Physics homework question revised?

Physics homework question revised? Topic: Physics homework question revised?
June 18, 2019 / By Alia
Question: A 4.5kg cat is sitting at rest on a level fence rail. 1. How large is the normal force on the cat? 2. Suppose the fence rail is tilted to 24 degree angle above horizontal, but the cat remains at rest because of friction. What is the normal force on the cat? 3. To what angle should the fence rail be tilted so that the normal force on the cat is exactly 6N? Here is what I have. 1. n=mg (m=0.23kg and g=9.8) n=0.23kg(9.8m/s^2) n= 2.254N 2. I think the second one is solved using a ramp vectors and the equation would be: m x g x cos θ If so it would be 0.23kg x 9.8m/s^2 x cos 24 = 2.05913N 3. If the last problem is set up right then the last problem should just be algebra right? But I'm not sure how to set it up/solve it. I am not feeling very good about this problem at all. Can you please walk me through this, especially parts 2 and 3. Thank you in advance! Sorry, I have been working on this homework for a couple hours and I'm just really sleepy. Thank you, that helps. I missed 2 classes from this chapter due to bad weather and while I am ok working through this stuff for the most part my brain is about to shut down (as you can see by my accidental blending of problems.) So 2) 4.5kg x 9.8m/s^2 cos 24 = 40.2874 3) 4.5kg x 9.8m/s^2 x cos θ = 6N 6N/44.1= 0.136054 cos-1(0.136054)=82.1804 Yes, the cat probably would fall at that angle. Perhaps it is using it's claws?
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Best Answers: Physics homework question revised?

Tyron Tyron | 8 days ago
Where did the frog come from ?? Parts one and two look OK. except the cat's mass seems to have changed ?? For 3 just put m g costheta = 6N and solve for theta 4.5 x 9.8 costheta = 6 so theta = arccos ( 6/44.1) or about 82 degrees (the cat would probably slide down by then!!)
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Tyron Originally Answered: PLEASE HELP! Physics homework question.?
m=mass=80000kg v=velocity=900km/h=250m/s h=height=10km=10000m Kinetic Energy=KE=(1/2)*m*v^2 :.KE=(1/2)*80000kg*(250m/s)^2 [900km/hr is 250 m/s which we need to get KE in joules] :.KE=2 500 000 000 joules Potential Energy =PE=m*g*h :.PE = 80000kg*9.80665 m/s^2*10000m [10km=10000m] :.PE = 7 845 320 000 joules :.Total energy =KE+PE=2 500 000 000 joules+7 845 320 000 joules Total energy=10 345 320 000 joules Seeing that the no. of significant digits in the question is 1 :.Total energy = 10^10 joules
Tyron Originally Answered: PLEASE HELP! Physics homework question.?
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Rigby Rigby
a million) Magnet will be in a position to carry something interior the absence of gravity, because the burden = Mass * Gravity 2) pressure will boost the density of each rely. 3) there is not any relation between density and gravity. until eventually you consider gravity as a severe pressure. 4) relies upon on the container of technology you're interpreting. Why: a) Glass enables electromagnetic waves to pass by. b) infrared is an electromagnetic wave. c) Infrared photon power is distinct than the glass's electron power that couldn't excite it. How: Refraction will happen. 5) Forgot the formula. You effective can locate it on your physics e book. wish I helped =)
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Merton Merton
You seem to be using .23 kg where the problem specifies 4.5 kg 1. Fn = m*g*cosΘ = 4.5*9.8*cos0° = 44.1 N 2 Fn = m*g*cosΘ = 4.5*9.8*cos24° = 40.287 N 3. 6 = m*g*cosΘ → cosΘ = 6/(m*g) = .136 → Θ = 82.18°
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Merton Originally Answered: Can someone explain this physics homework question?
The question is very badly worded. I believe it should be this (but I might be wrong!): Power is transmitted between 2 high voltage cables over a distance of 80km.(so each cable is 80km long). At the supply end, there is a voltage of 150kV between the ends of the cables. The current through each cable is 300A. Each cable has a resistance of 5x10^-5 ohms per metre. Which statements are incorrect: 1) Each cable has a voltage of 1.2 kV between its ends. 2) 0.8% of a cable's output is being converted into heat in the cable 3) The heat generation inside a cable is 45*10^16J per second 4) The user gets an output voltage of 149 kV ___________________________ The important point is that there are 4 different voltages here. Students often get confused by this and mix them up. Think of the 'circuit as a battery, 2 long wires, and a bulb. The 'battery' supplies 150kV. Call this V1. Due to the (small) resistance of each wire, there is a voltage drop from one end of a wire to the other end. Call this V2. Total voltage drop across both wires is 2V2 (think of the 2 wires as 2 resistors, each with a voltage V2 across it). There is a voltage V3 across the bulb ('user') V3 = V1 - 2V2. __________________ 1) The resistance of a single cable = length x resistance per metre = 80,000 x 5x10^-5 = 4Ω The voltage drop between the ends of a cable (separated by 80km) is V = IR = 300x4=1200V=1.2kV. So the 1st statement is correct. ___________________________ 2) The power into the cables from the supply = P=VI = 150,000 x 300 = 4.5x10^7 W =45MW The heating power in ONE cable = P = VI = 1200 x 300 = 360000W = 0.36MW The power output of the cable (at the user end) = 45 - 0.36 = 44.64MW % of a cable's output is being converted into heat in the cable = (0.36/44.64) x 100 = 0.806% The 2nd statement is correct (ignoring rounding) ____________ 3) Statement 3 is Incorrect. The heat generation in the cable is NOT 45x10^16W. It is 0.36MW (see above). ____________ 4) The user receives a voltage V3 - V1 - 2V2 = 150- 2x1.2 = 147.6kV Statement 4 is Incorrect. However, please note the question I answered is not quite the same question as your question - I may have misinterpreted your question.

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