Ask question

Ask question

All categories

3 years ago

12

in physics lab, a cube slides down a frictionless incline as shown in the figure below, and elastically strikes another cube at

the bottom that is only one-half its mass. if the incline is 30 cm high and the table is 90cm off the floor, where does each cube land [hint: both leave the incline moving horizontally]

1 answer:

Tema [17]3 years ago

3 0

<span>In the physics lab, a cube slides down a frictionless incline as shown in the figure below, check the image for the complete solution:

</span>

You might be interested in

Please can anybody tell me what are these lab equipments

olga nikolaevna [1]

Answer:

5 is the tripoid stand

Thanks have a bangtastic day

4 0

2 years ago

Read 2 more answers

At what speed, as a fraction of c , is a particle's total energy twice its rest energy

WINSTONCH [101]

The rest energy of a particle is
$E_0=m_0 c^2$
where $m_0$ is the rest mass of the particle and c is the speed of light.

The total energy of a relativistic particle is
$E=mc^2 = \frac{m_0 c^2}{ \sqrt{1- \frac{v^2}{c^2} } }$
where v is the speed of the particle.

We want the total energy of the particle to be twice its rest energy, so that
$E=2E_0$
which means:
$\frac{m_0c^2}{ \sqrt{1- \frac{v^2}{c^2} } }=2m_0 c^2$
$\frac{1}{ \sqrt{1- \frac{v^2}{c^2} } }=2$
From which we find the ratio between the speed of the particle v and the speed of light c:
$\frac{v}{c}= \sqrt{1- (\frac{1}{2})^2 } =0.87$
So, the particle should travel at 0.87c in order to have its total energy equal to twice its rest energy.

3 0

3 years ago

Write a hypothesis about the effect a change in temperature will have on the volume of a gas.

Licemer1 [7]

The colder the more likely it is to become a liquid

7 0

3 years ago

Read 2 more answers

If the net torque is zero, what does this imply about the clockwise and counterclockwise torques.

lapo4ka [179]

Answer:

Explanation:

When two forces acting on a line of action and they are equal in magnitude but opposite it direction, it forms a couple.

Torque is defined as the product of either force and the perpendicular distance between the two forces.

It is a vector quantity.

The net torque is zero, it means the anticlockwise torque is equal to the clockwise torque.

It means they balances each other.

6 0

3 years ago

A practical rule is that a radioactive nuclide is essentially gone after 10 half-lives. What percentage of the original radioact

ArbitrLikvidat [17]

Answer:

0.09 % of the original radioactive nucllde its left after 10 half-lives
It will take 241,100 years for 10 half-lives of plutonium-239 to pass.

Explanation:

The equation for radioactive decay its:

$N ( t) \ = \ N_0 \ e^{ \ - \frac{t}{\tau}}$ ,

where N(t) its quantity of material at time t, $N_0$ its the initial quantity of material and $\tau$ its the mean lifetime of the radioactive element.

The half-life $t_{\frac{1}{2}}$ its the time at which the quantity of material its the half of the initial value, so, we can find:

$N (t_{\frac{1}{2} }) \ = \ N_0 \ e^{ \ - \frac{t_{\frac{1}{2}}}{\tau}} \ = \frac{N_0}{2}$

so:

$\ N_0 \ e^{ \ - \frac{t_{\frac{1}{2}}}{\tau}} \ = \frac{N_0}{2}$

$e^{ \ - \frac{t_{\frac{1}{2}}}{\tau}} \ = \frac{1}{2}$

$- \frac{t_{\frac{1}{2}}}{\tau}} \ = - \ ln( 2 )$

$t_{\frac{1}{2}}\ = \tau ln( 2 )$

So, after 10 half-lives, we got:

$N ( 10 \ t_{\frac{1}{2}}) \ = \ N_0 \ e^{ \ - \frac{10 \ t_{\frac{1}{2}}}{\tau}}$

$N ( 10 \ t_{\frac{1}{2}}) \ = \ N_0 \ e^{ \ - \frac{10 \ \tau \ ln( 2 ) }{\tau}}$

$N ( 10 \ t_{\frac{1}{2}}) \ = \ N_0 \ e^{ \ - 10 \ \ ln( 2 ) }$

$N ( 10 \ t_{\frac{1}{2}}) \ = \ N_0 \ * \ 9.76 * 10^{-4}$

So, we got that a 0.09 % of the original radioactive nucllde its left.

Putonioum-239 has a half-life of 24,110 years. So, 10 half-life will take to pass

$10 \ * \ 24,110 \ years \ = \ 241,100 \ years$

It will take 241,100 years for 10 half-lives of plutonium-239 to pass.

7 0

3 years ago

Read 2 more answers