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2 years ago

A spaceship whose rest length is 350m has a speed of .82c

Physics

1 answer:

igomit [66]2 years ago

4 0

Answer:

$t'=1.1897*10^{-6} s$

t'=1.1897 μs

Explanation:

First we will calculate the velocity of micrometeorite relative to spaceship.

Formula:

$u=\frac{u'+v}{1+\frac{u'*v}{c^{2}}}$

where:

v is the velocity of spaceship relative to certain frame of reference = -0.82c (Negative sign is due to antiparallel track).

u is the velocity of micrometeorite relative to same frame of reference as spaceship = .82c (Negative sign is due to antiparallel track)

u' is the relative velocity of micrometeorite with respect to spaceship.

In order to find u' , we can rewrite the above expression as:

$u'=\frac{v-u}{\frac{u*v}{c^{2} }-1 }$

$u'=\frac{-0.82c-0.82c}{\frac{0.82c*(-0.82c)}{c^{2} }-1 }$

u'=0.9806c

Time for micrometeorite to pass spaceship can be calculated as:

$t'=\frac{length}{Relatie seed (u')}$

$t'=\frac{350}{0.9806c}$ (c = 3*10^8 m/s)

$t'=\frac{350}{0.9806* 3.0*10^{8} }$

$t'=1.1897*10^{-6} s$

t'=1.1897 μs

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A straight wire carries a current of 10 A at an angle of 30° with respect to the direction of a uniform 0.30-T magnetic field. F

Dimas [21]

Answer:

Explanation:

Given that,

Current in wire is

I = 10A

And the current makes an angle of 30° with respect to the magnetic field

Then, θ = 30°

And the magnetic field is

B = 0.3 T

Length of the wire is

L = 0.5m

Force on the wire F?

The force on the wire in calculated using

F = iL × B

Where

The magnitude of the cross produce of L and B is

L × B = LB•Sinθ

Then, force becomes

F = iLB•Sinθ

F = 10 × 0.5 × 0.3 × Sin30

F = 0.75 N

The force on the wire is 0.75 Newton

6 0

2 years ago

Read 2 more answers

A Canyon ball is fired from the ground level in angle of 35 from the ground at a speed of 150 m/s ignore air resistance calculat

Virty [35]

The velocity, 50 m/s, has two components - vertical and horizontal velocities.

The vertical component = 50 sin 30 = 25 m/s

The horizontal component = 50 cos 30 = 43.3 m/s

(a) Let t be the time taken for the vertical component to reach its peak from initial velocity = 25 m/s to its final velocity = 0.

Using the linear motion equation v = u - gt

0 = 25 - 10t

t = 2.5 s

Time taken to go up and down = 5 s

Time to hit the ground = 5 s

(b) Horizontal distance dealt x = 43.3 * 5 = 216.5 m

This is not the correct answer, but explains the problem thoroughly.

7 0

3 years ago

An ideal Otto cycle has a compression ratio of 8. At the beginning of the compression process, air is at 95 kPa and 27°C, and 75

katen-ka-za [31]

Answer:

Part A) 3899 kPa

Part B) 392.33 kJ/kg

Part C) 0.523

Part D) 495 kPa

Explanation:

Part A

First from the temperature at state 1 the relative specific volume and the internal energy at that state are determined from:

$u_{1}$ = 214.07 kJ/kg

$\alpha$ $r_{1}$ = 621.2

The relative specific volume at state 2 is obtained from the compression ratio:

$\alpha$ $r_{2}$ = $\frac{\alpha r_{1} }{r}$

=621.2/ 8

= 77.65

From this the temperature and internal energy at state 2 can be determined using interpolation with data from A-17(table):

$T_{2}$ = 673 K

$u_{2}$ = 491.2 kJ/kg

The pressure at state 2 can be determined by manipulating the ideal gas relations at state 1 and 2:

$P_{2}$ = $P_{1} r\frac{T_{2} }{T_{1} }$

= 95*8*673/300

= 1705 kPa

Now from the energy balance for stage 2-3 the internal energy at state 3 can be obtained:

$deltau_{2-3} =q_{in}\\ u_{3} -u_{2} =q_{in}\\u_{3}=u_{2}+q_{in}$

= 1241.2 kJ/kg

From this the temperature and relative specific volume at state 3 can be determined by interpolation with data from A-17(table):

$T_{3}$ = 1539 K

$\alpha r_{3}$ = 6.588

The pressure at state 3 can be obtained by manipulating the ideal gas relations for state 2 and 3:

$P_{3} =P_{2} \frac{T_{3} }{T_{2} }$

= 3899 kPa

Part B

The relative specific volume at state 4 is obtained from the compression ratio:

$\alpha r_{4}= r\alpha r_{3}$

= 52.7

From this the temperature and internal energy at state 4 can be determined by interpolation with data from A-17:

$T_{4}$ =775 K

$u_{4}$ = 571.74 kJ/kg

The net work output is the difference of the heat input and heat rejection where the heat rejection is determined from the decrease in internal energy in stage 4-1:

$w=q_{in}-q_{out}\\q_{in}-(u_{4} -u_{1} )\\=392.33 kJ/kg$

Part C

The thermal efficiency is obtained from the work and the heat input:

η= $\frac{w}{q_{in} }$

=0.523

Part D

The mean effective pressure is determined from its standard relation:

MEP= $\frac{w}{\alpha_{1}- \alpha_{2} }$

= $\frac{w}{\alpha_{1}(1- \frac{1}{r} }$

= $\frac{rwP_{1} }{RT_{1} (r-1) }$

=495 kPa

8 0

3 years ago

The temperature of a sample of silver increased by 24.0 °C when 269 J of heat was applied. What is the mass of the sample?

statuscvo [17]

Answer:

Mass of the silver will be equal to 46.70 gram

Explanation:

We have given heat required to raise the temperature of silver by 24°C is 269 J , so $\Delta T=24^{\circ}C$

Specific heat of silver = 0.240 J/gram°C

We have to find the mass of silver

We know that heat required is given by

$Q=mc\Delta T$ , here m is mass, c is specific heat of silver and $\Delta T$ is rise in temperature

So $269=m\times 0.240\times 24$

m = 46.70 gram

So mass of the silver will be equal to 46.70 gram

3 0

2 years ago

A wheel is rotating freely at angular speed 660 rev/min on a shaft whose rotational inertia is negligible. A second wheel, initi

sweet-ann [11.9K]

Answer:

(a) 110 rev/ min

(b) 5/6

Explanation:

As per the conservation of linear momentum,

L ( initial ) = L ( final )

I' ω' = ( I' + I'' ) ωf

I' is the rotational inertia of first wheel and I'' is the rotational inertia of second wheel which is at rest.

(a)

So, ωf = I' ω' / ( I' + I'' )

As I'' = 5I'

ωf = I' ω' / ( I' + 5I' )

ωf = ω'/ 6

now we know ω' = 660 rev / min

therefore ωf = 660/6

= 110 rev/ min

(b)

Initial kinetic energy will be K'

K' = I'ω'² / 2

and final K.E. will be K'' = ( I' + I'' )ωf² / 2

K'' = ( I' + 5I' ) (ω'/ 6)²/ 2

K'' = 6I' ω'²/72

K'' = I' ω'²/ 12

therefore the fraction lost is

ΔK/K' = ( K' - K'' ) / K'

= {( I'ω'² / 2) - (I' ω'²/ 12)} / ( I'ω'² / 2)

= 5/6

6 0

3 years ago