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3 years ago

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On a 10 kg cart (shown below), the cart is brought up to speed with 50N of force for 7m, horizontally. At this point (A), the ca

rt begins to experience an average frictional force of 15N throughout the ride.
Find:
a) The total energy at (A)
b) The velocity at (B)
c) The velocity at (C)
d) Can the cart make it to Point (D)? Why or why not?

1 answer:

pav-90 [236]3 years ago

3 0

Answer: Letter B! Is your answer

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At each corner of a square of side l there are point charges of magnitude Q, 2Q, 3Q, and 4Q.What is the magnitude and direction

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Answer:

$F_T=6k\frac{Q^2}{L}\hat{i}+10k\frac{Q^2}{L}\hat{j}=2k\frac{Q^2}{L}[3\hat{i}+5\hat{j}]$

$|F_T|=2\sqrt{34}k\frac{Q^2}{L}$

$\theta=tan^{-1}(\frac{5}{3})=59.03\°$

Explanation:

I attached an image below with the scheme of the system:

The total force on the charge 2Q is the sum of the contribution of the forces between 2Q and the other charges:

$F_T=F_Q+F_{3Q}+F_{4Q}\\\\F_T=k\frac{(Q)(2Q)}{R_1}\hat{i}+k\frac{(3Q)(2Q)}{R_2}\hat{j}+k\frac{(4Q)(2Q)}{R_3}[cos\theta \hat{i}+sin\theta \hat{j}]$

the distances R1, R2 and R3, for a square arrangement is:

R1 = L

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R3 = (√2)L

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$F_T=k\frac{2Q^2}{L}\hat{i}+k\frac{6Q^2}{L}\hat{j}+k\frac{8Q^2}{\sqrt{2}L}[cos(45\°)\hat{i}+sin(45\°)\hat{j}]\\\\F_T=k\frac{2Q^2}{L}\hat{i}+k\frac{6Q^2}{L}\hat{j}+k\frac{8Q^2}{\sqrt{2}L}[\frac{\sqrt{2}}{2}\hat{i}+\frac{\sqrt{2}}{2}\hat{j}]\\\\F_T=6k\frac{Q^2}{L}\hat{i}+10k\frac{Q^2}{L}\hat{j}=2k\frac{Q^2}{L}[3\hat{i}+5\hat{j}]$

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$|F_T|=2k\frac{Q^2}{L}\sqrt{3^2+5^2}=2\sqrt{34}k\frac{Q^2}{L}$

the direction is:

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3 years ago

A spring stretches 0.150 m when a 0.30 kg mass is hung from it. The spring is then stretched an additional 0.100 m from this equ

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Answer:

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b) $V_m=0.81m/s$

c) $a_m=6.561m/s^2$

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$V_m=A\omega\\\\V_m=8.1*0.1$

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