Let, time taken is t.
So,
( Here, us is vertical component of initial velocity )
Now, distance covered from the base of the cliff is :
![D=5\times 3.873\ m\\\\D = 19.365\ m](https://tex.z-dn.net/?f=D%3D5%5Ctimes%203.873%5C%20m%5C%5C%5C%5CD%20%3D%2019.365%5C%20m)
Therefore, time taken and distance covered is 3.873 s and 19.365 m.
Hence, this is the required solution.
Answer:
La tensión es 85.3 N.
Explanation:
Cuando el objeto gira en dirección horizontal, la sumatoria de fuerzas se puede calcular usando la segunda ley de Newton:
![\Sigma F_{x} =ma_{c}](https://tex.z-dn.net/?f=%20%5CSigma%20F_%7Bx%7D%20%3Dma_%7Bc%7D%20)
Dado que el movimiento es horizontal, el peso (que está en el eje y) no contribuye en la sumatoria de fuerzas en el eje x. Por lo que la única fuerza actuando sobre el objeto en la dirección del movimiento es la tensión.
En donde:
m: es la masa del objeto = 200 g = 0.200 kg
: es la aceleración centrípeta
La aceleración centrípeta viene dada por:
![a_{c} = \omega^{2} r](https://tex.z-dn.net/?f=a_%7Bc%7D%20%3D%20%5Comega%5E%7B2%7D%20r)
En donde:
ω: es la velocidad angular del objeto = 3 rev/s
r: es el radio = 1.20 m
Entonces, la tensión es:
![T = m\omega^{2} r = 0.200 kg(3\frac{rev}{s}*\frac{2\pi rad}{1 rev})^{2}*1.20 m = 85.3 N](https://tex.z-dn.net/?f=T%20%3D%20m%5Comega%5E%7B2%7D%20r%20%3D%200.200%20kg%283%5Cfrac%7Brev%7D%7Bs%7D%2A%5Cfrac%7B2%5Cpi%20rad%7D%7B1%20rev%7D%29%5E%7B2%7D%2A1.20%20m%20%3D%2085.3%20N)
Por lo tanto, la tensión es 85.3 N.
Espero que te sea de utilidad!
Answer:
32.5 m/s
Explanation:
The total momentum must be conserved before and after the collision:
![p_i = p_f\\m_c u_c + m_t u_t = m_c v_c + m_t v_t](https://tex.z-dn.net/?f=p_i%20%3D%20p_f%5C%5Cm_c%20u_c%20%2B%20m_t%20u_t%20%3D%20m_c%20v_c%20%2B%20m_t%20v_t)
where
m_c = 300 kg is the mass of the car
m_t = 600 kg is the mass of the truck
u_c = 60 m/s is the initial velocity of the car
u_t = 10 m/s is the initial velocity of the truck
v_c = 15 m/s is the final velocity of the car
v_f is the final velocity of the truck
Solving for v_f, we find:
![v_f = \frac{m_c u_c + m_t u_t - m_c v_c}{m_t}=\frac{(300 kg)(60 m/s)+(600 kg)(10 m/s)-(300 kg)(15 m/s)}{600 kg}=32.5 m/s](https://tex.z-dn.net/?f=v_f%20%3D%20%5Cfrac%7Bm_c%20u_c%20%2B%20m_t%20u_t%20-%20m_c%20v_c%7D%7Bm_t%7D%3D%5Cfrac%7B%28300%20kg%29%2860%20m%2Fs%29%2B%28600%20kg%29%2810%20m%2Fs%29-%28300%20kg%29%2815%20m%2Fs%29%7D%7B600%20kg%7D%3D32.5%20m%2Fs)
Answer:
The difference in atomic mass between the two isotopes is 1.002780942 atomic mass unit.
Explanation:
For an isotope-I (heavier)
Mass of an isotope-I=M
Number of neutrons = n+1
Number of protons = p
![\Delta m_1=((n+1)\times m_n)+(p\times m_p))-M](https://tex.z-dn.net/?f=%5CDelta%20m_1%3D%28%28n%2B1%29%5Ctimes%20m_n%29%2B%28p%5Ctimes%20m_p%29%29-M)
For an isotope-II
Mass of an isotope-II=M'
Number of neutrons = n
Number of protons = p
![\Delta m_2=((n)\times m_n)+(p\times m_p))-M'](https://tex.z-dn.net/?f=%5CDelta%20m_2%3D%28%28n%29%5Ctimes%20m_n%29%2B%28p%5Ctimes%20m_p%29%29-M%27)
Difference in binding energy:
(general binding energy expression)
Binding energy difference between two isotopes:
..(1)
![5.4810 MeV=(\Delta m_1-\Delta m_2)c^2](https://tex.z-dn.net/?f=5.4810%20MeV%3D%28%5CDelta%20m_1-%5CDelta%20m_2%29c%5E2)
![=([((n+1)\times m_n)+(p\times m_p))-M]-[((n)\times m_n)+(p\times m_p))-M'])c^2](https://tex.z-dn.net/?f=%3D%28%5B%28%28n%2B1%29%5Ctimes%20m_n%29%2B%28p%5Ctimes%20m_p%29%29-M%5D-%5B%28%28n%29%5Ctimes%20m_n%29%2B%28p%5Ctimes%20m_p%29%29-M%27%5D%29c%5E2%3C%2Fp%3E%3Cp%3E)
B.E-B.E'=5.4810 MeV
![=([((n+1)\times m_n)+(p\times m_p))-M]-[((n)\times m_n)+(p\times m_p))-M'])c^2](https://tex.z-dn.net/?f=%3D%28%5B%28%28n%2B1%29%5Ctimes%20m_n%29%2B%28p%5Ctimes%20m_p%29%29-M%5D-%5B%28%28n%29%5Ctimes%20m_n%29%2B%28p%5Ctimes%20m_p%29%29-M%27%5D%29c%5E2)
![5.4810 MeV=[1\times m_n-M+M']c^2](https://tex.z-dn.net/?f=5.4810%20MeV%3D%5B1%5Ctimes%20m_n-M%2BM%27%5Dc%5E2)
![\frac{5.4810}{931.50} u=m_n-M+M'](https://tex.z-dn.net/?f=%5Cfrac%7B5.4810%7D%7B931.50%7D%20u%3Dm_n-M%2BM%27)
![M-M'=1.008665 u -\frac{5.4810}{931.50} u=1.002780942 u](https://tex.z-dn.net/?f=M-M%27%3D1.008665%20u%20-%5Cfrac%7B5.4810%7D%7B931.50%7D%20u%3D1.002780942%20u)
Answer:
The answer is <em>D. In the direction of the wave</em>
Explanation:
I took the test, hope this helps! :)