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Suppose, a car is moving with constant velocity. Because there is no acceleration, there must be an equal and opposite force balancing the kinetic friction and thus net force becoming zero. What kinetic friction? Kinetic friction arises when two surfaces slide past each other. A car in a skid or a spin is experiencing kinetic friction. Under normal ...


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On level ground, with no strong wind behind the car, the drive train must still supply enough torque to the drive wheels to equal resistance, even if there is no forward acceleration. If a car runs out of fuel and the motor dies, it will eventually come to a stop.


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In an ideal case, the wheels do not slip, so at the point of contact there is no friction, that is why you do not need any extra forces. In a real case the wheels slow down a little in which case the friction will act in the back direction, slowing down the car. If you press the gas then the wheels will move faster and friction will point forward, which will ...


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Now supposing I throw the ball from height ℎ of tall building then why does he gets more hurt? Isn't the force still mg? The impact force of the ball falling on the man's head is not the same as the weight of the ball on the persons head. This is because it takes a force to perform work in order to absorb the kinetic energy of the ball at impact. The work ...


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The force is still $mg$, but note - it is the force that is applied on the ball, not on the man's head! As the ball falls from above, it picks up speed due to its constant acceleration $g$, which respectively increases its momentum, $p = mv$. Bigger momentum means a bigger force, so that's why the poor person's head suffers more when you drop the ball from ...


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mg is the force that the earth does on the ball, it is not the force between the head and the ball. This last force will be a function of the contact speed when they collide


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The force is no longer $mg$. In the second scenarion the ball has a speed $v(h)$ just before it hits the person's head. If the speed of the ball is slowed down from $v(h)$ to $0$ by the head in time $\tau$, then the total force felt by the head is: $$F_{head}= M\frac{v(h)}{\tau}.$$ In the first scenario there is no such stopping force. This is true for both ...


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Based on published values for hydrogen density in interstellar space one can estimate the heating on the leading surface of a ship moving at about 0.2 c to be several watts per square meter. (With radiators on other suraces, that might be collected and used to power devices within the ship.) (The drag forces wold still be quite small.)


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There are so few gas molecules in (orbital) space that the space shuttle was not frictionally heated by them. Instead, the shuttle was heated primarily by the sun's rays.


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Under norm conditions there are around $2.69 \times 10^{19}$ molecules in a cubic centimeter of air. As you can imagine this number is easily sufficient such that particles close to the wall are pressed against it (don't think of it as sticking) and left unable to move in any direction (You can see this is a question of diluteness and not viscosity: As soon ...


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‘Viscous’ doesn’t mean ‘sticky’. It means “the fluid exerts force against adjacent slip”. The viscosity of air or water is less than treacle, but it’s still non-zero. Air next to a non-moving surface will feel a force that’ll tend to make it stop. Smaller viscosity means less force, but that just means a less massive I.e. thinner layer is brought to a stop....


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In general, the effects of air resistance are rather complicated, and not all that accessible to high-school students except in a quantitative way. But here's a brief run-down of how air resistance works and how it affects the solutions of the equations of motion. For most everyday situations, the drag force from a fluid can be modeled in one of two ways: ...


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Effects are the same in relation to distance if we control for initial ball velocity and other launch variables like spin, launch angle, etc. With a draw you have an extra lever: your right hand is probably rotating the clubface through impact. With a fade this extra lever does not exist: right hand is led by left hand and holding clubface open. Thus ...


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This is a simple application of Newton's law $$ m \vec a = m \frac{d \vec u}{d t} = m \frac{d^2 \vec x}{d^2 t} = \sum\limits_i \vec F_i .$$ Systems of equations The equation system is thus given by $$ m \frac{d}{d t} \left( \begin{array}{} u_x\\ u_y\\ \end{array} \right) = m \left( \begin{array}{} \phantom{-}0\\ -g\\ \end{array} \right) - k \sqrt{u_x^2 + ...


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