Answer
Assuming
east is the positive x direction
north is the positive y direction
initial velocity , u = 19 j m/s
a)
acceleration , a = 1.6 j m/s^2
Using first equation of motion
v = u + a × t
v = 19 + 5.6× 1.6
v = 28 j m/s
the velocity of the car after 5.6 s is 28 m/s north
b)
acceleration , a = -1.5 j m/s^2
Using first equation of motion
v = u + a × t
v = 19 - 5.6 ×1.5
v = 10.6 j m/s
the velocity of the car after 5.6 s is 10.6 m/s north
The frictional force is given by F = μmg
<span>where μ is the coeficient of friction. </span>
<span>Work done by frictional force = Fd = μmgd </span>
<span>Kinetic energy "lost" = 1/2 mv² </span>
<span>Fd = μmgd = 1/2 mv² </span>
<span>The m's cancel μgd = v² / 2 </span>
<span>d = v² / 2μg </span>
<span>d = 8² / 2(0.41)(9.8) </span>
<span>d = 32 / (0.41)(9.8) </span>
<span>d = 7.96 </span>
<span>Player slides 8 m . </span>
<span>Note. In your other example μ = 0.46 and v = 4 m/s </span>
<span>d = v² / 2μg </span>
<span>= 4² / 2(0.46)(9.8) </span>
<span>= 8 / (0.46)(9.8) </span>
<span>= 1.77 or 1.8 m.
</span>
Hope i Helped :D
5.6 g/ml. That is the density.
Answer:
I'm not sure if I know whatever the answer is
Answer:
c. 2 MeV.
Explanation:
The computation of the binding energy is shown below
![= [Zm_p + (A - Z)m_n - N]c^2\\\\=[(1) (1.007825u) + (2 - 1 ) ( 1.008665 u) - 2.014102 u]c^2\\\\= (0.002388u)c^2\\\\= (.002388) (931.5 MeV)\\\\=2.22 MeV](https://tex.z-dn.net/?f=%3D%20%5BZm_p%20%2B%20%28A%20-%20Z%29m_n%20-%20N%5Dc%5E2%5C%5C%5C%5C%3D%5B%281%29%20%281.007825u%29%20%2B%20%282%20-%201%20%29%20%28%201.008665%20u%29%20-%202.014102%20u%5Dc%5E2%5C%5C%5C%5C%3D%20%280.002388u%29c%5E2%5C%5C%5C%5C%3D%20%28.002388%29%20%28931.5%20MeV%29%5C%5C%5C%5C%3D2.22%20MeV)
= 2 MeV
As 1 MeV = (1 u) c^2
hence, the binding energy is 2 MeV
Therefore the correct option is c.
We simply applied the above formula so that the correct binding energy could come
And, the same is to be considered
