Two protons are 1 × 10−10 m apart (about one atomic radius). Which interaction between two protons is stronger, the gravitationa
l attraction or the electric repulsion? Note that since both the gravitational and electric forces depend on the inverse square distance, this comparison holds true at all distances, not just at a distance of 1 × 10−10 m. Use the accurate values of G = 6.67 × 10−11 N·m2/kg2, proton mass of 1.67 × 10−27 kg, and electron mass of 9.11 × 10−31 kg.
a) First, calculate the electric force |Felec| = ___ N
b) Next, calculate the gravitational force |F grav| = ___ N
c) Calculate the ratio of the two forces |Felec|/|Fgrav| = ___
a) First, calculate the electric force |Felec| = ___ N
b) Next, calculate the gravitational force |F grav| = ___ N
c) Calculate the ratio of the two forces |Felec|/|Fgrav| = ___
1 answer:
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Answer:
34.6 m/s
Explanation:
From conservation of momentum, the sum of initial and final momentum are equal. Momentum is a product of mass and velocity. Initial mass will be 42.8+31.5+25.9=100.2 kg
Final mass will be 31.5+25.9=57.4 kg
From formula of momentum
M1v1=m2v2
Making v2 the subject of the formula then
Substitute 100.2 kg for M1, 19.8 m/s fkr v1 and 57.4 kg for m2 then
Answer:
h=12.41m
Explanation:
N=392
r=0.6m
w=24 rad/s
So the weight of the wheel is the force N divide on the gravity and also can find momentum of inertia to determine the kinetic energy at motion
moment of inertia
Kinetic energy of the rotation motion
Kinetic energy translational
Total kinetic energy
Now the work done by the friction is acting at the motion so the kinetic energy and the work of motion give the potential work so there we can find height