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4 years ago

Have a blessed day christains!

Physics

1 answer:

LenaWriter [7]4 years ago

3 0

Answer:

i say amen to that brother a big amen

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It takes 9 sec for a 10 newton force to move an object 4 meters to the right. What is direction & magnitude of the force

lana66690 [7]

Answer:

the question is wrong

Explanation:

M is not given
after 9 second the acceleration multiply by the time divided by two then multiplied by the time is equal to 4 meter
((10/Mm/s *9s-)

3 0

4 years ago

Identify one force acting on the cart.

Artist 52 [7]

Answer:

pulling force

Explanation:

because the person is pulling that cart.

5 0

3 years ago

A car moves along a horizontal road with constant velocity v0 = v0xi until it encounters a smooth inclined hill, which it climbs

Serga [27]

Answer:

a ave = ( v1x - v0x ) i/Δt + v1y j/Δt

Explanation:

8 0

3 years ago

A muon is traveling at 0.995

LenaWriter [7]

The momentum of a relativistic particle is given by
$p= \gamma m_0 v$
where
$\gamma= \frac{1}{ \sqrt{1- \frac{v^2}{c^2} } }$ is the relativistic factor
$m_0$ is the rest mass of the particle
v is the speed particle

The rest mass of the muon is 207 times the rest mass of the electron:
$m_0 = 207 m_e = 207 \cdot 9.1 \cdot 10^{-31} kg=1.88 \cdot 10^{-28} kg$
The muon is moving at speed 0.995 c, therefore its velocity is
$v=0.995 c=0.995 \cdot 2.998 \cdot 10^8 m/s =2.983 \cdot 10^8 m/s$
And the relativistic factor is
$\gamma = \frac{1}{ \sqrt{1- (\frac{0.995 c}{c})^2 } } =10.01$

If we plug these numbers into the first equation, we find the muon momentum:
$p= \gamma m_0 v=(10.01)(1.88 \cdot 10^{-28} kg)(2.983 \cdot 10^8 m/s)=$
$=5.61 \cdot 10^{-19} kgm/s$

3 0

3 years ago

Uranium-238 has a half-life of 4.5 billion years. Given that scientists estimate Earth's age to be 4.6 billion years, what is th

anyanavicka [17]

Answer:

103.3 %

Explanation:

For a radioactive isotope, the number of radioactive nuclei left (parent nuclei) after a time t, N(t), is

$N(t)=N_0 (\frac{1}{2})^{\frac{t}{\tau}}$

where

N0 is the initial number of radioactive nuclei

t is the time

$\tau$ is the half-life of the isotope

Here we have

$t=4.5 \cdot 10^9 y\\\tau = 4.6\cdot 10^9 y$

So we find

$\frac{N(t)}{N_0}=(\frac{1}{2})^{\frac{4.5\cdot 10^9}{4.6\cdot 10^9}}=0.508$

Which means that the fraction of parent nuclei left after this time is 0.508 (50.8% of the initial value). So the fraction of daugther nuclei at this time is

$\frac{N_d}{N_0}=1-0.508=0.492$

So the percentage of parent to daughter isotopes is

$\frac{N(t)}{N(d)}=\frac{0.508}{0.492}=1.033$

Which corresponds to 103.3 %.

5 0

3 years ago