A.) For letter a, we use the law of universal gravitation using the constant G = 6.674×10−<span>11 m3</span>⋅kg−1⋅s−<span>2
Grav. F = G*m1*m2*(1/d^2)
m1 is mass of electron = </span>9.11 × 10-31<span> kg
m2 is mass of proton = </span>1.67 × 10<span>-27 kg
d = 4.5 nm = 4.5 x 10^-9 m
Grav F = 5.01 x 10^-51 N
b.) </span>For letter b, we use the Coulomb's using the constant k = 9×10^9 N
Electric force = k*Q1*Q2*(1/d^2)
Q1 is charge of electron = -1.6 × 10-19 C
Q2 is charge of proton = +1.6 × 10-19 C
Electric force = 1.14 x 10^-11 N
E=274J
h=140cm=1,4m
g≈9,8m/s²
m=?
E=mgh
m=E/gh=274J/9,8m/s²*1,4m≈20kg
"Non nobis Domine, non nobis, sed Nomini tuo da gloriam."
Regards M.Y.
Answer:
14.25 m/s
Explanation:
In this problem, we need to find the horizontal component of a ball thrown at a 27 degree angle at 16 m/s.
It can be given by :
So, the horizontal component of the ball is 14.25 m/s.
