How much force is needed to push down a nearly weightless but rigid 1-L carton beneath the water?
If u calculate the average force needed to push it under water it wud be different from the total force needed to keep it stationary under the water. Initially no upthrust is actin' in the upward direction but slowly it reaches to its maximum when the whole carton is inside the water, Umax=V(volume of carton)*d(density of water)*g=1l*1kg/l*9.8ms^-2=9.8kgms^-2 Uavg=4.9kgms^-2 I hav assumed all atmospheric pressure change to be zero. GOOD QUESTION.
If a hole is made right through Earth so it reaches the other side, what will happen when one jumps into it?
Earth’s center of mass is pretty much at its core. It is therefore clear that the attractive force exerted by the Earth towards all bodies is directed towards its center.Why is this important? Say the particular hole looks like thisSay the center is at B (it’s not exactly B, but slightly below B). When the person is in the first leg of the journey, i.e. not yet reached the center, he is falling to the center, or basically towards the point of attraction. So, his speed naturally increases.Once he passes B, though, he is now falling away from the center, which is kind of equivalent to jumping up. The force he now experiences is against the direction of his motion, so he now decelerates. Once he completely stops, he falls back towards to the center and the entire cycle repeats.If you’ve noticed, this is basically an SHM (simple harmonic motion). He will keep oscillating about the center with a time period of about 43 minutes (if I’m not wrong).This, however, is assuming that you initially simply dropped into the hole or had low start speed. If you were actively launched into the hole at a high speed, it is possible you may not decelerate fast enough to stop within the hole. You may fly right out the other side. If the speed is high enough, you could even reach space.
How much force is needed?
F=pVg F=force p=density of the water (1000 kg/m^3) V=volume of carton (1L) g=acceleration due to gravity (9.8 m/s^2)
Physics: Two masses M and m are connected by a weightless string. They are pulled by a force F on a frictionless horizontal surface. The tension in the string will be?
Let the tension in the String be T .Now , since the string is supposed to be inextensible, both blocks have same acceleration . Say a(=F/(M+m))Eqns of motion formed are,F-T = Ma( supposing the M mass block to be acted upon by the force)T=ma.=> T=mF/(M+m).
A block of mass 3kg is placed on a rough surface. The coefficient of static friction between two surfaces is 0.2 then what is the minimum horizontal force required to move the block?
As stated, there is not sufficient information to answer.What is the acceleration due to gravity whereas the experiment is performed?If you can assume 1 Earth standard gravity, the the block is pressed down onto the surface by a force of 9.8 * 3 N. The coefficient of friction presents a retarding force of 0.2 * 29.4, or 5.88N
A rope of length L is pulled by a constant force F. What is the tension in the rope at a distance x from the end where the force is applied?
At a distance x from its end we took a small element dxMass of dx= Mdx/l =dmRope is moving with an acceleration aEquation of motion for the small element dx:{T + dT} - T = dm × adT = [M dx/l] × F/MdT = F dx /lIntegrating both the sides with limit of T being F to F(x) and limits of x being x to 0 we get,F(x) -F = - Fx/lF(x) = F[1-x/l]