<span> attraction between the relative abundance of electrons in one object and protons in the other
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For this problem, we use the derived equations for rectilinear motion at constant acceleration. The equations used for this problem are:
a = (v - v₀)/t
2ax = v² - v₀²
where
a is the acceleration
x is the distance
v is the final velocity
v₀ is the initial velocity
t is the time
The solution is as follows;
a = (60mph - 30 mph)/(3 s * 1 h/3600 s)
a = 36,000 mph²
2(36,000 mph²)(x) = 60² - 30²
Solving for x,
x = 0.0375 miles
<span>1.0x10^3 Joules
The kinetic energy a body has is expressed as the equation
E = 0.5 M V^2
where
E = Energy
M = Mass
V = Velocity
Since the shot was at rest, the initial energy is 0. Let's calculate the energy that the shot has while in motion
E = 0.5 * 7.2 kg * (17 m/s)^2
E = 3.6 kg * 289 m^2/s^2
E = 1040.4 kg*m^2/s^2
E = 1040.4 J
So the work performed on the shot was 1040.4 Joules. Rounding the result to 2 significant figures gives 1.0x10^3 Joules</span>
Answer:
Opposite to the direction that you are pulling
Explanation:
Static friction acts in the opposite direction to the acceleration.
Kinetic friction acts in the opposite direction to the velocity.
