The ball only accelerates during the brief time that the club is in contact
with it. After it leaves the club face, it takes off at a constant speed.
If it accelerates at 20 m/s² during the hit, then
Force = (mass) x (acceleration) = (0.2kg) x (20 m/s²) = <em>4 newtons</em> .
Answer:
The tangential speed of the tack is 6.988 meters per second.
Explanation:
The tangential speed experimented by the tack (
), measured in meters per second, is equal to the product of the angular speed of the wheel (
), measured in radians per second, and the distance of the tack respect to the rotation axis (
), measured in meters, length that coincides with the radius of the tire. First, we convert the angular speed of the wheel from revolutions per second to radians per second:
Then, the tangential speed of the tack is: (, 
)
The tangential speed of the tack is 6.988 meters per second.
That is correct. Or so I believe. Either more or less than the other on the amount of protons and electrons, you can get either an unstable or a stable atom of an element.
Answer: The electric field is given in three regions well defined; 0<r<2; 2<r<4; 4<r<5 and r>5
Explanation: In order to solve this problem we have to use the gaussian law in the mentioned regions.
Region 1; 0<r<2
∫E.ds=Qinside the gaussian surface/ε0
inside of the solid conducting sphere the elevctric field is zero because the charge is located at the surface on this sphere.
Region 2; 2<r<4;
E.4*π*r^2=8,84/ε0
E=8,84/(4*π*ε0*r^2)
Region 3; 4<r<5
E=0 because is inside the conductor.
Finally
Region 4; r>5
E.4*π*r^2=(8,84-2.02)/ε0
Answer:
v = K √(E / ρ)
Explanation:
Modulus of elasticity has units of N/m², or kg/m/s².
Density has units of kg/m³.
Velocity has units of m/s.
If we divide modulus of elasticity by density, we can eliminate kg:
E / ρ = [kg/m/s²] / [kg/m³]
E / ρ = [m²/s²]
Taking the square root gets us units of velocity:
√(E / ρ) = [m/s]
Multiply by the constant K:
v = K √(E / ρ)
