Answer:
a)
, b)
, c)
, d)
, e)
, f)
![\beta = 0.877](https://img.qammunity.org/2021/formulas/physics/college/su83nbp8rh4g1iwrvc2rb4ght6qumremrc.png)
Step-by-step explanation:
From relativist physics we know that
is the symbol for the speed of light, which equal to approximately 300000 kilometers per second. (300000000 meters per second).
a) A car traveling 120 kilometers per hour:
At first we convert the car speed into meters per second:
![v = \left(120\,(km)/(h) \right)* \left(1000\,(m)/(km) \right)* \left((1)/(3600)\,(h)/(s) \right)](https://img.qammunity.org/2021/formulas/physics/college/sfb104zm5nkbinien9t8827c9ak237vcjz.png)
![v = 33.333\,(m)/(s)](https://img.qammunity.org/2021/formulas/physics/college/g28i6p1b1u56ihwlb8humufx1bwqty19yy.png)
The ratio
is now calculated: (
,
)
![\beta = (33.333\,(m)/(s) )/(3* 10^(8)\,(m)/(s) )](https://img.qammunity.org/2021/formulas/physics/college/qsu83cpfv7wqk986kvmdthtuz0cl3echfx.png)
![\beta = 1.111* 10^(-7)](https://img.qammunity.org/2021/formulas/physics/college/1p0c7vf2owklidol8o0ityk6eru0ddmi4n.png)
b) A commercial jet airliner traveling 270 meters per second:
The ratio
is now calculated: (
,
)
![\beta = (270\,(m)/(s) )/(3* 10^(8)\,(m)/(s) )](https://img.qammunity.org/2021/formulas/physics/college/us3bb3cahq4yo55h6jkhbk5nqr4o2x7n8d.png)
![\beta = 9* 10^(-7)](https://img.qammunity.org/2021/formulas/physics/college/z669ho97m1czyqlmi5kw06voqcmhf7ldel.png)
c) A supersonic airplane traveling Mach 2.7:
At first we get the speed of the supersonic airplane from Mach's formula:
![v = Ma\cdot v_(s)](https://img.qammunity.org/2021/formulas/physics/college/rfvk7t7nyn7pwl28d9rqnng1jvrxmscqd6.png)
Where:
- Mach number, dimensionless.
- Speed of sound in air, measured in meters per second.
If we know that
and
, then the speed of the supersonic airplane is:
![v = 2.7\cdot \left(343\,(m)/(s) \right)](https://img.qammunity.org/2021/formulas/physics/college/jk8ifmj9iy8aujd7aeb10enqq6ucx84nwd.png)
![v = 926.1\,(m)/(s)](https://img.qammunity.org/2021/formulas/physics/college/iu5t92leoe8q4fdum13yqo2d7och1hyixw.png)
The ratio
is now calculated: (
,
)
![\beta = (926.1\,(m)/(s) )/(3* 10^(8)\,(m)/(s) )](https://img.qammunity.org/2021/formulas/physics/college/cimwjmm7df83xyoww1v69vasxnchkcqjvm.png)
![\beta = 3.087* 10^(-6)](https://img.qammunity.org/2021/formulas/physics/college/ranw9k4ilvoadkwpe50su9qz83vzz3r9cm.png)
d) The space shuttle, travelling 27000 kilometers per hour:
At first we convert the space shuttle speed into meters per second:
![v = \left(27000\,(km)/(h) \right)* \left(1000\,(m)/(km) \right)* \left((1)/(3600)\,(h)/(s) \right)](https://img.qammunity.org/2021/formulas/physics/college/8y0pqodrkg4r1s07k1n8pimki4wqp3bx7s.png)
![v = 7500\,(m)/(s)](https://img.qammunity.org/2021/formulas/physics/college/83tker9pxkxk2ko5vrbcwtch9fs51ve88v.png)
The ratio
is now calculated: (
,
)
![\beta = (7500\,(m)/(s) )/(3* 10^(8)\,(m)/(s) )](https://img.qammunity.org/2021/formulas/physics/college/i143vb4b6rmkqqiwsxaegw32ctbvhls06q.png)
![\beta = 2.5* 10^(-5)](https://img.qammunity.org/2021/formulas/physics/college/7w76obh5v9uv9dbp98abtnuf3nxmya60m9.png)
e) An electron traveling 30 centimeters in 2 nanoseconds:
If we assume that electron travels at constant velocity, then speed is obtained as follows:
![v = (d)/(t)](https://img.qammunity.org/2021/formulas/physics/high-school/92gvb93xk1w1vccd1h7w180tqnk5ef2mqv.png)
Where:
- Speed, measured in meters per second.
- Travelled distance, measured in meters.
- Time, measured in seconds.
If we know that
and
, then speed of the electron is:
![v = (0.3\,m)/(2* 10^(-9)\,s)](https://img.qammunity.org/2021/formulas/physics/college/24cwn00qv4957qfijkg3e9e5oeky0uanks.png)
![v = 1.50* 10^(8)\,(m)/(s)](https://img.qammunity.org/2021/formulas/physics/college/fbcmsgkipupi7mu5hu008mrs6lmw3aw63h.png)
The ratio
is now calculated: (
,
)
![\beta = (1.5* 10^(8)\,(m)/(s) )/(3* 10^(8)\,(m)/(s) )](https://img.qammunity.org/2021/formulas/physics/college/ksir5pn53ocyfaj74js65wsapd42j1igs8.png)
![\beta = 0.5](https://img.qammunity.org/2021/formulas/physics/college/371nouccfiwva74uv2vwmxzk5och0oqsp5.png)
f) A proton traveling across a nucleus (10⁻¹⁴ meters) in 0.38 × 10⁻²² seconds:
If we assume that proton travels at constant velocity, then speed is obtained as follows:
![v = (d)/(t)](https://img.qammunity.org/2021/formulas/physics/high-school/92gvb93xk1w1vccd1h7w180tqnk5ef2mqv.png)
Where:
- Speed, measured in meters per second.
- Travelled distance, measured in meters.
- Time, measured in seconds.
If we know that
and
, then speed of the electron is:
![v = (10^(-14)\,m)/(0.38* 10^(-22)\,s)](https://img.qammunity.org/2021/formulas/physics/college/kpzet412mqqtxloecrsnyi156n9e7dux20.png)
![v = 2.632* 10^(8)\,(m)/(s)](https://img.qammunity.org/2021/formulas/physics/college/qud5rvoqjf4uv7exiz4h6xi38tq3cdbz6x.png)
The ratio
is now calculated: (
,
)
![\beta = (2.632* 10^(8)\,(m)/(s) )/(3* 10^(8)\,(m)/(s) )](https://img.qammunity.org/2021/formulas/physics/college/ecwcjy2s6tragu48cfmkhajwy0cnw34h42.png)
![\beta = 0.877](https://img.qammunity.org/2021/formulas/physics/college/su83nbp8rh4g1iwrvc2rb4ght6qumremrc.png)