Physics test tomorrow, I DO NOT UNDERSTAND?
Force = Mass times acceleration. If the block is not to move, then the total force = 0. The incline "splits" the gravitational force into two parts. One is downhill along the wedge, and the other is perpendicular right into the face of the wedge. You use trigonometry to figure out what each of these "components" are. The total gravity is the hypotenuse of a right triangle, the along and into components are the legs of this triangle. The wedge itself resists the block--it doesn't penetrate. So the "into" part is automatically cancelled. Hence you need only worry about the "along" part. BUT there is friction. This is an additional along part that points uphill if the block wants to slide downhill. You can figure this out because the frictional force is related to the "into" force. The larger the into force, the greater the friction force. More succinctly, "Friction force = Into force times coefficient of friction." So now you have the total force tending to slide the block downhill: the "along" part of the gravity, minus the friction force. If this comes out positive, the block will start to slide, and you have to add an opposite (negative) "uphill" force to make up the difference. If it comes out negative, then the friction alone is enough to keep the block at rest.
Physics test tomorrow and i don't understand this question.. please help?
tension = Mass circumstances acceleration. If the block is to no longer flow, then the full tension = 0. The incline "splits" the gravitational tension into 2 factors. One is downhill alongside the wedge, and the different is perpendicular wonderful into the face of the wedge. you employ trigonometry to be certain what each and every of those "factors" are. the full gravity is the hypotenuse of a fabulous triangle, the alongside and into factors are the legs of this triangle. The wedge itself resists the block--it would not penetrate. So the "into" area is immediately cancelled. consequently you prefer basically hassle appropriate to the "alongside" area. yet there is friction. it truly is one greater alongside area that factors uphill if the block desires to slip downhill. you are able to parent this out because of the fact the frictional tension is on the subject rely of the "into" tension. the better the into tension, the greater desirable the friction tension. greater succinctly, "Friction tension = Into tension circumstances coefficient of friction." So now you have the full tension tending to slip the block downhill: the "alongside" area of the gravity, minus the friction tension. If this comes out advantageous, the block will start to slip, and you are going to be able to desire to function an opposite (unfavourable) "uphill" tension to make up the version. If it comes out unfavourable, then the friction on my own is sufficient to maintain the block at relax.
,HELP!Physics Test don't understand, tomorrow !!?
initial velocity = stationary position = 0 final velocity = 30 km/h acceleration = delta v /delta t F=ma F=m(delta v) / delta t F=m(vf-vo)/(tf-to) delta v = (30 10^3 m/h)(1h/60 min)(1 min/ 60 secs) delta t = do you have the time? F=1.5 x 10^3 m (30 10^3 secs/ 3600) / (time final - time orig) secs F is in Newtons = kg m/s^2 ***************** In response to below: a= change in velocity / change in time so units are m/s/s = m/s^2 so 30 km/h cannot be the A b/c you are missing the second time component ... meters per second PER SECOND meters per second squared You must be given another variable. Convert time from 1 hour to secs: 1 h (60 mins/h)(60 s/min)= 3600 s = 3.6 x 10^3 s =1 hr. F= m(a) F=1.5 x 10^3 (30 10^3 / 3600) / delta time
Physics study help. Test tomorrow!?
Basically you need to understand the gravity formula. [(m1 * m2) * 6.67 E-11]/D^2. It should be in your book. OK, what's that mean? Basically it says you take the mass of two objects, multiply it together, multiply by 6.67 E-11, then divide by the distance squared between the center of the objects, to come up with the newtons of gravitational force. What does this mean to you? All you need to know for these two questions is the bottom of the equation. Gravity is inversely proportional to the square of the distance. So if you double the distance, the force of gravity is divided by 4. That's it. So for part 1, the distance between the center of the earth and the surface of the earth is 1 earth radius (remember distance is measured from the center of an object). Move away by 1 earth radius, and you're now 2 earth radii away from the center. THe distance doubled, so divide gravity's force by 4. 9.8 m/s^2/4 ~ 2.49 m/s^2 For part 2, it's the same thing. You've got the same mass as the earth but double the distance from the center. 800/4 = 200 newtons. Got it? Edit: looks like the person above me might have it better than I do, I assumed question 2 "same materials" meant same amount of mass, which after re-reading it probably doesn't.
Physics study help. Test tomorrow!?
Basically you need to understand the gravity formula. [(m1 * m2) * 6.67 E-11]/D^2. It should be in your book. OK, what's that mean? Basically it says you take the mass of two objects, multiply it together, multiply by 6.67 E-11, then divide by the distance squared between the center of the objects, to come up with the newtons of gravitational force. What does this mean to you? All you need to know for these two questions is the bottom of the equation. Gravity is inversely proportional to the square of the distance. So if you double the distance, the force of gravity is divided by 4. That's it. So for part 1, the distance between the center of the earth and the surface of the earth is 1 earth radius (remember distance is measured from the center of an object). Move away by 1 earth radius, and you're now 2 earth radii away from the center. THe distance doubled, so divide gravity's force by 4. 9.8 m/s^2/4 ~ 2.49 m/s^2 For part 2, it's the same thing. You've got the same mass as the earth but double the distance from the center. 800/4 = 200 newtons. Got it? Edit: looks like the person above me might have it better than I do, I assumed question 2 "same materials" meant same amount of mass, which after re-reading it probably doesn't.
