F.B.D Of first block
(I have shown only the horizontal Forces)
f1(max) = μ (1kg)(g) = 0.5 * 10 = 5N
F.B.D Of the second Block
f2(max) = μ (3kg)(g) = 15N
Now the string will become taut and the tension will start acting when f = t = 5N
But for 0<f<5N there will be no motion between the 1 kg...
Since the question says that "velocity along the cylinder axis" and "magnetic field perpendicular to the cylinder axis". So cross product of velocity and magnetic field becomes their magnitude.
##\vec v\times \vec B=||v|| \\ ||B||##
So
##\vec F=qvB##
##mg=qv\frac{\mu_0 nI}{4\pi r}##
At first...
My attempt :
##\frac{\vec Ft^2}{2}=m\vec s##
##s=\frac{Ft^2}{2m}##
##P=\frac{W}{t}##
##k=\frac{\vec F\cdot \vec s}{t}##
##k=\frac{F^2t^2}{2mt}##
##k=\frac{F^2t}{2m}##
##F=\sqrt{\frac{2mk}{t}}##
But there was an option which was ##2\sqrt{\frac{mk}{t}}##. And my assumption was that it was...
I don’t know what is contact force. Are friction and normal forces called contact forces? And we have to take the resultant of the two to get the net contact force?
$$a= -40/(10+6+4)$$
$$a=-2 m/s^2$$
Taking one mass of 10 kg.
$$T-40=10(-2)$$
$$T=20 N$$
This is correct.
But if I make the eqn of the system then
$$-40+T-T+T-T=20(-2)$$
I have also drawn the diagram. It looks like the second body m2 is subject to no force. But it’s accelerating. How?
Q1a)
- My current wrong answer is <-5.9e7, -3.3e8, -2.17e8> I used the Fnet = Gm1m2 / d^2 <unit vector> But i keep getting a dif answer each time
Q2a) - I thought i could find net force and then divide it by the mass, and multiply it by the time interval. However I got the answer <-4.5, 4.5...
Hello All, :)
Please see the picture , input force F1 and area A1 and second area A2 are all same for 3 cases .
in case 2 and 3 , red open outline represents a free-to-move , ie , "T" piece move upward
for the 1st case , output force F2 should be F2=A2/A1 x F1 , by pascal's principle ...
Hello All,
pls see picture , input force F1 and area A1 and second area A2 are all same for 3 cases
for the 1st case , output force F2 should be F2=A2/A1 x F1 , by pascal's principle ,
what about output forces of F3 and F4 ? can be same as F2 ? how i calculate them ?
thanks
The solution is :
The pressure inside the solid is zero and outside it equals atmospheric pressure, 1.01 × 10^5 Pa.
Thus, the force is given by: F =pA = 1.01 × 10^5 × (0.25)^2 = 6.3 × 10^3 N
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How did you find PF?: Google
Hi 👋
I have a question 🙋♀️ that’s needs help
Can I calculate the following ?
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May I ask if the following process is correct?
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Centrifugal Force Calculations:
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Here is the link to the question.
https://www.khanacademy.org/science/physics/forces-Newtons-laws/Newtons-laws-of-motion/a/what-is-Newtons-second-law
What is the magnitude of the force F1 ?
What is the magnitude of the force F2 ?
Here is my drawing I made At this step I am lost.
## ay...
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Hi
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Hi!
Given three voltages as follows;
Q1 = 1C,Q2 = 1C,Q3 = 2C
The distance a is 1m and b = 2m
a) Find the values of the forces that are acting on Q2
I did that like this;
$$ F_{12} = \frac{Q1*Q2}{4\pi\epsilon r^2} $$
$$ F_{32} = \frac{Q1*Q3}{4\pi\epsilon r^2} $$
The results are ...
This is the initial setup of the problem:
The electric field due to the ring is:
$$E = \int\frac{k(dq)}{(\sqrt{R^2 + x^2})^2}\frac{x}{\sqrt{R^2 + x^2}} = \frac{kqx}{(R^2 + x^2)^{3/2}}$$
the force on the rod due to this Electric field produced by the ring is:
Consider a differential element...
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Hey,
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Gauss's Law...