.49 105 J
, c) vf = 3.75 m / s
Explanation:
The mechanical energy of a body is the sum of its kinetic energy plus the potential energies it has
Em = K + U
a) Let's look for the initial mechanical energy
Em₀ = K + U
Em₀ = ½ m v2 + mg and
Em₀ = ½ 50.0 (1.20 102) 2 + 50 9.8 142
Em₀ = 36 104 + 6.96 104
Em₀ = 42.96 104 J
b) The work of the friction force is equal to the change in the mechanical energy of the body
= Em₂ -Em₀
Em₂ = K + U
Em₂ = ½ m v₂² + m g y₂
Em₂ = ½ 50 85 2 + 50 9.8 427
Em₂ = 180.625 + 2.09 105
Em₂ = 1,806 105 J
= Em₂ -Em₀
= 1,806 105 - 4,296 105
= -2.49 105 J
The negative sign indicates that the work that force and displacement have opposite directions
c) In this case the work of the friction going up is already calculated in part b and the work of the friction going down would be 1.5 that job
We have that the work of friction is equal to the change of mechanical energy
= ΔEm
= Emf - Emo
-1.5 2.49 10⁵ = ½ m vf² - 42.96 10⁴
½ m vf² = -1.5 2.49 10⁵ + 4.296 10⁵
½ 50.0 vf² = 0.561
vf = √ 0.561 25
vf = 3.75 m / s