Answer:
79.2 m/s
Explanation:
θ = angle at which projectile is launched = 29.7 deg
a = initial speed of launch = 130 m/s
Consider the motion along the vertical direction
v₀ = initial velocity along the vertical direction = a Sinθ = 130 Sin29.7 = 64.4 m/s
y = vertical displacement = - 108 m
a = acceleration = - 9.8 m/s²
v = final speed as it strikes the ground
Using the kinematics equation
v² = v₀² + 2 a y
v² = 64.4² + 2 (-9.8) (-108)
v = 79.2 m/s
It falls down and I hate that this makes me type a lot
Answer:
Explanation:
Given
charge on alpha particle=+2e
mass of alpha particle=
kg
Charge on gold nucleus=+79e
Velocity at r=1m is 
Using Energy conservation
Kinetic energy of particle will be converting to Potential energy as it approaches to nucleus
therefore


![\frac{1}{2}\left ( 6.64\times 10^{-27}\right )\left ( 1.9\times 10^{7}^2\right )=\frac{9\times 10^9\times 158\times \left ( 1.6\times 10^{-19}\right )}{y}\left [\frac{1}{r_0}-\frac{1}{1}\right ]](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%5Cleft%20%28%206.64%5Ctimes%2010%5E%7B-27%7D%5Cright%20%29%5Cleft%20%28%201.9%5Ctimes%2010%5E%7B7%7D%5E2%5Cright%20%29%3D%5Cfrac%7B9%5Ctimes%2010%5E9%5Ctimes%20158%5Ctimes%20%5Cleft%20%28%201.6%5Ctimes%2010%5E%7B-19%7D%5Cright%20%29%7D%7By%7D%5Cleft%20%5B%5Cfrac%7B1%7D%7Br_0%7D-%5Cfrac%7B1%7D%7B1%7D%5Cright%20%5D)
on solving we get


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Answer:
doubled
Explanation:
F=ma1----------(1)
2F = ma2-------(2)
Divide 2nd equation by 1st one
we get a1×2=a2