So we want find the coefficient of friction, which comes from F = (mu)*N, where F is the force of static friction, mu is the coefficient of friction and N is the normal force of the object. We can talk about how the penny would move after this force is reached with F = ma, where a is acceleration and m is the mass of the penny. The maximum acceleration of an oscillatory motion is a = Aω², where A is the amplitude and ω is the angular frequency. To turn ω into the regular frequency we're given, we can use ω = 2πf. Using this, our max acceleration becomes a = A(2πf)², which we can put into F=ma, so it becomes F = m(4Aπ²f²), then we can set it equal to F = (mu)N, so 4Amπ²f² = (mu)N. N, the normal force, will equal the weight force of the penny (draw a free body diagram), W = mg, where m is the mass of the penny and g is the gravitational constant (9.8m/s²). So N = W = mg, so we sub mg in for N, and we get 4Amπ²f² = (mu)mg, we solve for mu, cancelling our m's (good thing too since we don't know the mass of the penny) and we get mu = 4Aπ²f²/g, then we plug stuff in (convert mm to m by multiplying by 10⁻³) mu = 4(40*10⁻³)(3.14)²(1.75)²/(9.8) = 0.49 for our coefficient of friction. Lots of equations, but I hope that made sense!
F = k(Q₁Q₂)/r²
where k is Coulomb's constant 8.99e9
F = 8.99e9(2.8e-6)²/0.5² = 0.2819N
Answer:
Exothermic reactions increase the entropy of the surroundings. Simply put, entropy measures the dispersal of energy. Since ΔH is negative in an exothermic reaction, this must mean that ΔS will take on a positive value, indicating an increase in entropy.
Answer:
159 N
Explanation:
The force of friction, Fr is a product of coefficient of feiction and the normal force. Therefore, Fr=uN where N is the normal force and u is coefficient of friction. Here, we have two coefficients of friction but since it is sliding, then we use coefficient of kinetic energy. Substituting 0.25 for u and 636 N for N then
Fr=0.25*636=159 N
Therefore, the force of friction is equivalent to 159 N
Displacement is the area under the velocity/time graph. So for example this object's displacement in the first 3 seconds is (1/2)(3sec)(12.5 m/s)= 18.75m. (and then it starts backing up, displacement decreasing, after 3sec when velocity is negative).
But This object is never speeding up. Its velocity is smoothly decreasing at (25/6) m/s^2 (the slope of the graph). So the answer to the question is actually zero.
