What is the formula for time? (something like speed *distance=time??)?

Physics

1 answer:

butalik [34]3 years ago

4 0

It would be: Time = Distance / Speed

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A block slides down a frictionless inclined ramp. If the ramp angle is 17.0° and its length is find the speed of the block as it

SashulF [63]

Answer:

$2.4\sqrt{L}$ where L is the length of the ramp

Explanation:

Let L (m) be the length of the ramp, and g = 9.81 m/s2 be the gravitational acceleration acting downward. This g vector can be split into 2 components: parallel and perpendicular to the ramp.

The parallel component would have a magnitude of

$gsin\theta = 9.81 sin17^o = 2.87 m/s^2$

We can use the following equation of motion to find out the final velocity of the book after sliding L m:

$v^2 - v_0^2 = 2a\Delta s$

where v m/s is the final velocity, $v_0$ = 0m/s is the initial velocity when it starts from rest, a = 2.87 m/s2 is the acceleration, and $\Delta s = L$ is the distance traveled:

$v^2 - 0 = 2*2.87*L$

$v = \sqrt{5.74L} = 2.4\sqrt{L}$

6 0

3 years ago

The oxygen molecule, O2, has a total mass of 5.30×10-26 kg and a rotational inertia of 1.94×10-46 kg-m2 about an axis perpendicu

Nady [450]

Answer:

$\omega=2.85*10^{13}\frac{rad}{s}$

Explanation:

The translational kinetic energy depends on the mass and speed of the body, as follows:

$K_T=\frac{mv^2}{2}\\K_T=\frac{5.30*10^{-26}kg(1.49*10^3\frac{m}{s})^2}{2}\\\\K_T=1.18*10^{-19}J$

While rotational kinetic energy depends on the moment of inertia and the angular velocity of the body, as follows:

$K_R=\frac{I\omega^2}{2}(1)$ . We know that:

$K_R=\frac{2}{3}K_T(2)$

Replacing (1) in (2):

$\frac{I\omega^2}{2}=\frac{2}{3}K_T\\\\\omega=\sqrt{\frac{4}{3}\frac{K_T}{I}}\\\omega=\sqrt{\frac{4}{3}\frac{1.18*10^{-19}J}{1.94*10^{-46}kg\cdot m^2}}\\\omega=2.85*10^{13}\frac{rad}{s}$