W=-qEd
=-(1.6*10^-19)(23)(0.75)
= -2.76*10^-18 J
However, the answer is 2.76*10^-18 J. Why is the word done positive and not negative? Since it's traveling in the same direction as the electric field, shouldn't it be negative work?
Summary:: Confused on correction for contunuity
I'm confused about one step for questions like this where you have to determine to accept or reject the null hypothesis.
Sample size: 50
Number of successes: 23
Significance level: 10%
Null hypothesis: p = 0.4
Accepted hypothesis: p > 0.4...
Question:
In finance, the strong form of the efficient market hypothesis states that studying financial information about stocks is a waste of time since all public and private information that might affect the stock price is already reflected in the price of the stock. However, a study of 450...
e.g. 10.00cm = 0.0010 m, 16.38cm = 0.1638 m, 29.10 cm = 0.2910 m
Or, would the sig figs for the meters be to 2 digits after the decimal (e.g. 16.38 cm = 0.16 m)
I tried using the equation E(k)=1/2mv^2 and isolating for v but no mass was given. Then, I tried W=fd but there is no distance given. I don't know how to solve this.
F(net)=F(app)-F(resistance)
ma=F(app)-F(resistance)
(1.5*10^3)(16.67)=F(app)-600
25005=F(app)-600
P=ΔE/Δt
P=25005/6
P=4167.5 W
I got the incorrect answer (because I didn't take into account the resistance?). I am also confused about which Force value to use for the...
I still wouldn't be able to calculate velocity for part b by using kinematics as I have 3 missing values (no acceleration, no final velocity, and no displacement). Would I have to assume that the velocity is constant?
a) E(mech)=E(k)+E(g)
E(mech)=1/2mv^2+(0.2)(9.8)(0)
E(mech)=19.6 J
b) E(mech1)=E(mech2)
E(k)+E(g)=E(k)+E(g)
E(g)=E(k)+E(g)
0=1/2mv^2+(mgh)
*No height is given so I can't solve using this method. It says instantaneous velocity meaning the velocity at...