Can someone help me with my IGCSE Physics Question?
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The formula for universal gravitation is F = G m₁ m₂ /r² where G is the universal gravitational constant, m₁ and m₂ are the masses of two bodies and r is the distance between them. 1. On the surface of the asteroid, F = m₁ g So you know that g = G m₂ /r² where the m₂ is the mass of the asteroid and r is its radius. Escape velocity v = √(2Gm₂ /r ) Now substitute from above, and you find that v = √(2gr) You know v, and you know g, so solve for r. 2. Use the formula for g given above, but now m₂ = mass of the earth and r is the sum of the earth's radius plus 200 km. 3. For a simple pendulum, the relationship is T = 2π √(l/g) where T is the period, l is length. Solve for g.
Can someone help me explain how to do this AP Physics 1 question?
I assume that this problem involves firing a cannon in a frictionless vacuum under earth’s standard gravity and having the cannnonball hit the ground 7.5 seconds later and 70 meters away.If that assumption is right, then they want to know the angle above horizontal that the cannon fired..I get 1.322938593 radians, which is about 75 degrees.Here’s how I got that.… and since I didn’t have my calculator handy, I used excel for the arctangent function.I have not checked this answer. Apologies for that; lunch time is almost over.P.s. A better answer is 1.3 radians, because we only have two decimal places given in the question. All of those extra decimal places make it look like we have more precise measurements than we do. Prof. Lewin would drive from Cambridge to Atlanta and hit me with a stick if I didn’t add “Plus an unknown amount of measurement uncertainty.”HTH.P.s.s. If you need help with this subject, I found Rice’s AP Physics classes to be worth the time investment: AdvPHY1.1x Course Info | edX(Not a paid endorsement, I sat for this class (online)).
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The pendulum continues in an arc of reduced radius of L-D (L being the length of string).Okaay, so if it can get to the "highest point of the new reduced arc", let's write an energy balance for it: the starting energy (gravitational potential energy mgL) should equal the sum of kinetic energy and potential energy at the "highest point of the new reduced arc".mgL = 2mg(L-D) + 0.5 m v^2 (1) (2mg(L-D) because the ball rises to a height of 2(L-D) or twice the reduced radius)So if you want the D, you need the V. (I'm so sorry.) We can find v from the condition that D is the MINIMUM peg distance for which the string remains juuust taut, so in this limiting case, the tension in the string is juuust zero. Write a force balance for the ball when it is just at the "highest point of the new reduced arc" for this limiting case. Why, the only force seems to be gravity and the only acceleration of the ball is the centripetal acceleration. From Newton's law,m v^2/(L-D) = mg (centripetal acceleration is mv^2/r remember?)or, m v^2 = mg(L-D) (2)Substituting the above in Eqn. 1, mgL = 2.5 mg(L-D)or, D = 3/5 L Sic transit gloria mundi.
Im so confused, can someone help me with this physics question?
A particle executes simple harmonic motion with an amplitude of 2.54 cm. At what positive displacement from the midpoint of its motion does its speed equal one half of its maximum speed? Answer in units of cm.
Can someone help me with physics?
This is the questionIn the figure, two particles, each with mass m = 0.85 kg, are fastened to each other, and to a rotation axis at O, by two thin rods, each with length d = 5.4 cm and mass M = 1.3 kg. The combination rotates around the rotation axis with angular speed ω = 0.28 rad/s. Measured about O, what is the combination's (a) rotational inertia and (b) kinetic energy?
Can someone help me with a physics question about electricity?
Yes. Try Physics Forums - The Fusion of Science and Community .There are lots of specialist physics help forums available. Quora, otoh, is not well set up for it.
Can someone help me solve this simple physics question?
A 6.0 kg object undergoes an acceleration of 2.0 m/s^2. (a) What is the magnitude of the resultant force acting on it? (b) If this same force is applied to a 4.0 kg object, what acceleration is produced?
Can anyone help me answer this physics question?
No, there is not enough information. The position the clown maintains in the air would change his air resistance and thus flight time and distance. However if you assume no air resistance, you can:resolve the vector for the vertical componant (sin23*51), to give the vertical speed initially,divide this value by the acceleration incurred by gravity (9.8ms^-2) and this will give you the time it takes for the vertical velocity to reach 0.Then you calculate the vertical distance travelled during this time (((initial vertical velocity+ final vertical velocity)/2)*time)((sin23*51+0)/2)*((sin23*51)/9.8)=d. This will give you the clowns height.Sub this value into (distance= initial velocity*time+1/2*acceleration*time^2), and hence using the vertical distancetravlled calculated in the previous step d=0*t+1/2*9.8*t^2, you can re-arrange to solve for t. (This should be equal to the value you obtained in step 2)Add the value of t to (sin23*51)/9.8 and you have the time spent in the air.Assuming no air resistance the horizontal speed does not change, so resolve for the vertical component (cos23x51) and multiply this by the time spent in the air and you have distance travelled from the canon.