Physics block help- test tomorrow :(?
tension = Mass circumstances acceleration. If the block is to no longer flow, then the full tension = 0. The incline "splits" the gravitational tension into 2 aspects. One is downhill alongside the wedge, and the different is perpendicular impressive into the face of the wedge. you employ trigonometry to make sure what each and each of those "aspects" are. the full gravity is the hypotenuse of a impressive triangle, the alongside and into aspects are the legs of this triangle. The wedge itself resists the block--it would not penetrate. So the "into" area is immediately cancelled. subsequently you % in common terms hardship with regards to the "alongside" area. yet there is friction. it extremely is an extra alongside area that factors uphill if the block desires to slip downhill. you could discern this out because of the fact the frictional tension is with regards to the "into" tension. the better the into tension, the extra beneficial the friction tension. extra succinctly, "Friction tension = Into tension circumstances coefficient of friction." So now you have the full tension tending to slip the block downhill: the "alongside" area of the gravity, minus the friction tension. If this comes out constructive, the block will start to slip, and you will desire to upload an opposite (unfavourable) "uphill" tension to make up the version. If it comes out unfavourable, then the friction on my own is sufficient to maintain the block at relax.
Help me study for tomorrow's physics test?
Take horizontal as x axis and vertical as y axis with upward as positive. Let t = time taken to reach the height of the other roof. In y direction: - Initial velocity uy = 5.0 sin(15 deg) = 1.29 m/s Acceleration = -g = -9.8 m/s^2 (Negative because g is downward) Displacement Y = -2.5 m (negative because other roof is below the first roof) Let vy = final velocity in y direction vy^2 = uy^2 - 2gY Or vy^2 = 1.29^2 - 2*9.8*(-2.5) Or vy^2 = 1.6641 + 49 Or vy^2 = 50.6641 Taking square root on both sides. We should take negative square root because fnally the velocity is downward. vy = - sqrt(50.6641) Or vy = -7.12 m/s Average velocity in y direction = vyavg = (vy + uy)/2 = (-7.12 + 1.29)/2 = -5.83/2 = -2.915 m/s Time taken t = Y/vyavg Or t = (-2.5)/(-2.915) Or t = 0.858 s (Note: If you want, you can find t by using formula Y = uy * t - 1/2 * gt^2 and solve for t. But I wanted to avoid solving quadratic equation. But you can use whichever method you find easier.) Horizontal component of velocity vx = 5.0 cos(15 deg) = 4.83 m/s vx is constant because there is no force in X direction. Let X = displacement in time t X = vx * t Or X = 4.83 * 0.858 Or X = 4.14 m This is more than distance of the other roof. So, he will make it. ______________________________ If you want, you can directly use the projectile formula if it helps you: - y = x tan(theta) - gx^2/{2(v cos theta)}^2 Put y = -2.5 m, g = 9.8 m/s^2, theta = 15 deg Note: In the projectile formula, use 9.8 m/s^2 and not -9.8 because effect of g being downward has already been taken care in the formula. Use the projectile formula to find x. If it is more than 4.0 m, then he will make it.
Help me study for tomorrow's physics test?
Take horizontal as x axis and vertical as y axis with upward as positive. Let t = time taken to reach the height of the other roof. In y direction: - Initial velocity uy = 5.0 sin(15 deg) = 1.29 m/s Acceleration = -g = -9.8 m/s^2 (Negative because g is downward) Displacement Y = -2.5 m (negative because other roof is below the first roof) Let vy = final velocity in y direction vy^2 = uy^2 - 2gY Or vy^2 = 1.29^2 - 2*9.8*(-2.5) Or vy^2 = 1.6641 + 49 Or vy^2 = 50.6641 Taking square root on both sides. We should take negative square root because fnally the velocity is downward. vy = - sqrt(50.6641) Or vy = -7.12 m/s Average velocity in y direction = vyavg = (vy + uy)/2 = (-7.12 + 1.29)/2 = -5.83/2 = -2.915 m/s Time taken t = Y/vyavg Or t = (-2.5)/(-2.915) Or t = 0.858 s (Note: If you want, you can find t by using formula Y = uy * t - 1/2 * gt^2 and solve for t. But I wanted to avoid solving quadratic equation. But you can use whichever method you find easier.) Horizontal component of velocity vx = 5.0 cos(15 deg) = 4.83 m/s vx is constant because there is no force in X direction. Let X = displacement in time t X = vx * t Or X = 4.83 * 0.858 Or X = 4.14 m This is more than distance of the other roof. So, he will make it. ______________________________ If you want, you can directly use the projectile formula if it helps you: - y = x tan(theta) - gx^2/{2(v cos theta)}^2 Put y = -2.5 m, g = 9.8 m/s^2, theta = 15 deg Note: In the projectile formula, use 9.8 m/s^2 and not -9.8 because effect of g being downward has already been taken care in the formula. Use the projectile formula to find x. If it is more than 4.0 m, then he will make it.
Urgent Physics help, Exam tomorrow?
This question is on my revision sheet and will almost definitely be on my test. Could someone please help me with it. Draw a transistor circuit to investigate voltage gain. If the supply voltage used is Vcc, draw a table that gives typical values of the collector current, Ic, and the base-emitter voltage, Vbe. Complete the table and calculate the voltage gain for this circuit. Any help is really appreciated thanks.
Physics Question. Big Test tomorrow please help.?
I have a big physics question tomorrow. And I really need some help. I tried doing the questions but i couldn't seem to find the right answers. Please If you could do them all please do them. But Its okay if you do just a couple or one. But I would really appreciate if you did them all. Please explain it to me too I dont wanna flunk physics its just too freakin hard heeeeeeelp. I can't seem to find the right formula to use A 755 N diver drops from a board 10. 0 m above the water's surface.Find the dvier's speed 5 m above the water's surface. Then find the diver's speed just before striking the water. An Olympic runner leaps over a hurdle. If the runner's initial vertical speed is 2.2m/s how much will the runner's center of mass be raised during the jump? A pendulum bob is released from some initial height such that the speed of the bob at the bottom of the swing is 1.9m/s. What is the initial height of the bob? And please put some explanation to make me understand.