Recent content by Mirole

How do you know their in parallel?

Homework Statement A 100 μμF capacitor, C1, is charged to a potential difference of 50 Volts. The charging battery is disconnected and the capacitor is connected to a second uncharged capacitor, C2. If the potential across C1 drops to 35 volts, what is the capacitance of C2? Homework...

So, for the first part of the question where the point is in the middle, it would be 0, right? Since you have the same charges, opposite signs; if you add them up, you'll get 0.

We have two charges, and then one in the middle. Which one goes into Q?

Homework Statement What is the difference in electric potential between a point midway between the charges and a point 0.05 meters away from the positive charge but along the line between the charges? Homework Equations UE = K * (Q1*Q2)/r The Attempt at a Solution K = 9.0x109Nm2/C2 UE =...

Using F = G * (m1*m2)/(r2) We know that G = 6.6673x10-11 N*m2/kg2 m1(earth) = 6x1024 kg m2(moon) = 7.4x1022 kg the Earth and moon are an average of 3.9x108m apart would r2 be [3.9x108]2 or [3.9x108 / 2]2? -Thanks!

A bacteria culture initially contains P(o) cells and grows at the rate dP/dt = kP where k is a growth constant. After an hour the population has doubled. (a) Determine an expression for the number of bacteria present after t hours. (b) Computer the number of bacteria present, and the rate of...

Nevermind, I got it to be -x+\pi/2, which is correct!

f(x) = cosx, a = \pi/2 since, L(x)=f'(x)(x-a) -f(a) f'(x) = -sinx = -sin(\pi/2)(x-\pi/2) - f(a) I'm stuck as to where to go next, is this even right?

No, because fk = μkn. Where n is the normal force. Using the two vertical forces from your FBD: (Fnet)y = n - Fg = 0(since your object is not moving up or down.) (Fnet)y = n = Fg => n = mg So, fk = (μk)(mg)

Yes, Fpush - fk is Fnet, and since the box is moving at a constant speed, ma = 0. So, Fpush - fk = 0, you know the rest. mu_k is the co-efficient of friction, μk. μ is mu.

Are you sure it's 20kg and not 2.0kg? Because: Fpush - fk = 0 3 = mu_k*mg 3 = mu_k*(2.0)(9.8) mu_k = .15, which is your answer. Now, using 20: Fpush - fk = 0 3 = mu_k*mg 3 = mu_k*(20)(9.8) mu_k = .015 Also, 20kg(44.1 pounds) would need a bit more than 3 Newtons of force to push it constantly...

Did you draw a FBD?

Wow, I did that by mistake and didn't even catch it. Thanks, so would 1.97m/s2 be correct?

Homework Statement You, a 75-kg skier, glide straight down a snow-covered slope inclined at 15 degrees to the horizontal over spring break. Let's be realistic, what is your acceleration(magnitude and direction) Assume your skis are wood and the snow is dry. mu_k on snow is 0.060. Homework...